ELECTRIC CHARGE, FORCE, AND FIELD

20.1

ELECTRIC CHARGE, FORCE, AND FIELD

EXERCISES

Section 20.1 Electric Charge

14. Nearly all of the mass of an atom is in its nucleus, and about one half of the nuclear mass of the light elements in

living matter (H, O, N, and C) is protons. Thus, the number of protons in a 65 kg average-sized person is

approximately27 281

2(65 kg)/(1.67 10 kg) 2 10 , < which is also the number of electrons, since an average

person is electrically neutral. If there were a charge imbalance of 9

proton electron| | 10 ,q q e a person’s net charge

would be about28 9 192 10 10 1.6 10 C 3.2 C, or several coulombs (huge by ordinary standards).

15. INTERPRET This problem deals with quantity of charge in a typical lightning flash. We want to express the

quantity in terms of elementary charge e.

DEVELOP Since the magnitude of elementary charge e is191.6 10 C,e the number of electrons involved is

given by / .Q e

EVALUATE Substituting the values given in the problem statement, we find the number to be

20

19

25 C/ 1.56 10

1.6 10 CN Q e

ASSESS Since 1 coulomb is about18

6.25 10 elementary charges, our result has the right order of magnitude.

16. (a) The proton’s charge is 2 2 1

3 3 31 ,e e e e corresponding to a combination of uud quarks; (b) for neutrons, 0

2 1 1

3 3 3 corresponds to udd. (See Chapter 39, or Chapter 45 in the extended version of the text.)

Section 20.2 Coulomb’s Law

17. INTERPRET In this problem we are asked to compare the gravitational and electrical forces between a proton and

an electron.

DEVELOP The gravitational force and the electrostatic force between a proton and an electron separated by a

distance r are, respectively,

grav 2

p eGm mF

r

and

2

elec 2

keF

r

EVALUATE The ratio of the two forces is

2 2 9 2 2 19 2

elec

2 11 2 2 27 31

grav

39

(9 10 N m /C )(1.6 10 C)

(6.67 10 N m /kg )(1.67 10 kg)(9.11 10 kg)

2.3 10

p e

F ke r

F Gm mr

Note that the spatial dependence of both forces is the same, and cancels out.

ASSESS At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is

about40

10 times stronger than the gravitational force.

20

20.2 Chapter 20

18. 0 52.9a pm is called the Bohr radius. For a proton and electron separated by a Bohr radius,2 2

Coulomb 0/F ke a . 9 2 2 19 11 2 8

(9 10 N m /C )(1.6 10 C/5.29 10 m) 8.23 10 N.

19. INTERPRET This problem is about comparing the gravitational and electrical forces.

DEVELOP The electric force between a proton and an electron has magnitude2 2

elec / ,F ke r while the weight of an

electron is .g eF m g

EVALUATE When the two forces are equal, elec ,gF F the distance between the proton and the electron is

2 9 2 2 19 2

31 2

(9 10 N m /C )(1.6 10 C)5.08 m

(9.11 10 kg)(9.8 m/s )e

ker

m g

ASSESS The distance is almost fifty billion atomic diameters (or angstroms). This demonstrates that gravity is

unimportant on the molecular scale.

20. The mass could be suspended at the Earth’s surface if the electric force were repulsive (q same sign as qE) and 2

/ .E Ekqq R mg Thus,

3 2 6 2

9 2 2 5

(10 kg)(9.8 m/s )(6.37 10 m)103 C

(9 10 N m /C )( 4.3 10 C)q

21. INTERPRET This problem is about finding the unit vector associated with the electrical force one charge exerts on

the other.

DEVELOP A unit vector from (1 m, 0),qr r

the position of charge q, to any other point ( , )r x yr

is

2 2

( ) ( 1 m, )ˆ

| | ( 1 m)

q

q

r r x yn

r r x y

r r

r r

EVALUATE (a) When the other charge is at position (1 m,1 m),r r

the unit vector is

2

(0,1 m) ˆˆ (0,1)0 (1 m)

n j

(b) When (0,0),r r

2

( 1 m,0)

( 1 m) 0

ˆˆ ( 1,0) .n i

(c) Finally, when (2 m,3 m),r r

the unit vector is

2 2

(1 m,3 m) (1,3) 늿ˆ 0.316 0.94910(1 m) (3 m)

n i j

The sign of q doesn’t affect this unit vector, but the signs of both charges do determine whether the force exerted

by q is repulsive or attractive, i.e., in the direction of n or ˆ.n

ASSESS The unit vector always points away from the charge q located at (1 m,0).

22. The magnitude of the force is

2 9 2 2 19 210

2 2 2 18 2

(9 10 N m /C )(1.6 10 C)| | 7.74 10 N

(0.41 0.36 ) 10 mp

keF

r

r

and its direction is from the proton (at 0)pr r

to the electron (at 늿(0.41 0.36 ) nm),er i j r

for an attractive force,

at an angle1

tan (0.36/0.41) 41.3 to the x axis.

Section 20.3 The Electric Field

23. INTERPRET This problem is about calculating the electric field strength due to a source, when the force

experienced by the electron is known.

DEVELOP Equation 20.2a shows that the electric field strength (magnitude of the field) at a point is equal to the

force per unit charge that would be experienced by a charge at that point:

FE

q

EVALUATE With | |,q e we find the field strength to be

Electric Charge, Force, and Field 20.3

109

19

6.1 10 N3.81 10 N/C

| | 1.6 10 C

FE

e

ASSESS Since the charge of electron is negative, the force experienced by the electron is in the opposite direction

of the electric field.

24. From Equation 20.2b,4

(2 C)(100 N/C) 2 10 N.F qE

25. INTERPRET This problem is about calculating the electric field strength due to a source, when the force

experienced by a charge is known.

DEVELOP Equation 20.2a shows that the electric field strength (magnitude of the field) at a point is equal to the

force per unit charge that would be experienced by a charge at that point:

FE

q

The equation allows us to calculate E. For part (b), the force experienced by another charge q in the same field is

.F q E

EVALUATE (a) With 68 nC,q we find the field strength to be

6150 mN2.21 10 N/C

| | 68 nC

FE

e

(b) The force experienced by another charge, 35 C,q in the same field is

6(35 C)(2.21 10 N/C) 77.2 NF q E

ASSESS The force a test charge particle experiences is proportional to the magnitude of the test charge. In our

problem, since 35 C 68 nC,q q we find .F F

26. The electric field is 늿10 N/( 1 C) 10 MN/C.E i i r

The force on a proton is19(1.6 10 C)eE

r

늿( 10 MN/C) 1.6 pN.i i

27. INTERPRET This problem is about the electric field strength due to a point source charge—the proton.

DEVELOP The electric field strength at a distance r from a point source charge q is given by Equation 20.3:

2ˆ

kqE r

r

r

The proton in a hydrogen atom behaves like a point charge.

EVALUATE At a distance of one Bohr radius (a0 = 0.0529 nm) away, the electric field strength is

9 2 2 1911

2 11 2

0

(9 10 N m /C )(1.6 10 C)5.15 10 N/C

(5.29 10 m)

keE

a

ASSESS The field strength at the position of the electron is enormous because of the close proximity.

Section 20.4 Fields of Charge Distributions

28. Take the origin of x-y coordinates at the midpoint, as indicated, and use Equation 20.4. Let ˆ(2.5 cm)r j r

denote

the positions of the charges, and rr

that of the field point. A unit vector from one charge to the field point is

( )/| |,r r r r r r r r

so the spacial factors in Coulomb’s law are2 3 3/ ( )/ | | .i i i ir r r r r r r r

r r r r r r(a) For

1ˆ(5.0 cm) , r j r r r

r r r r 늿 ?(5.0 cm) (2.5 cm) (2.5 cm) ,j j j and2 2

ˆ(7.5 cm) .r r r r j r r r r

Then

291 1 2 2

3 3 2 2 2

1 2

늿N m ˆ9 10 (2 C) (25.6 MN/C)C (2.5 cm) (7.5 cm)

q r q r j jE k j

r r

r rr

(b) For ˆ(5.0 cm) ,r ir

29

2 2 2 2 3/ 2 2 2 3 / 2

늿 늿N m 2 C (5.0 2.5 ) (5.0 2.5 ) ˆ9 10 (5.15 MN/C)C cm (5 0 ( 2.5) ) (5.0 2.5 )

i j i jE j

r

(c) For 0,r r

20.4 Chapter 20

29

2 2 2 2

늿N m 2 C ˆ9 10 (57.6 MN/C)C cm (2.5) (2.5)

j jE j

r

29. INTERPRET Given the magnitude of the dipole moment, we are asked to calculate the distance between the pair

of opposite charges that make up the dipole.

DEVELOP As shown in Equation 20.5, the electric dipole moment p is the product of the charge q and the

separation d between the two charges making up the dipole:

p qd

EVALUATE Using the equation above, the distance separating the charges of a dipole is

30

19

6.2 10 C m38.8 pm 0.0388 nm

1.6 10 C

pd

q

ASSESS The distance d has the same order of magnitude as the Bohr radius0( 0.0529 nm).a

30. For a very long wire ( 38 cm),L >> Example 20.7 shows that the magnitude of the radial electric field falls off like

1 .r Therefore, (38 cm)/ (22 cm) 22 cm/38 cm;E E or (38 cm) (22/38)1.9 kN/C 1.10 kN/C.E

31. INTERPRET In this problem we are asked about the line charge density, given the field strength at a distance from

the wire.

DEVELOP If the electric field points radially toward the long wire ( 45 cm),L the charge on the wire must be

negative. The magnitude of the field is given by the result of Example 20.7, 2 .kr rE

EVALUATE Using the equation above, we find the line charge density to be

9 2 2

( 260 kN/C)(0.45 m)6.50 C/m

2 2(9 10 N m /C )

rE r

k

ASSESS The electric field strength due to a line charge density decreases as1/ .r This can be compared to the 2

1/r dependence due to a point charge.

32. INTERPRET We use Coulomb’s law and the definition of electric field to find the electric field at a point on the

axis of a charged ring.

DEVELOP From Example 20.6, which is done for a general distance x, we see that2 2 3/ 2( )

.kQx

x aE

We want to know

the field E at position .x a

EVALUATE

2 2 3 / 2 2 3 / 2 2( ) (2 ) 8

kQx kQa kQE

x a a a

ASSESS The units are2(distance),

kQwhich are correct for electric field.

Section 20.5 Matter in Electric Fields

33. INTERPRET This problem is about the Millikan oil drop experiment. Two forces are involved, gravitational and

electrical.

DEVELOP In equilibrium, the gravitational and electrostatic forces cancel: ,g eF F r r

or .mg qE The equation can be

used to compute the mass m.

EVALUATE Using the equation above, we find the mass to be


19 712

2

(10 1.6 10 C)(2 10 N/C)3.27 10 kg

(9.8 m/s )

qEm

g

ASSESS Because this mass is so small, the size of such a drop may be better appreciated in terms of its

radius,1/3

oil(3 /4 ) .R m Millikan used oil of density3

oil 0.9199 g/cm , so 9.46 mR for this drop.

34. For uniform acceleration, / ,a eE m electrons, starting from rest, reach speed2 2 ,v ax traversing a region of length x.

Therefore,2

( /10) 2( / ) ,c eE m x or

2 31 8 24

19

(9.11 10 kg) (3 10 m/s)5.12 10 N/C

200 200(1.6 10 C)(0.05 m)

mcE

ex

35. INTERPRET This problem is about the motion of a proton, a charged particle, in an electric field that points to the left.

DEVELOP Choose the x axis to the right, in the direction of the proton, so that the electric field is negative to the

left. If the Coulomb force on the proton is the only important one, the acceleration is

( )x

e Ea

m

The negative sign means that the proton decelerates as it enters the electric field. The motion is one-dimensional

kinematics.

EVALUATE (a) Using Equation 2.11, with5

3.8 10 m/soxv and 0,xv we find the maximum penetration into

the field region to be

2 2 27 5 2

0 19 3

(1.67 10 kg)(3.8 10 m/s)1.35 cm

2 2 2(1.6 10 C)(56 10 N/C)

ox ox

x

v mvx x

a eE

(b) The proton subsequently moves to the left, with the same constant acceleration in the field region, until it exits

with the initial velocity reversed.

ASSESS The deceleration of the proton increases with the field strength E. In addition, when E is large, the

penetration is small, and the proton reverses its path rather quickly.

36. From the analysis in Example 20.8,2 27 3 2 19

0 / (1.67 10 kg)(84 10 m/s) /(1.6 10 C)(0.075 m)E mv eb

982 N/C.

PROBLEMS

37. INTERPRET The problem asks for an estimate of the fraction of electrons removed from rubbing.

DEVELOP Suppose that half the ball’s mass is protons (the other half comes from neutrons). Their number is

0 1 g/ .p pN m This is equal to the original number of electrons,0.eN

EVALUATE The number of electrons removed is 1 C/ ,eN e so the fraction removed is

6 1911

24

0

(1 C/ ) (10 C)/(1.6 10 C)1.04 10

(1 / ) 1 g/(1.67 10 g)

e

e p

N e

N g m

ASSESS The fraction is about a hundred billionth. Thus, only a very small amount of electrons has been removed

by rubbing.

38. The product of the charges is2 2 9 2 2 10 2

1 2 Coulomb / (0.15 m) (95 N)/1(9 10 N m /C ) 2.38 10 C .q q r F k If one

charge is twice the other,1 22 ,q q then

2 10112

2.38 10 Cq and1 21.8 C.q

39. INTERPRET In solving this problem we follow Problem Solving Strategy 20.1 and identify the source charges as

the proton and the electron.

DEVELOP The unit vector from the proton’s position to the origin is ˆ.i Using Equation 20.1, the Coulomb force

of the proton on the helium nucleus is

9 2 2He

P,He 2 2

p,He

(9 10 N m /C )( )(2 )늿 ?( ) ( ) ( 0.180 nN)(1.6 nm)

pkq q e eF i i i

r

r

Similarly, the unit vector from the electron’s position to the origin is ˆ,j so its force on the helium nucleus is

20.6 Chapter 20

He

e,He 2 2

e,He

( )(2 )늿 ?( ) ( ) (0.638 nN)(0.85 nm)

ekq q k e eF i j j

r

r

EVALUATE The net Coulomb force on the helium nucleus is the sum of these:

net P,He e,He늿( 0.180 nN) (0.638 nN)F F F i j

r r r

ASSESS In situations where there are more than one source charge, we apply the superposition principle and add

the electric forces vectorially. In the above, since the electron is closer to the He nucleus than the proton

( e,He p,Her r ), we expect P,He e,He| | | |.F Fr r

40. Denote the positions of the charges by1

늿(16 5 ) cmr i j r

for1 9.5 C,q and

2늿(4.4 11 ) cmr i j

rfor

2 3.2 C.q The vector from q1 to q2 is 2 1,r r r r r r

and a unit vector in this direction is2 1 2 1

ˆ ( )/| |.r r r r r r r r r

The vector form of Coulomb’s law for the electric force of q1 on q2 is3

12 1 2 2 1 2 1( )/| | .F kq q r r r r r r r r r

(This gives the

Coulomb force between two point charges, as a function of their positions, and is a convenient form to memorize

because of its direct applicability.) Substituting the given values for this problem, we find:

9 2

12 2 2 2 3 / 2 3

늿 늿9 10 N m (4.4 11 16 5 ) cm(9.5 C)( 3.2 C)

C [(4.4 16) (11 5) ] cm

늿(14.2 7.37 ) N,

i j i jF

i j

r

with magnitude 16.0 N and direction 27.3 to the x axis (negative angle measured CW).

41. INTERPRET Coulomb’s law applies here. Since more than one source charge is involved, we make use of the

superposition principle.

DEVELOP For the force on a third charge Q to be zero, it must be placed on the x axis to the right of the (smaller)

negative charge, i.e., at .x a The net Coulomb force on a third charge so placed is

2 2

(3 ) ( 2 )

( )x

kQ q kQ qF

x x a

We set 0xF to solve for x.

EVALUATE The condition 0xF implies that2 2

3( ) 2 ,x a x or2 2

6 3 0.x xa a Thus,

2 23 9 3 (3 6)x a a a a

Only the solution (3 6) 5.45x a a is to the right of .x a

ASSESS At (3 6)x a the forces acting on Q from 3q and 2q exactly cancel each other. Notice that our

result is independent of the sign and magnitude of the third charge Q.

42. By symmetry, the negative charge must be at the midpoint between the two positive charges (the force on it is zero

there) such that its attractive force on one positive charge cancels the repulsive force of the other. Thus,

2 2 2(4 ) /(2 ) (4 ) | |/ ,k q a k q q a

which holds for any a. The equilibrium is unstable, since if q is displaced slightly toward one charge, the net

force on it will be in the direction of that charge.

43. INTERPRET More than one source charge is involved in this problem. Therefore, we use Coulomb’s law and

apply the superposition principle to find the force on q3.

DEVELOP We denote the positions of the charges by1 2

늿 m , 2 m) ,r j r i r r

and3

늿2 m) 2 m) .r i j r

The unit

vector pointing from q1 toward q3 is

3 1

13

3 1

( )ˆ

| |

r rr

r r

r r

r r


Similarly, the unit vector pointing from q2 toward q3 is3 2

3 2

( )

23 | |ˆ .

r r

r rr

r r

r r The vector form of Coulomb’s law and the

superposition principle give the net electric force on q3 as:

1 3 3 1 2 3 3 2

3 13 23 3 3

3 1 3 2

( ) ( )

| | | |

kq q r r kq q r rF F F

r r r r

r r r rr r r

r r r r

EVALUATE Substituting the values given in the problem statement, we find the force acting on q3 to be

1 3 3 1 2 3 3 2

3 13 23 3 3

3 1 3 2

6 69 2 2 6

22

( ) ( )

| | | |

늿 ?(68 10 C)(2 ) ( 34 10 C)2(9 10 N m /C )(15 10 C) =

8 m5 5 m

늿(1.64 0.326 ) N

kq q r r kq q r rF F F

r r r r

i j j

i j

r r r rr r r

r r r r

or2 2

3 3 3 1.67 Nx yF F F at an angle of1

3 3tan ( / ) 11.2y xF F to the x axis.

ASSESS The force between q1 and q3 is repulsive (1 3 0q q ) while the force between q2 and q3 is attractive

(2 3 0q q ). The two forces add vectorially to give the net force on q3.

44. The positions of the charges are the same as in the Problem 43, so the net force on q1 is

3 1 3 32 1 2 21 1 13 3 3/ 2 2 3 / 2 2

1 2 1 3

늿늿( ) ( 2 )( ) ( 2 )

| | | | 5 m m

q r r q i jq r r q i jF kq kq

r r r r

r rr rr

r r r r

(a) If1 0,yF then

2 3 0,q q or3 20 C.q (b) Then

9 2 3 / 2

1늿(9 10 N)(25 20 C )( 4 )5 1.61 N.F i i

r

45. INTERPRET Since more than one source charge is involved, we use Coulomb’s law and apply the superposition

principle to find the force on Q.

DEVELOP The magnitudes of the forces on Q from each of the four charges are equal to

0 22

2

( 2 /2)

kqQ kqQF

aa

To determine the net force and its direction, we note that the forces from the two positive charges on the same

diagonal are in opposite directions, and cancel, while the forces from the positive and negative charges on the other

diagonal are in the same direction (depending on the sign of Q) and add.

EVALUATE (a) The net force on Q has magnitude

net 0 2

42

kqQF F

a

(b) The direction ofnetFr

is toward the negative charge for 0,Q and away from the negative charge for 0.Q

ASSESS Even though we have four charges acting on charge Q, only two need to be considered. By the

superposition principle, the force on a charge placed midway between two identical charges must add to zero.

46. The electric field from a point charge at the origin is2 3ˆ( ) / / ,E r kqr r kqr r

r r rsince ˆ / .r r r

r(a) For

ˆ0.5 mr ir

and29 2 2늿65 C, (9 10 N m /C )(65 C) /(0.5 m) 2.34 MN/C.q E i i

r(b) At 늿0.5 m ( ),r i j

r

3 2 3늿 늿(9 65 10 N m /C)(0.5 m)( )/(0.5 2 m) (827 kN/C)( ).E i j i j r

(The field strength is 1.17 MN/C at

20.8 Chapter 20

45 to the x axis.) (c) When5 2늿 늿( 0.25 0.75 ) m, (5.85 10 N m /C)( 0.25 0.75 )m/r i j E i j

rr

2 2 3 / 2 3 늿[( 0.25) (0.75) ] m ( 296 888 ) kN/C (| | 936 kN/C, 108 ).xi j E r

47. INTERPRET Coulomb’s law applies here. Since more than one source charge is involved, we use the

superposition principle to find the point where the field strength vanishes.

DEVELOP We first note that the field can be zero only along the line joining the charges (the x axis). To the left

or right of both charges, the fields due to each are in the same direction, and cannot add to zero. Between the two, a

distance 0x from the1 C charge, the electric field is

1 2

2 2

늿 ( )=

(10 cm )

q i q iE k

x x

r

EVALUATE With1 1.0 Cq and

2 2.0 C,q the field vanishes when2 21 C/ 2 C/(10 cm ) ,x x or

10 cm/( 2 1) 4.14 cm.x

ASSESS Since 0E r

at 10 cm/( 2 1),x a charge placed at that point does not experience any force.

48. The proton, charge e, is at 0,pr r

and the ion, charge q, is at ˆ5 nm.Ir ir

The field at point ˆ5 nmr i r

is given by

Equation 20.4, with spacial factors written as in the solutions to Problem 15 or Exercise 28:

3 3 3

늿 ?( ) ( 5 nm) ( 5 nm 5 nm)( )

| | (5 nm) (10 nm)

i

i

i i

r r i i iE r kq ke kq

r r

r r

r rr r

Therefore, 0E r

implies3 32 /(10) /(5) ,q e or 4 .q e (Note how we used the general expression for the electric

field, at position 0,r r

due to a distribution of static point charges at positions .)irr

49. INTERPRET Coulomb’s law applies here. With two source charges, we use the superposition principle to find the

field strength at a point on the y axis.

DEVELOP As in Example 20.2, we apply the symmetry argument to show that the x components of the electric

field due to both charges cancel, and the net electric field points in the +y direction.

EVALUATE (a) Since electric field is the force per unit charge, from Example 20.2, we obtain

net, 2 2 2 3 / 2

22 cos

( )y

kq kqyE

r a y

(b) The magnitude of the field, a positive function, is zero for 0y and ,y hence it has a maximum in between.

Setting the derivative equal to zero, we find

2 2 3/2 2 2 5/230 ( ) ( ) (2 )

2a y y a y y

or2 2 23 0.a y y Thus, the field strength maxima are at

2.ay

ASSESS By symmetry, we expect the directions of the electric field at2

ay to be opposite.

50. We can use the result of Example 20.5, with y replaced by x, and x by y (or equivalently, 늿 by ,j i and 늿 by ).i j

Then2 2 3/ 2ˆ( ) 2 ( ) ,E x kqa j a x

r

where191.6 10 Cq e and 0.6 nm.a (Look at Figure 20.12 rotated 90

CW.) The constant9 2 2 19 2

2 2(9 10 N m /C )(1.6 10 C) (2.88 GN/C)(nm) .kq (a)

At2ˆ0, (0) 2 /x E kqj a

r

2늿(2.88 GN/C) /(0.6) (8.00 GN/C) .j j (b)

For 2 nm,x 2 2 3 / 2늿(2.88 GN/C) (0.6)(0.6 2 ) (190 MN/C) .E j j r

(c) At2 2 3 / 2늿20 nm, (2.88 GN/C) (0.6)(0.6 20 ) (216 kN/C) .x E j j

r

51. INTERPRET We find the electric field on the axis of a dipole, and show that Equation 20.6b is correct. To do this

we will use the equation for electric field.

DEVELOP The spacing between the + and – charges is 2a. We will use 2

q

rE k for each charge to find the total

field at a point .x a

EVALUATE


2 2

2 2

2 2

2

늿 ?[( ) ( ) ]( ) ( )

ˆ 1 1

q qE k i k i kq x a x a i

x a x a

kq a aE i

x xx

r

r

For .x a 2(1 ) 1 2 ,a a

x x

m so 2 2 3

(2 )늿[(1 2 ) (1 2 )] [4 ] 2 .kq kq k qaa a a

x x xx x xE i i r

But 2 ,p qd qa so 3

2 ˆ.kp

xE ir

ASSESS We have shown what was required.

52. Taking the hint, we suppose that the field strength varies with a power of the distance, .n

E r< Then

282/119 (1.5/2) ,n or ln(282/119)/ln(0.75) 3.00.n A dipole field falls off like3,r

hence the net charge is

zero.

53. INTERPRET Coulomb’s law applies here. With three source charges, we use the superposition principle to find

the field strength at a point on the y axis.

DEVELOP The electric field on the y axis ( 3 /2)y a due to the two charges on the x axis follows from

Example 20.2:

1 2 2 3/2

2 ˆ( /4)

kqyE j

y a

r

On the other hand, using Equation 20.3, we find the electric field due to the charge on the y axis to be

2 2ˆ

( 3 /2)

kqE j

y a

r

EVALUATE (a) For 3 /2,y a the total field is simply the sum of both terms:

1 2 2 2 3/2 2

2 1 ˆ( /4) ( 3 /2)

yE E E kq j

y a y a

r r r

(b) For ,y a the electric field may be approximated as

2 3 /2 2 2

2 1 3늿( )

y kqE kq j j

y y y

The field is like that due to a point charge of magnitude 3q.

ASSESS At large distance, the charge distribution looks like a point charge located at the origin.

54. The charges initially attract, so q1 and q2 have opposite signs, and2

1 22.5 N /1 m .kq q When the spheres are

brought together, they share the total charge equally, each acquiring 11 22

( ).q q The magnitude of their repulsion is

2 211 24

2.5 N ( ) /1 m .k q q Equating these two forces, we find a quadratic equation21

1 2 1 24( ) ,q q q q or

2 2

1 1 2 26 0,q q q q with solutions1 2( 3 8) .q q Both solutions are possible, but since

13 8 (3 8) ,

they merely represent a relabeling of the charges. Since2 9 2 2 2

1 2 2.5 N m /(9 10 N m /C ) (16.7 C) ,q q the

solutions are1 3 8 (16.7 C) 40.2 Cq and

2 40.2 C/(3 8) 6.90 C,q m m or the same values

with q1 and q2 interchanged.

20.10 Chapter 20

55. INTERPRET Two forces are involved in this problem: Coulomb force and spring force. The spring is stretched

due to Coulomb repulsion.

DEVELOP Suppose that the Coulomb repulsion is the only force stretching the spring. When balanced with the

spring force, ,e sF F or

2

2

0( )s

kqk x

L x

where 0L is the equilibrium length. This cubic equation can be solved by iteration or by Newton’s method.

EVALUATE Substituting the values given in the problem statement gives

2 9 2 2 22 2 2 3

0

(9 10 N m /C )(34 C)( ) (0.5 m ) 6.94 10 m

150 N/ms

kqx L x x x

k

Newton’s method yields 15.95 cm.x

ASSESS Our result makes sense since the amount stretched is seen to decrease with increasing spring constant ,sk

and increase with the magnitude of the charge q.

56. INTERPRET We show that the electric field magnitude a distance x from either end of a uniformly charged rod of

charge Q and length L is as shown. We will use the integral form for electric field, Equation 20.7.

DEVELOP For simplicity, we will align our coordinate axis with the right end of the rod, and integrate from –L to 0.

In this case, x will be not only the x coordinate, but it will be the distance from one end of the rod as well. We will

find the field at x using 2ˆ.

k dq

rE r r

EVALUATE The infinitesimal charge dQ is the charge per length times the infinitesimal length: .Q

LdQ dx We

substitute this into the equation for electric field:

00 0

2 2 2

( ) 1 1 1

( ) ( )

( ) ( )

Q

L

L LL

k drk dq kQ dr kQ kQE

L L x r L x x Lr x r x r

kQ x L x kQE

L x x L x x L

ASSESS We have found the field a distance x from one end of the rod. By symmetry, the field at the other end

must be the same.

57. INTERPRET The electron undergoes circular motion, and the centripetal force is provided by the Coulomb force.

DEVELOP The electric field of the wire is radial and falls off like1/ ( 2 / ,r E k r see Example 20.7). For an

attractive force (negative electron encircling a positively charged wire), this is the same dependence as the

centripetal acceleration. For circular motion around the wire, the Coulomb force provides the electron’s centripetal

acceleration:

22

r

eE ke va

m mr r

The equation can be used to deduce the speed of the electron.

EVALUATE Substituting the values given, we find the speed to be

9 2 2 19 96

31

2 2(9 10 N m /C )(1.6 10 C)(2.5 10 C/m)2.81 10 m/s

9.11 10 kg

kev

m

ASSESS We find the speed of the electron to be independent of r, the radial distance from the long wire. This is

because both the electric field and the centripetal acceleration fall off as1/ ;r hence, the r dependence cancels out.

58. When the device is in operation, an isotope, of nuclear charge q and mass m, is accelerated from rest to a speed v,

in a distance d, by the field E1, where2

1 12 2( / ) .v a d qE m d This will be the proper speed to pass through the

analyzer if2

2 2 1/ / 2( / ) / ,a qE m v r qE m d r or2 12 / .E E d r This condition depends on the fields and the

geometry, but not on / ,q m so different isotopes cannot be separated. (Essentially, the device compares two

accelerations, both of which are proportional to / .)q m


59. INTERPRET The charge undergoes circular motion, and the centripetal force is provided by the Coulomb force.

We are asked to find the line charge density, given the particle’s speed.

DEVELOP The electric field of the wire is radial and falls off like1/ ( 2 / ,r E k r see Example 20.7). For an

attractive force (positive charge encircling a negatively charged wire), this is the same dependence as the

centripetal acceleration. For circular motion around the wire, the Coulomb force provides the centripetal

acceleration:

22

r

qE kq va

m mr r

The equation can be used to deduce the line charge density, once the speed is known.

EVALUATE The above equation gives

2 9 2

9 2 2 9

(6.8 10 kg)(280 m/s)14.1 C/m

2 2(9 10 N m /C )(2.1 10 C)

mv

kq

ASSESS For the force to be attractive, we require the line charge density to be negative.

60. (a) The torque on an electric dipole in an external electric field is given by Equation 20.9; | | sinP E pE r r

(1.5 nC m)(4.0 MN/C)sin 30 3.0 mN m. (b) The work done against just the electric force is equal to the

change in the dipole’s potential energy (Equation 20.10); ( ) ( ) (cos30f iW U p E p E pE r rr r

cos180 ) (1.5 nC m) (4.0 MN/C)(1.866) 11.2 mJ.

61. INTERPRET This problem is about an electric dipole placed in an external electric field.

DEVELOP Using Equation 20.10, the energy required to reverse the orientation of such a dipole is 2 .U pE

EVALUATE Using the equation above, the electric dipole moment is

2730

3

3.1 10 J1.29 10 C m

2 2(1.2 10 N/C)

Up

E

ASSESS An electric dipole tends to align itself in the direction of the external electric field. Thus, energy is

required to change its orientation.

62. All the forces are along the same line, so take the origin at the center of the left-hand dipole and the positive x axis in

the direction of the right-hand dipole in Figure 20.29. The right-hand dipole has charges q at /2,x a q at /2,x a

each of which experiences a force from both charges of the left-hand dipole, which are q at /2a and q at /2.a

(There are forces between four pairs of changes.) The Coulomb force on a charge in the right-hand dipole, due to one

in the left-hand one, is3ˆ( ) | |r L r r Lkq q x x i x x (see solution to Problem 15), so the total force on the right-hand

dipole is

2 2 2 22

2 2 2 2 2 2 2 2

1 1 1 1 2 (3 )늿( ) ( ) ( )

x

kq a x aF kq i i

x x a x a x x x a

(a) In the limit2 2 2 6 2 2 4 2 4늿 ?, 2 (3 ) / 6 / 6 / ,xa x F kq a x i x kq a i x kp i x where p qa is the dipole moment of

both dipoles. (b) The force on the right-hand dipole is in the negative x direction, indicating an attractive force.

63. INTERPRET This problem is about the interaction between a dipole and the electric field due to a source charge.

DEVELOP With the x axis in the direction from Q to pr

and the y axis parallel to the dipole in Figure 20.30,

we have ˆ(2 )p qa jr

and2 ˆ( / ) .E kQ x i In the limit ,x a the torque on the dipole is given by Equation

20.9, ,p E rr r

where Er

is the field from the point charge Q, at the position of the dipole.

EVALUATE (a) Using Equation 20.9, we find the torque to be

20.12 Chapter 20

2 2

2 ˆ늿(2 )kQ kQqa

p E qaj i kx x

rr r

The direction is into the page, or clockwise, to align pr

with .Er

(b) The Coulomb force obeys Newton’s third law. The field of the dipole at the position of Q is (Example 20.5

adapted to new axes)

dipole 3

2 ˆkqaE j

x

r

Thus, the force on Q due to the dipole is

on dipole 3

2 ˆQ

kQqaF QE j

x

r r

The force on the dipole due to Q is the opposite of this:

on dipole on 3

2 ˆQ

kQqaF F j

x

r r

The magnitude of on dipoleFr

is3

2 / .kQqa x

(c) The direction of on dipoleFr

is in ˆ,j or parallel to the dipole moment.

ASSESS The net force on dipoleFr

will cause the dipole to move in the ˆ direction.j In addition, there is a torque that

tends to align pwith .E So, the motion of the dipole involves both translation and rotation.

64. The electron’s field is directed toward the electron (a negative charge) and the ion’s field is directed away from the

ion (a positive charge). Therefore, the fields can cancel only at points on the negative x axis ( 0),x since the

directions are opposite there and the smaller charge is closer. The field from one point charge is ( )qE x r

3ˆ( )/| | ,q qkqi x x x x where , 0qq e x for the electron, and 5 , 10 nmqq e x for the ion. The total field is

zero when30 [( ) | |k e x x 35 ( 10 nm) | 10 nm| ].e x x (See note to solution of Problem 48.) Since

0, | |x x x and | 10 nm| 10 nm ,x x so this implies2 2

5(10 nm ) 0,x x or

24 2(10 nm)x x

2(10 nm) 0. The negative solution to this quadratic is

2 2[ 10 nm (10 nm) 4(10 nm) ]

2.5 nm(1 5) 8.09 nm4

x

65. INTERPRET This problem is about the electric field due to a charged ring.

DEVELOP The electric field on the axis of a uniformly charged ring of radius a is calculated in Example 20.6:

2 2 3 / 2( )

kQxE

x a

Knowing the field strengths at two different values of x allows us to deduce a and Q.

EVALUATE The data given in the question imply

1 2 2 3 /2

2 2 2 3/ 2

(5 cm)380 kN/C

[(5 cm) ]

(15 cm)160 kN/C

[(15 cm) ]

kQE

a

kQE

a

Dividing these two equations and taking the 2

3root, we get

2/3 2 2

2 2

380 15 (15 cm)3.70

160 5 (5 cm)

a

a


which when solved for the radius, gives

2 2(15 cm) (3.70)(5 cm)7.00 cm

2.70a

To calculate Q, we substitute for a in either one of the field equations. This leads to

2 2 3/ 2

9 2 3

(380 kN/C)[(5 cm) (7 cm) ]538 nC

(9 10 N m /C )(5 cm)Q

ASSESS To check that our results are correct, we may substitute the values obtained for a and Q into the field

equation to calculate 1E and 2E at 1 5 cmr and 2 15 cm.r Note that the field strength decreases as x is increased.

66. (a) The electric field, from the three point charges shown, at points on the x axis with x a is:

2 22

2 2 2 2 2 2

2 (3 )늿( ) 2( ) ( ) ( )

q q q x aE x ki kqa i

x a x x a x x a

r

(b) For2 4ˆ, ( ) 6 / .x a E x kqa i x

r(The quadrupole moment of this “linear quadrupole” is

24 .)xxQ qa

67. INTERPRET This problem is about the electric field due to a 10-m long straight wire, which is our source charge.

DEVELOP For a uniformly charged wire of length L and charge Q, the line density is / .Q L Under the

assumption that the wire is approximately infinitely long, the electric field due to the line charge can be written as

(see Example 20.7):

2kE

r

EVALUATE (a) The charge density is

25 C2.5 C/m

10 m

Q

L

(b) Since 15 cm <<10 mr L and the field point is far from either end, we may regard the wire as approximately

infinite. Then Example 20.7 gives

9 2 22 (2 9 10 N m /C )(2.5 C/m)300 kN/C

0.15 m

kE

r

(c) At 350 m 10 mr L , the wire behaves approximately like a point charge, so the

field strength is

ASSESS The finite-size, line charge distribution looks like a point charge at large distances.

68. (a) / .Q L (b) The x components of the fields from symmetrically placed elements of charge, dq dx at ,x

cancel, so the net field is along the y axis (see Fig. 20.16). (c) Proceed exactly as in Example 20.7, except that the

limits of integration are from /2 to /2.L L Thus,

/ 2

/ 2

2 2 3 / 2 2/ 2 2 2 2 2

/ 2

( ) /4

L

L

yL

L

dx x k LE k y k y

x y y x y y y L

(d) For ,y L we can neglect L in the square root, so we obtain2 2/ /yE k L y kQ y as for a point charge.

(Of course, for ,L the result of Example 20.7 is recaptured.)

69. INTERPRET In this problem we want to find the electric field due to a uniformly charged disk of radius R.

DEVELOP We take the disk to be consisted of a large number of annuli. With uniform surface charge density ,

the amount of charge on an area element dA is .dq dA Our strategy is to first calculate the electric field dE due

to dq at a field point on the axis, simplify with symmetry argument, and then integrate over the entire disk to get E.

EVALUATE (a) The area of an annulus of radii1 2R R is just

2 2

2 1( ).R R For a thin ring,1R r and

2 ,R r dr

so the area is2 2 2[( ) ] (2 ).r dr r rdr dr When dr is very small, the square term is negligible, and

9 6 2

2 2

(9 10 25 10 N m /C)1.84 N/C

(350 m)

kQE

r

20.14 Chapter 20

2 .dA rdr (This is equal to the circumference of the ring times its thickness.) (b) For surface charge density

, dq dA 2 .rdr (c) From Example 20.6, the electric field due to a ring of radius r and charge dq is

2 2 3/2 2 2 3/2

2

( ) ( )x

xdq k xrdE k dr

x r x r

which holds for x positive away from the ring’s center. (d) Integrating from 0r to R, one finds

2 2 3 / 2 2 2 1/20 0 2 20

12 2 2

| |( ) ( )

RR R

x x

rdr x xE dE k x k x k

xx r x Rx r

For 0, | |x x x and the field is

2 22 1x

xE k

x R

On the other hand, for 0, | | ,x x x the electric field is

2 2

| |2 1x

xE k

x R

This is consistent with symmetry on the axis, since ( ) ( ).x xE x E x

ASSESS One may readily verify that (see Problem 71), for ,x R 2 .kQ

x xE In other words, the finite-size charge

distribution looks like a point charge at large distances.

70. An infinite flat sheet is the same as an infinite flat disk (as long as the dimensions are infinite in all directions, the

shape is irrelevant). Thus, we can find the magnitude of the electric field from a uniformly changed infinite flat

sheet by letting R in the result of Problem 69. Then, the limit of the second term is zero, and the magnitude is

constant, 2 .E k (The direction is perpendicularly away from (towards) the sheet for positive (negative) .)

71. INTERPRET In this problem we want to show that at large distances, the electric field due to a uniformly charged

disk of radius R reduces to that of a point charge.

DEVELOP The result of Problem 69 for the field on the axis of a uniformly charged disk, of radius R, at a distance

0x along the axis (away from the disk’s center) is

2 22 1x

xE k

x R

For2 2/ 1,R x we use the binomial expansion in Appendix A and write

1/ 22 2

2 2

11 1

2

R R

x x

EVALUATE Substituting the above expression into the first equation, we obtain

1/ 22 2 2

2 2 2 2

1 22 1 1 2 1 1

2 2x

R R k R kQE k k

x x x x

L

which is the field from a point charge2

Q R at a distance x.

ASSESS The result once again demonstrates that any finite-size charge distribution looks like a point charge at

large distances.

72. Begin by establishing a coordinate system with origin at point P, vertical y axis, and horizontal x axis. Then each

charge element dq creates an electric field of magnitude2

/dE kdq a in the direction 늿ˆ cos sin .r i j The total

electric field at P is then ˆ( ) .E P dEr r

The loop has a uniform charge Q along its length of a so its linear charge

density is / ,Q a and we can write .dq dl Expressing the line element as ,dl ad we have .dq ad We

can now formulate the integral for the total electric field in terms of starting with / :dE k d a


20 0 0

2 2늿 늿ˆ( ) cos sink k k kQ

E P d r d i d j j ja a a a

73. INTERPRET We find the position of the charge Q in Example 20.2 for which the force is a maximum. At large

distances, the force will be small because of the inverse-square nature of the force. At close distances, the net force

will be small because the forces from the two charges tend to cancel. Somewhere between near and far will be a

maximum. The equation for force is found in the example, so we will differentiate to find the value of y where

force is a maximum.

DEVELOP We are given the equation for force:2 2 3/ 2

2

( )

ˆ.kqQy

a yF j

We find the value of y at which this force is

a maximum by setting 0.dF

dy

EVALUATE

2 2 3 / 2 2 2 3 / 2 2 2 3 / 2

2 1 32 2

( ) ( ) ( ) 2

dF d kqQy d ykqQ kqQ

dy dy dya y a y a y

22

2 2 5 / 2

2 2 22 2 2 2 2

2 2 5 / 2

0( )

( ) 30 ( ) 3 0 2

( ) 2

y

a y

a y y aa y y a y y

a y

ASSESS This is a bit less than one and a half times the distance from the center to one charge.

74. INTERPRET We find the electric field at any point (x,y) due to a uniformly charged rod of length L and charge Q.

We will check our answer by showing that the result matches the special cases of Problem 68 and Problem 56.

This is an electric field problem, which we will solve by direct integration.

DEVELOP We will find the field at point ( , ),P x y and for the integration variable we’ll use .x The infinitesimal

charge dq is the charge per length times the length ,dx so .Q

Ldq dx The distance from each bit of charge dq to the

point (x,y) is2 2

( ) .r x x y The unit vector in the direction of r is 2 2 1/ 2

늿( )

| | [( ) ]ˆ .

x x i yjr

r x x yr

We will find the electric

field by integrating 2ˆ.

k dq

rdE r

EVALUATE

(a)

/ 2 / 2

2 2 2 2 2 1/ 2 2 2 3 / 2/ 2 / 2

/ 2

2 2 2 2

/ 2

2 22 2

2 2

늿 늿( ) ( )ˆ

[( ) ] [( ) ] [( ) ]

1 늿

( ) ( )

1 1

QL L

L

L L

L

L

L L

k dxk dq x x i yj kQ x x i yjdE r E dx

Lr x x y x x y x x y

kQ x xE i j

L x x y y x x y

kQE

L x y x y

r

r

2 2

2 22 2

2 2

늿L L

L L

x xi j

y x y y x y

When 0,x the electric field in part (a) reduces to

2 2 22

2

2늿04L

kQ L kQE i j

L y L yy y

r

(b) When 0,y the electric field reduces to

20.16 Chapter 20

2

02 2

1 1 ˆ lim

L

L L y

xkQE i

L x x

r

2Ly x

2Lx

2

Ly x

2 2

22 0

2

ˆ

1 1ˆ limL L

yL

j

x xkQE i

L y yx

rˆ kQj

L

L

2 2 22

2

4ˆ4L

kQi

x Lx

ASSESS The electric field in part (a) is not simple, but we have shown that for some simple cases it reduces to

simpler forms.

75. INTERPRET We find the electric field near a line of non-uniform charge density. This is an electric field

calculation, and we will integrate to find the field.

DEVELOP The rod has charge2

0 ( ) ,x

L and extends from 0x to .x L We want to find the electric field at

.x L We will use 2ˆ,

k dq

rdE rr

with dq dx and ( ).r x L

EVALUATE

22

0 0

2 2 200

2

0 0 0

2 2

늿ˆ 2 ln( )( )

늿 ?2 12 ln 2 ln(2) 2ln 2

2 2 2

LxL

L kk dq LdE r E ki dx i x L x L

x Lr x L L

k i k i k iL L LE L L L L

L L LL L

r

r

ASSESS Since0 is charge per length, the units are correct.

76. INTERPRET We find the electric field at points along the axis of a rod carrying non-uniform charge density. This

is an electric field calculation, and we will integrate to find the field. We will also show that at large distances the

field looks like that of a dipole, and we’ll find the dipole moment.


0 ( ) ,x

L and extends from x L to .x L We want to find the electric field

at some point .x L We will use 2ˆ,

k dq

rdE r with dq dx and ( ).r x x

EVALUATE

(a)

0 0

2 2

0

0

2 2

0

2 2

늿ˆ ln( )( )

ˆln

ˆ ( ) ( )ln

ˆ 2ln

LxL

L

LL

kk dq xdE r E ki dx i x x

L x xr x x

k i x x x LE

L x L x L x L

k i x x L x x L x LE

L x LL x

k i xL x LE

L x Lx L

r

r

r

(b) At ,x L


2

2

022

0

12

0

2

2

0

2

ˆ 2ln( ) ln( )

ˆ 2ln 1 ln 1

1

ˆ 21 ln 1 ln 1

ˆ 2 11

2

L

x

k i xLE x L x L

L x L

k i L L LE x x

L x xx

k i L L L LE

L x x xx

k i L L L LE

L x xx

r

r

r

r 2

2x

2

2

1

2

L L

x x

0ˆ 2k i L

EL x

r

3

3

2 2L L

xx

2

0

3

2 ˆk LE i

x

r

We compare this with the field of a dipole in the same orientation, 3

2 ˆ,kp

xE ir

and see that the dipole moment of this

charged rod is2

0 .p L

ASSESS We have shown what was required.

77. INTERPRET We estimate the electrostatic force between two people if the charge on the electron differed from

the charge on the proton by one part in a billion. We use Coulomb’s law.

DEVELOP We will assume that about half a person’s mass consists of protons and the rest consists of neutrons.

We will also assume that the number of electrons is equal to the number of protons, so the person’s net charge is

just one billionth of his total proton charge. One proton has a mass271.673 10 kg,pm

and the charge on a

proton is191.602 10 C.q The magnitude of the force between two charges is 1 2

2 .q q

rF k

EVALUATE The number of protons in a 65-kg person is about

28

27

1 65 kg1.9 10

2 1.673 10 kgN

We multiply this by the charge of a proton, and then divide by a billion to find the charge on a

person: 9103.11 C.

Nqq The force becomes

2

2 870 MN,q

rF k which does not qualify as “negligible.”

ASSESS This is a repulsive force nearly equal to the weight of a Nimitz-class aircraft carrier.

78. INTERPRET We find the minimum speed necessary for a droplet to traverse the region between two plates

without touching the plates. We solve this by finding the force on the droplet due to the electric field, then finding

the acceleration.

DEVELOP The force on the charge is .F qE The acceleration is then .qEF

m ma We can use this, and the

constant-acceleration equations, to find the time it would take the droplet to reach the plates, assuming that the

droplet starts in the middle of the gap. The minimum speed is such that it travels a distance L in this time.

EVALUATE First we find the time:

0x x 0v 2 221 2

2

d mx mdt at t

a qE qE

The speed is

L qEv L

t md

ASSESS Actually, the speed should be a bit higher than this because the drop has a non-zero diameter.

79. INTERPRET We are asked to find the electric field necessary to levitate small charged objects. We will assume

that the only forces acting on the beads are the force of gravity and the electric force.

DEVELOP To levitate the beads, we must arrange so that the force on the beads due to the field is equal to the

force due to gravity .qE mg The mass is 0.010 kg,m and the charge is65.0 10 C.q

20.18 Chapter 20

EVALUATE 319.6 10 N/C.

mg

qE The field should be upwards, so that the positively charged beads feel a

force opposite the direction of gravity.

ASSESS One complication to this plan is trying to keep the beads from accelerating sideways! It is very difficult

to actually make a large-scale uniform field as described here.

80. INTERPRET We find the electric field at the end of a rod with a non-uniform charge density. This is an electric

field calculation, and we will integrate to find the field.


0 ( ) sin( ),x x

L L

and extends from 0x to .x L We want to find the electric

field at 0.x We will use 2ˆ,

k dq

rdE rr

with dq dx and .r x

EVALUATE

2

0 0

2 2 20 0

0 0 0

2 2

0

sin늿ˆ sin

늿 22 ˆcos

x xL L

L L

L

kk dq xdE r E ki dx i dx

Lr x L

k i k i kL x LE i

L LL L

r r

r

ASSESS Since0 is charge per length, the units are correct.

21.1

GAUSS’S LAW

EXERCISES

Section 21.1 Electric Field Lines

18. The number of lines of force emanating from (or terminating on) the positive (or negative) charges is the same

(14 in Fig. 21.31), so the middle charge is 3 C and the outer ones are 3 C. The net charge shown is

therefore 3 3 3 3 C. This is reflected by the fact that 14 lines emerge from the boundary of the figure.

19. INTERPRET This problem is about drawing field lines to represent the field strength of a charge configuration.

DEVELOP We follow the methodology illustrated in Figure 21.3. There are 16 lines emanating from charge +2q

(eight for each unit of +q). Similarly, we have 8 lines ending on .q

EVALUATE The field lines of the charge configuration are shown below.

ASSESS Our sketch is similar to Fig. 21.3 ( f ) with twice the number of lines of force.

20. (The sketch shown follows the text’s convention of eight lines of force per charge magnitude q.)

21. INTERPRET In this problem we are asked to identify the charges based on the pattern of the field lines.

DEVELOP From the direction of the lines of force (away from positive and toward negative charge) one sees that

A and C are positive and B is a negative charge. Eight lines of force terminate on B, eight originate on C, but only

four originate on A, so the magnitudes of B and C are equal, while the magnitude of A is half that value.

EVALUATE Based on the reasoning above, we may write 2 .C B AQ Q Q The total charge isAQ Q

,B C AQ Q Q so.2C BQ Q Q

ASSESS The magnitude of the charge is proportional to the number of field lines emerging from or terminating at

the charge.

21

21.2 Chapter 21

Section 21.2 Electric Flux and Field

22. (a) When the surface is perpendicular to the field, its normal is either parallel or anti-parallel to .Er

Then Equation 21.1

gives2 2cos(0 or 180 ) (850 N/C)(2 m ) 1.70 kN m /C.E A EA

rr(b)

cos(45 or 135 )E A EA rr

2 2(1.70 kN m /C)(0.866) 1.20 kN m /C. (c) cos90 0.EA

23. INTERPRET This problem is about finding the electric field strength, given the flux through a surface.

DEVELOP The magnitude of the flux through a flat surface perpendicular to a uniform field is

| | EA

EVALUATE From the equation above, we find the field strength to be

2

4 2

| | 65 N m /C650 kN/C

10 mE

A

ASSESS The general expression for the electric flux is given by Equation 21.1: cos ,E A EA rr

where is

the angle between the normal vector Ar

and the electric field .Er

We see that when 0, EA is a maximum, and

when 180 , EA is a minimum.

24. The surface can be represented by a vector area2 ˆ(0.14 m )( ).A k

r(Since the surface is open, we have a choice of

normal to the x-y plane.) Then2 2 2ˆ (0.14 m ) (0.14 m ) (3.5 kN/C)(0.14 m )zE A E k E

rr r

2490 N m /C. (Only the z component of the field contributes to the flux through the x-y plane.)

25. INTERPRET This problem is about the electric flux through the surface of a sphere.

DEVELOP The general expression for the electric flux is given by Equation 21.1: cos ,E A EA rr

where

is the angle between the normal vector Ar

and the electric field .Er

The magnitude of the normal vector Ar

is 24 ,A r the surface area of the sphere.

EVALUATE For a sphere, with Er

parallel or anti-parallel to ,Ar

Equation 21.1 gives

2 2 24 4 (0.1 m/2) (47 kN/C) 1.48 kN m /Cr E

ASSESS The flux is positive if Ar

points outward, and is negative if Ar

points inward.

26. INTERPRET We find the total electric flux through a surface surrounding a charge q. This is the simplest

application of Gauss’s law, integration is not required.

DEVELOP According to Gauss’s law, the flux through a closed surface is proportional to the charge enclosed by

that surface: enclosed

0.

q

We will take the total charge to be12

enclosed 10 ,eq q where19

1.60 10 Ceq is the

charge on a single electron.

EVALUATE

19 123 2

12 2

1.6 10 10 C18 10 Nm

8.85 10 C/N m

ASSESS We could also use Gauss’s law to find the approximate electric field in the area surrounding this sock.

Section 21.3 Gauss’s Law

27. INTERPRET This problem is about applying Gauss’s law to find the electric flux through a closed surface.

DEVELOP Gauss’s law given in Equation 21.3 states that the flux through any closed surface is proportional to the

charge enclosed:

enclosed

0

qE dA

rrr

EVALUATE For the surfaces shown, the results are as follows:

(a) enclosed ( 2 ) ,q q q q 0/ .q

(b) enclosed ( 2 ) ( ) 3 ( 3 ) 2q q q q q q q and

02 / .q

(c) enclosed 0q and 0.

Gauss’s Law 21.3

(d) enclosed 3 ( 3 ) 0q q q and 0.

ASSESS The flux through the closed surface depends only on the charge enclosed, and is independent of the shape

of the surface.

28. Since the sphere encloses both charges (and none other) the flux through it isenclosed 0/ (6.8 4.7) C/(8.85q

12 2 210 C /N m) 237 kN m /C.

29. INTERPRET This problem is about applying Gauss’s law to find the electric flux through the surface of a cube

which encloses a charge.

DEVELOP Gauss’s law given in Equation 21.3 states that the flux through any closed surface is proportional to the

charge enclosed:

enclosed

0

qE dA

rrr

The symmetry of the situation guarantees that the flux through one face is1/6 the flux through the whole cubical

surface.

EVALUATE The flux through one face of a cube is

cube

2enclosed

face 12 2 2

0

1 1 2.6 C49.0 kN m /C

6 6 6(8.85 10 C /N m )

qE dA

rrr

ASSESS Since the flux through each surface is the same, the total flux through the cube is simply equal to

face6 , and is proportional toenclosed .q

Section 21.4 Using Gauss’s Law

30. The electric field due to a uniformly charged sphere is like the field of a point charge for points outside the sphere,

i.e.,2( ) 1/E r r for .r R Thus, at 10 cm from the surface, 15 cmr and

2(15 cm) (5/15) (5 cm)E E

(90 kN/C)/9 10 kN/C.

31. INTERPRET This problem is about applying Gauss’s law to calculate electric field. Our charge distribution has

spherical symmetry.

DEVELOP In computing the electric field strength, we make use of the result obtained in Example 21.1, or

Equations 21.4 and 21.5:

3

0

2

0

1,

4

1,

4

Qrr R

RE

Qr R

r

EVALUATE (a) At 15 cm ( 25 cm),r r R the electric field is

9 2 2

3 3

0

1 (9 10 N m /C )(14 C)(15 cm)1.21 MN/C

4 (25 cm)

QrE

R

(b) At ,r R

9 2 2

2 2

0

1 (9 10 N m /C )(14 C)2.02 MN/C

4 (25 cm)

QE

R

(c) Similarly, at 2 ,r R R we have

9 2 2

2 2

0

1 (9 10 N m /C )(14 C)504 kN/C

4 (2 ) (50 cm)

QE

R

ASSESS Inside the solid sphere where ,r R the electric field increases linearly with r. On the other hand, outside

the sphere where ,r R the field strength decreases as2

1/ .r Gauss’s law can be applied in this problem because the

charge configuration possesses spherical symmetry.

21.4 Chapter 21

32. The total electric field, the superposition of the fields due to the point charge and the spherical shell, is spherically

symmetric about the center. Inside the shell ( 8 cm),r R its field is zero, so the total field is just due to the10 nC

point charge. Outside ( ),r R the shell’s field is like that of a point charge of 20 nC at the same central location as

the10 nC charge. (This situation is described in Example 21.3.) (a) and (b) For 2 cm or 6 cm ,r R 2 9 2 2 2

pt/ (9 10 N m /C )(10 nC)/(2 cm or 6 cm) 225 kN/CE kq r or 25.0 kN/C, respectively, directed

radially outward. (c) For 2

pt shell15 cm , ( )/r R E k q q r 9 2 2 2(9 10 N m /C )(10 nC 20 nC)/(15 cm) 4.00 kN/C, directed radially inward.

33. INTERPRET In this problem we are given the field strength at two different points outside the charge distribution

and asked to determine the symmetry possessed by the configuration.

DEVELOP The symmetry of the charge distribution can be determined by noting that the electric field strength

decreases as21/r for a spherically symmetric charge distribution, and as1/r for a line charge (see Examples 21.4

and 21.1).

EVALUATE Let us write ,nE Cr for some constant C. This gives

2 1 2 1

1 2 1 2

ln ln

n

E r E rn

E r E r

Substituting the values given, we obtain

2 1

1 2

ln( / ) ln(55/43)1.00

ln( / ) ln(23/18)

E En

r r

Thus, we conclude that the charge distribution possesses line symmetry.

ASSESS The1/r dependence characteristic of line symmetry can be readily verified by taking the field strength to

be of the form 2 / .E k r

34. INTERPRET We are given the force on an electron near a sheet of charge, and are asked to find the surface charge

density on the sheet. This problem has planar symmetry, and we will use the result for the electric field near an

infinite plane of charge.

DEVELOP The force on a charge is ,F qE and02

E near a plane of charge. We note that the electron (a negative

charge) is repelled, so the charge on the plane must be negatively charged as well. The charge on the electron

is19

1.6 10 C,q and the magnitude of the force felt by the electron is

121.8 10 N.F We solve for .

EVALUATE 0

0

2 4 2

22.0 10 C/m .

Fq

qF qE

The charge on the plate must be negative, so

4 22.0 10 C/m .

ASSESS We could also get the sign of the charge sheet by being more careful about F and E. They are both vectors!

35. INTERPRET We find the electric field produced by a uniform sheet of charge. This problem has planar symmetry,

and we will use the result for the electric field near an infinite plane of charge.

DEVELOP The electric field near a uniform sheet of charge with charge density is02.E

The charge density

given in the problem is2 12 287 pC/cm 87 10 C/(0.01 m) .

EVALUATE

3

0

49 10 N/C2

E

ASSESS Although this seems at first to be a small charge per area, the resulting field is quite large. Remember that

a Coulomb is a very large charge.

36. INTERPRET We find the surface charge density that will produce a given electric field near a flat sheet of charge.

This problem has planar symmetry, and we will use the result for the electric field near an infinite plane of charge.

DEVELOP The electric field near a uniform sheet of charge with charge density is02.E

The electric field

given in the problem is31.4 10 N/C,E and we solve for .

EVALUATE

Gauss’s Law 21.5

8 2

0

0

2 2.5 10 C/m2

E E

ASSESS This is a small charge density in terms of the Coulomb, but the Coulomb is a relatively large charge.

Section 21.5 Field of Arbitrary Charge Distribution

37. INTERPRET The charge distribution has approximate line symmetry, and we apply Gauss’s law to compute the

electric field.

DEVELOP Close to the rod, but far from either end, the rod appears infinite, so the electric field strength is (see

Example 21.4): 2 / .E k r

EVALUATE (a) Substituting the values given in the problem statement, we obtain

9 2 262 2 ( / ) 2(9 10 N m /C )(2 C/0.5 m)

5.14 10 N/C0.014 m

k k Q lE

r r

(b) Far away ( ),r L the rod appears like a point charge, so

9 2 2

2 2

(9 10 N m /C )(2 C)34.0 N/C

(23 m)

kqE

r

ASSESS Far from the finite charge distribution (line charge in this case), the field always resembles that of a point

charge.

38. INTERPRET We determine the approximate field strength near a charged piece of paper with a given surface

charge density. We will assume that the paper is flat and is large compared to the distance involved.

DEVELOP The electric field due to an infinite sheet of charge is02.E

A piece of paper is not infinite, but the

problem states that it is an ordinary piece of paper, so we’ll assume that the paper is large compared to the 1-cm

distance from the paper. This approximation allows us to use the equation for the infinite case. The surface charge

density on the paper is245 nC/m .

EVALUATE 0

2500N/C2

E

<

ASSESS The infinite sheet of charge is not a useless construct. If you are close enough to any flat surface, it looks

infinite, and you can use the infinite case as an approximation.

39. INTERPRET We approximate the electric field strength near a charged disk and far from the charged disk. In both

cases, we will choose appropriate approximations for how the disk looks.

DEVELOP The area of the disk is given as2

0.14 m ,A so the radius of the disk must be

2 21 cm.Ar A r At a point 0.001 mr from the center of the disk, the disk will appear to be an

infinite plane, so we can use0 02 2

.Q

AE

At a point 2.5 mr from the disk, the disk will look more like a point

charge, so we can use 2 .kQ

rE The charge on the disk is 5.0 C.Q

EVALUATE

(a) 6

0

2.0 10 N/C2

QE

A

(b) 3

27.2 10 N/C

kQE

r

ASSESS If you are far enough from anything, it looks like a point charge; and if you are close enough, it looks like

an infinite plane.

Section 21.6 Gauss’s Law and Conductors

40. At the surface of a conductor,0/E (positive away from the surface), or

2(1.4 C/m )(8.85 12 2 2 110 C /N m ) 158 kN/C in this problem.

41. INTERPRET This problem involves finding the charge distribution of s a conductor using Gauss’s law.

21.6 Chapter 21

DEVELOP To answer the questions, we note that the electric field within a conducting medium, in electrostatic

equilibrium, is zero. In addition, the net charge must reside on the conductor surface.

EVALUATE (a) Gauss’s law implies that the net charge contained in any closed surface, lying within the metal,

is zero.

(b) If the volume charge density is zero within the metal, all of the net charge must reside on the surface of the sphere.

If the sphere is electrically isolated, the charge will be uniformly distributed (i.e., spherically symmetric), so

3 2

2 2

5 C3.98 10 C/m

4 4 (1 cm)

Q

R

(c) Spherical symmetry for depends on the proximity of other charges and conductors.

ASSESS Since charges are mobile, the presence of other charge near the conductor will cause the charges on the

surface to move and the new equilibrium configuration will not be spherically symmetric.

42. The field inside the shell is just due to the point charge (8 lines of force radiating outward). The field outside is like

that of a point charge 3 5

2 2q q q (20 lines of force radiating outward). (Note: There is a charge, ,q spread

uniformly over the inner surface of the shell, and the field inside the conducting material is zero.)

43. INTERPRET This problem is about finding the electric field near the surface of a conducting plate. The approximate

plane symmetry of the system allows us to make use of Gauss’s law.

DEVELOP The net charge of 18 CQ must distribute itself over the outer surface of the plate, in accordance

with Gauss’s law for conductors. The outer surface consists of two plane square surfaces on each face, plus the

edges and corners. Symmetry arguments imply that for an isolated plate, the charge density on the faces is the same,

but not necessarily uniform because the edges and corners also have charge. If the plate is thin, we could

assume that the edges and corners have negligible charge and that the density on the faces is approximately uniform.

Then the surface charge density is the total charge divided by the area of both faces,

2

18 C16.0 C

2 2(75 cm)

Q

A

EVALUATE Using Equation 21.8, we find the field strength near the plate (but not near an edge) to be

0

1.81 MN/CE

ASSESS Note the distinction between a charged conducting plate and a uniformly charged plate. In the latter,

charges are not free to move and the electric field is (see Example 21.6)0/2 .E

PROBLEMS

44. All of the lines of force going through the hemisphere also go through an equatorial disk covering its edge in

Fig. 21.34. Therefore, the flux through the disk (normal in the direction of Er

) equals the flux through the

hemisphere. Since Er

is uniform, the flux through the disk is just2

.R E (Note: Gauss’s law gives the same result,

since the flux through the closed surface, consisting of the hemisphere plus the disk, is zero. See Section 21.3.)

Gauss’s Law 21.7

45. INTERPRET This problem is about finding the electric flux through a given surface.

DEVELOP The electric flux through a surface is given by Equation 21.2:

E dA rr

Since the electric field depends only on y, we break up the square into strips of area ˆdA a dy k r

of length a

parallel to the x axis and width dy. The normal to the surface could be ˆ.k

EVALUATE The integral of Equation 21.2 gives

2

0 0 00 0

1늿( / ) ( )2

a a

E dA E y a k ady k E ydy E a rr

ASSESS Our result can be compared to the case where the field strength is constant. In that case, the flux through

the surface would be2

0 .E a

46. INTERPRET From the electric field, we determine the volume charge density in a region. We use Gauss’s law.

DEVELOP The electric field in the region is given as ,E axir

where 40 N/C m.a Since the field does not depend

on y or z, we must conclude that there is no variation in the y or z directions. We use a Gaussian surface that is a right

prism with a 1-m square base, with the base A of the prism at 0.x Gauss’s law states that enclosed

0.

qE dA

rr

EVALUATE For this particular electric field, the field is only in the i direction so the only surfaces on the prism

that have a flux are the surfaces at 0x and .x x The flux through the surface at 0x is zero, since the field

there is zero. The flux through the surface at x x is ( ) ( ) .E x dA E x A axA rr

The flux through this surface

is also the flux through the entire Gaussian surface, since the rest of the surface has zero flux, so E dA axA rr

enclosed

0enclosed 0 .

qq a Ax The volume of the prism is ,V Ax so the charge density is enclosed

0

q

Va

10 33.54 10 C/m .

ASSESS This is the electric field and charge density we would have inside a thick slab of charged material, such as

in Figure 21.35.

47. INTERPRET The charge distribution has spherical symmetry, so we can apply Gauss’s law to find the electric field.

DEVELOP The balloon can be regarded as a spherical shell with charge residing on the outer surface. For (b), we

note that outside the shell, the field is like that of a point charge, with total charge at the center. Therefore,

2 2

1 1 2 2E r E r

EVALUATE (a) Since 70 cm,R the point at 50 cmr is inside the uniformly charged spherical shell. Therefore,

enclosed 0q and by Gauss’s law, the electric field is zero (see Example 21.2).

(b) Let the electric field at 70 cmR be0 .E The electric field at 190 cmr is

22

0

70 cm(26 kN/C) 3.53 kN/C

190 cm

RE E

r

21.8 Chapter 21

(c) Using the given field strength at the surface, we find a net charge

2 2

0

9 2 2

(26 kN/C)(0.70 m)1.42 C

9 10 N m /C

E RQ

k

ASSESS The electric field inside a spherical shell is identically zero. Outside the shell, the field decreases as2

1/ .r

48. (a) Referring to Example 21.1, we see that at3 21 1

2 2, ( )/ /2 .r R E kQ R R kQ R This is also the field strength

outside the sphere at a distance 2 2(2 cm) 2.83 cm.r R (b) Using the given field strength at 1

2,r R we

find2 2 9 2 2

2 / 2(2 cm) (39 kN/C)/(9 10 N m /C ) 3.47 nC.Q R E k


DEVELOP The total electric field is the superposition of the fields due to the point charge and the spherical shell.

The field is spherically symmetric about the center.

EVALUATE (a) At /2r R R (inside shell), the electric field is

pt shell 2 2

( 2 ) 8 0

( /2)

k Q kQE E E

R R

Note that the minus sign means the direction is radially inward.

(b) At 2r R R (outside shell), the field strength is

pt shell 2 2

( 2 )

(2 ) 4

k Q Q kQE E E

R R

Again the direction of the field is radially inward.

(c) Ifshell 2 ,Q Q the field inside would be unchanged, but the field outside would be zero since

enclosed shell pt 2 2 0.q q q Q Q

ASSESS By Gauss’s law, the shell produces no electric field in its interior. The field outside a spherically

symmetric distribution is the same as if all the charges were concentrated at the center of the sphere.

50. (a) As in the previous solution, the field strength at the surface of the shell ( )r R is2

pt shell pt shell( )/ ,E E k q q R

hence2 9 2 2

pt [(15 cm) (750 kN/C)/(9 10 N m /C )] 4.8 C 2.93 C.q (b) Just inside the shell, at

15 cmr (where 15 cm), the field is due to the point charge only: 2( 2.93 C)/(15 cm )E k

1.17 MN/C, directed radially inward.


DEVELOP The total electric field is the superposition of the fields due to the point charge and the spherical shell.

The field is spherically symmetric about the center.

EVALUATE (a) The field due to the shell is zero inside, so at 5 cm,r the field is due to the point charge only. Thus,

9 2 26

2 2

(9 10 N m /C )(1 C)3.60 10 N/C

(0.05 m)

kqE

r

The field points radially outward.

(b) Outside the shell, its field is like that of a point charge, so at 45 cm,r the field strength is

9 2 26

2 2

( ) (9 10 N m /C )(86 C)(3.82 10 N/C)

(0.45 m)

k q QE

r

The direction of the field is radially outward.

(c) If the charge on the shell were doubled, the field inside would be unaffected, while the field outside would be

2

(1.0 C 2 85 C)7.60 MN/C

(45 cm)

kE

which is approximately doubled.

ASSESS By Gauss’s law, the shell produces no electric field in its interior. The field outside a spherically

symmetric distribution is the same as if all the charges were concentrated at the center of the sphere.

52. Use the result of Gauss’s law applied to a spherically symmetric distribution,2

enclosed 0/4 .E q r For a r b in a

spherical shell with charge density3 34

enclosed 3, ( ) ,q r a so

3 3 2 3 2

0 0( )/ 3 ( /3 )( / ).E r a r r a r If

Gauss’s Law 21.9

0,a Equation 21.5 for a uniformly charged spherical volume is recovered.


electric field.

DEVELOP With the assumption that the electric field is approximately that from an infinitely long, line symmetric

charge distribution, we have

enclosed2kE

r

where 104

k andenclosed is the charge inside a unit length of cylindrical Gaussian surface of radius r about the

symmetry axis.

EVALUATE (a) For 0.5 cm 0.005 m,r between the wire and the pipe, the enclosed charge per unit length is

enclosed wire, and

9 2 2

enclosed2 2(9 10 N m /C )(5.6 nC/m)20.2 kN/C

(0.005 m)

kE

r

The field is (positive) radially away from the axis of symmetry, i.e., the wire.

(b) For 1.5 cm 0.015 m,r the enclosed charge per unit length is enclosed wire pipe, and

9 2 2

enclosed2 2(9 10 N m /C )(5.6 nC/m 4.2 nC/m)1.68 kN/C

(0.015 m)

kE

r

The field is in the same direction as the field in part (a).

ASSESS Between the wire and the pipe, the enclosed charge per unit length isenclosed wire, while outside the

pipe, the enclosed charge isenclosed wire pipe. Since the pipe and the wire carry opposite

charges,enclosed enclosed, and ,E E as we have shown.

54. The charge distribution has line symmetry, so the flux, 2 ,rLE through a coaxial cylindrical surface of radius r

equalsenclosed 0/ ,q from Gauss’s law. For r R (outside the rod),

2

enclosed ,q R L hence2

out 0/2E R L rL 2

0/2 .R r For r R (inside the rod),2

enclosed ,q r L hence2

in 0 0/2 /2 .E r L rL r (The field direction is

radially away from the symmetry axis if 0, and radially inward if 0.)


electric field.

DEVELOP We assume that the rod is long enough and approximate its field using line symmetry:

enclosed2kE

r

where 104

k andenclosed is the charge inside a unit length of cylindrical Gaussian surface of radius r about the

symmetry axis. Since the charge is uniformly distributed throughout the solid rod, the line charge density is simply

equal to the volume charge density times the cross-sectional area of the rod:2

enclosed .A r Combining the

two equations yields

2

enclosed2 2 ( )2

k k rE k r

r r

This equation (valid for ,r R the radius of the rod) allows us to solve for .

EVALUATE With the electric field at r R given, we find the volume charge density to be

21.10 Chapter 21

12 2 2302 2(8.85 10 C /N m )(16 kN/C)

6.29 C/m2 0.045 m

EE

kR R

This is the magnitude of , since the direction of the field at the surface, radially inward or outward, was not

specified.

ASSESS In case we are given the field strength at a point ,r R then

2

enclosed2 2 ( )k k RE

r r

and the volume charge density would be 0

2 2

2

2.

ErEr

kR R

56. A positive surface charge density , on the floor, would produce an approximately uniform electric field upward

of0/2 ,E at points near the floor and not near an edge. The field needed to balance the weight of a particle, of

mass m and charge q, is given by ,mg qE therefore12 2 2

0 02 2 / 2(8.85 10 C /N m )(5E mg q 3 2 6 210 kg) (9.8 m/s )/(15 10 C) 57.8 nC/m .

57. INTERPRET The infinitely large slab has plane symmetry, and we can apply Gauss’s law to compute the electric field.

DEVELOP When we take the slab to be infinitely large, the electric field is everywhere normal to it (the x direction, as

shown in the figure) and symmetrical about the center plane. We follow the approach outlined in Example 21.6 to

compute the electric field everywhere.

EVALUATE (a) For points inside the slab | | /2,x d the charge enclosed by our Gaussian cylinder is

enclosed enclosed (2 )q V A x

Thus, Gauss’s law gives

enclosed

0 0

(2 )(2 )

q A xE dA E A

rr

or0.

xE

The direction of E

ris away from (toward) the central plane for positive (negative) charge density.

(b) For points outside the slab | | /2,x d the enclosed charge is

enclosed enclosedq V Ad

Thus, applying Gauss’s law leads to

0

(2 )Ad

E A

or02.

dE

Again, the direction of E

ris away from (toward) the central plane for positive (negative) charge density.

ASSESS Inside the slab, the charge distribution is equivalent to a sheet with .x On the other hand, outside the

slab, it is equivalent to a sheet with .d

58. INTERPRET We are given some charged concentric spheres, and are asked to find the electric field at various

distances from the center. We will use Gauss’s law.

Gauss’s Law 21.11

DEVELOP The central solid sphere has radius 10 cm,R and has a uniformly distributed charge 40 C.Q The

outer shell has radius 20 cm, and holds the same charge. Gauss’s law tells us that the field outside a spherical charge

distributionµ is04

,Q

rE inside a spherical shell is 0,E and inside a uniformly charged sphere is 3

0

1

4.

Qr

RE

EVALUATE (a) At 5 cm,r we are inside of the inner sphere. 30

61

418 10 N/C.

Qr

RE

(b) At 15 cm,r we are outside the inner sphere. The outer sphere does not contribute to the field yet.

0

6

42.4 10 N/C.

Q

rE

(c) At 30 cm,r we are outside both spheres. The total charge within our spherical Gaussian surface is now 2 ,Q

and the electric field is0

2 6

42.4 10 N/C.

Q

rE

ASSESS The answers to (b) and (c) are the same! The charge in (c) is twice as much, but we are also twice as

far away.

59. INTERPRET The square plate has approximate plane symmetry, and we can apply Gauss’s law to compute the

electric field.

DEVELOP The electric field strength close to the plate ( 1 cm << 75 cm )x a has approximate plane symmetry,

and is given by (see Equation 21.7)

02E

Therefore, the charge on the plate is2

0 02 2 ,q A EA Ea where a is the length of the square plate. At

a point sufficiently far from the plate ( )r a the field strength will resemble that from a point charge,2

/ .E kq r

EVALUATE Substituting the values given, we find the charge on the plate to be

2 12 2 2 2

02 2(8.85 10 C /N m )(45 kN/C)(0.75 m) 448 Cq Ea n

The field strength at 15 m >> 0.75 mr is like that from a point charge:

9 2 2

2 2

(9 10 N m /C )(448 nC)17.9 N/C

(15 m)

kqE

r

ASSESS Far from the finite charge distribution (plane charge in this case), the field always resembles that of

a point charge.

60. (a) There is a non-zero field outside the shell, because the net charge within is not zero. Therefore, there is

a surface charge density0 E on the outer surface of the shell, which is uniform, if we ignore the possible

presence of other charges and conducting surfaces outside the shell. Gauss’s law (with reasoning similar to

Example 21.7) requires that the charge on the shell’s outer surface is equal to the point charge within, so 2 2 2

/4 250 nC/4 (0.20 m) 497 nC/m .q R (b) Then the field strength at the outer surface is

0/ 56.2 kN/C.E

61. INTERPRET This problem involves a conductor in electrostatic equilibrium.

DEVELOP To answer the questions, we note that the electric field within a conducting medium, in electrostatic

equilibrium, is zero. In addition, the net charge must reside on the conductor surface.

EVALUATE (a) When there is no charge inside the cavity, the flux through any closed surface within the cavity

1( )S is zero, hence the electric field is also zero. Note that the argument depends on the conservative nature of the

electrostatic field, for then positive flux on one part of1S canceling negative flux on another part is ruled out.

(b) If the surface charge density on the outer surface (and also the electric field there) is to vanish, then the net

charge inside a Gaussian surface containing the conductor2( )S is zero. Thus, the point charge in the cavity must

equal ,Q so thatenclosed ( ) 0.q Q Q

21.12 Chapter 21

ASSESS An alternative approach is to say that the total charge on the conductor isinner outer .cq q q Q

Requiringouter 0q means that

inner .q Q Since electric field inside the cavity vanishes, the point charge placed

inside must beinner .q q Q

62. Assume line symmetry, and apply Gauss’s law to the outer cylindrical conducting surface,

surf enclosed 02 / .rLE q Since the conductors in the cable carry opposite charges of equal magnitude, there is zero

charge enclosed, so the field and the charge density there0 surf( )E vanish.

63. INTERPRET This is a spherically symmetric charge distribution, so Gauss’s law can be applied in this problem.

DEVELOP The field from the given charges is spherically symmetric, so (from Gauss’s law) is like that of a point

charge, located at the center, with magnitude equal to the net charge enclosed by a sphere of radius equal to the

distance to the field point.

EVALUATE Thus, the electric field is 2

kq

rE inside the first shell (8 lines radially inward), 2

kq

rE between the

first and second shells (8 lines radially outward), and 22

kq

rE outside the second shell (4 lines radially inward).

ASSESS The direction of the electric field depends on the net charge enclosed. Whenenclosed 0,q E

ris radially

outward. On the other hand, Er

is radially inward whenenclosed 0.q The discontinuity in electric field across the shell

is due to a net surface charge density on the shell.

64. As explained in Example 21.3, (a) for 1enclosed2

, r R R q q and2 21

2/( ) 4 / ,E kq R kq R and (b) for 2 ,r R R

enclosed 2q q q and2 23 /(2 ) 3 /4 .E kq R kq R

65. INTERPRET The charge distribution has spherical symmetry, so we can apply Gauss’s law. Also, since the charge

distribution is non-uniform, integration is needed to find the electric field.

DEVELOP The charge inside a sphere of radius r a is0

( ) .r

q r dV For volume elements, take concentric

shells of radius r and thickness dr, so2

4 .dV r dr

EVALUATE (a) Integrating over r, the amount of charge enclosed by a Gaussian sphere of radius r is

4

2 3 0

00 0

( ) 4 4 ( / ) r r r

q r r dr a r dra

For ,r a the total charge is3

0 .Q a

(b) For spherical symmetry, Gauss’s law and Equation 21.5 give

4 2

2 0 0

0 0 0

( ) 14 ( ) ( )

4

r rq rr E r E r

a a

Gauss’s Law 21.13

ASSESS The2r dependence of E inside the sphere can be contrasted with the r dependence in the case (see

Example 21.1) where the charge distribution is uniform.

66. The electric field on the top surface of the box has magnitude 0/2 ,r and direction radially away from the line of

charge. The flux through a strip of width dx at position x is0cos ( /2 )( / ) ,d E L dx r a r L dx where

2 2 ,r x a so

1

top 2 2

0 0 0 0

1tan

2 2 2 2 4( )

aa

aa

aL dx aL x aL

a a aa x

From symmetry, the flux through the whole box is four times this, or0/ ,L which equals

enclosed 0.q

67. INTERPRET The charge distribution has spherical symmetry, so we can apply Gauss’s law. Also, since the charge

distribution is non-uniform, integration is needed to find the condition which gives zero electric field outside the sphere.

DEVELOP Using Gauss’s law, we see that the field outside the sphere will be zero if the total charge within the

volume is zero.

EVALUATE Thus, using thin concentric shells of thickness dr and volume2

4dV r dr as our charge elements,

we require that

2 2 3 5

enclosed 0 0sphere 0

1 10 ( )4 4

3 5

R

q dV ar r dr R aR

or 0

2

5

3.

Ra

ASSESS The charge density starts out from the center of the sphere as0 and decreases as

2.r At ,r R the

density is

0 0

5( ) 1 2 /3

3R

Note that ( )r must change sign (from positive to negative) in order for enclosedq to be zero at .r R

68. INTERPRET We calculate—directly, by integration—the electric field inside and outside a uniformly charged

spherical shell of radius R and total charge Q.

DEVELOP We will divide the spherical surface up into rings around an axis that includes the field point. Each ring

has radius sina R and width ,Rd and holds a charge2

/4 (2 ) .dq Q R a Rd The electric field due to each ring

is2 2 3/2( )

,kdqx

x adE

where x is the distance from the field point to the ring, cos .x r R We will integrate

from 0 to .

EVALUATE

2 2 3 /20 0

( cos )sin .

2(( cos ) ( sin ) )

k r R QE dE d

r R R

Use of an integral table or computational package shows that for , 0.r R E For ,r R the integral becomes

2/ .E kQ r

ASSESS In both cases, this somewhat complicated integral gives us the same results as is obtained by use of

Gauss’s law.

69. INTERPRET Given a charge density in a sphere, we find the electric field at the surface of the sphere. We use

Gauss’s law, and integrate the charge density to find the charge enclosed.

21.14 Chapter 21

DEVELOP The charge density is given by/

0 ,r Re where r is the distance from the center and R is the radius of

the sphere. Gauss’s law tells us that enclosed

0

qE dA

rrr

EVALUATE

enclosed

0 0

22 / 2

00 0 0

0

2 / / 2 20

0 00

0 0

2 2 30 0

0 0

1

1(4 ) sin

14 ( ( 2 2 ))

4

[ ( ) (2 )] ( 2)

V

Rr R

Rr R r R R

qE dA E dA dq

E R e r drd d

E r e dr e R r rR R

E eR R R R R e

rrr r

ASSESS This answer is positive: the field points away from the sphere. It is also linear with0, as we would expect.

70. INTERPRET We are asked to find the charge needed on a square plate in order to create a given electric field. We

use the result from Gauss’s law for an infinite sheet of charge.

DEVELOP The square plate is 4.5 meters on a side, and your friend needs the electric field 430 N/CE at a point

only 10 cm from the center of the plate. This is a huge plate compared to the distance, so we can approximate it as

an infinite plate and use02,E

where .Q

A

EVALUATE

7

0 0

0

2 2 1.54 10 C2

QE E Q EA

A

ASSESS Notice that we didn’t really use the distance from the plate, other than to note that it was small compared

to the size of the plate.

71. INTERPRET We use the gravitational equivalent of Gauss’s law to find the gravitational field inside the Earth.

DEVELOP We are told thatenclosed4 .g dA GM

rrr We can use the same approach to solve for g

rthat we use to

solve for Er

in enclosed

0.

qE dA

rrr

EVALUATE Assume that the Earth is spherically symmetric, so gr

is parallel to .dAr

We will also assume that the

density is approximately constant, so34

enclosed 3,M r where

343

.E

E

m

R Then at some internal radius r,

2

enclosed

44 (4 ) 4

3g dA GM g r G r 3

43

Emr

3

3

E

E

E

R

m rg G

R

ASSESS Coulomb’s law and Newton’s law of universal gravitation have the same form, so it’s not surprising that

we can use some of the same mathematics on both. Compare this result with the result for the electric field inside

a uniformly charged sphere, 3 .Qr

RE k

22.1

ELECTRIC POTENTIAL

EXERCISES

Section 22.1 Electric Potential Difference

16. The potential difference and the work per unit charge, done by an external agent, are equal in magnitude, so

(50 C)(12 V) 600 J.W q V (Note: Since only magnitudes are needed in this problem, we omitted the

subscripts A and B.)

17. INTERPRET This problem is about the energy gained by an electron as it moves through a potential difference .V

DEVELOP We assume that the electron is initially at rest. When released from the negative plate, it moves toward

the positive plate, and the kinetic energy gained is | | .U q V

EVALUATE As the electron moves from the negative side to the positive side (i.e., against the direction of the

electric field), the kinetic energy it gains is

19 17|( ) | 120 eV (1.6 10 C)(120 V) 1.92 10 JK e V

ASSESS Moving a negative charge through a positive potential difference is like going downhill; potential energy

decreases. However, the kinetic energy of the electron is increased.

18. The work done by an external agent equals the potential energy change, 45 J ,AB ABU q V henceABV

45 J/15 mC 3 kV. (Since the work required to move the charge from A to B is positive, andB A ABV V V is

positive.)

19. INTERPRET This problem is about conversion of units.

DEVELOP By definition,1 volt 1 joule/coulomb (1 V 1 J/C). On the other hand,1 joule 1 newton-meter

(1 J 1 N m).

EVALUATE Combining the two expressions gives1 V 1 J/C 1 N m/C. It follows that1 V/m 1 N/C.

ASSESS These are the units for the electric field strength.

20. For rv

in the direction of a uniform electric field, Equation 22.2b gives | | (650 N/C)(1.4 m) 910 V.V E r v

(See note in solution to Exercise 16. Since ,dV E dr r v

the potential always decreases in the direction of the

electric field.)

21. INTERPRET This problem is about the work done by the battery to move the charge from the positive terminal to

the negative terminal.

DEVELOP The work done by the battery is equal to the kinetic energy gained by the charge, and is equal to

| |.W q V

EVALUATE Substituting the values given, we have

(3.1 C) (9.0 V) 27.9 JW q V

ASSESS The charged particle gains kinetic energy as it moves toward the negative plate (in the direction of the

electric field). The battery is needed to maintain the potential difference between the plates.

22. The energy gained is q V (see Example 22.1). The proton and singly-ionized helium atom have charge e, so they

gain19 17

100 eV (1.6 10 C)(100 V) 1.6 10 J, while the -particle has charge 2e and gains twice this energy.

Section 22.2 Calculating Potential Difference

22

22.2 Chapter 22

23. INTERPRET In this problem we are given a uniform electric field and asked to calculate the potential difference

between two points.

DEVELOP For a uniform field, the potential difference between two points a and b is given by Equation 22.1b:

ab b aV V V E r r r

where rr

is a vector from a to b.

EVALUATE With ˆ,b ar r r yj r r r

we obtain

0 0늿( ) (0) ( ) ( ) ( )V y V V y E r E j yj E y

r r

ASSESS The electric potential decreases in the direction of the electric field. In other words, electric field lines

always point in the direction of decreasing potential.

24. The potential of the proton, at the position of the electron (both of which may be regarded as point-charge atomic

constituents) is (Equation 22.3) 0/ ,V ke a where 0a is the Bohr radius. Numerically,9 2 2

(9 10 N m /C )V 19 11

(1.6 10 C)/(5.29 10 m) 27.2 V. (The energy of an electron in a classical, circular orbit, around a

stationary

proton, is one half its potential energy, or 1 1

2 2( ) 13.6 eV.U e V The excellent agreement with the ionization

energy of hydrogen was one of the successes of the Bohr model.)

25. INTERPRET We are asked to find the charge on a sphere, given the potential at the surface. We will assume that

the charge is spherically symmetric, and use the equation for potential of a point charge.

DEVELOP The equation for a point charge is ( ) .kq

rV r Since any spherically symmetric charge distribution looks

like a point charge from outside, we will solve this for q. The potential at the surface of the sphere is 4.8 kV,V

and the radius is 0.10 m.r

EVALUATE

( ) 53.3 nCkq Vr

V r qr k

ASSESS The key is the recognition that spherically symmetric charge distributions look like point charges from the

outside. This is the same as the case for gravitational potential.

26. For an isolated metal sphere, the potential at the surface is / ,V kQ R while the electric field strength at the surface

is2/ / .kQ R V R Thus, 1

2/ 3 MV/m implies (3 MV/m)( 0.05 m) 75 kV.V R V

27. INTERPRET This problem is about the electric potential of a spherically symmetric charge distribution.

DEVELOP Since the electric field outside the spherical charge distribution is the same as that of a point charge, the

electric potential outside the metal sphere ( )r R is given by Equation 22.3:

( )kQ

V rr

Note that we have taken the zero of the potential to be at infinity.

EVALUATE (a) An isolated metal sphere has a uniform surface charge density, so the potential at its surface is

9 2 2(9 10 N m /C )(0.86 C)( ) 442 kV

(0.035 m)/2

kQV R

R

(b) The work done by the repulsive electrostatic field (the negative of the change in the proton’s potential energy)

equals the proton’s kinetic energy at infinity:

21( ( )) (R)

2W e V V R eV mv

Thus, the speed of the proton far from the sphere is

196

27

2 ( ) 2(1.6 10 C)(442 kV) 9.21 10 m/s

1.67 10 kg

eV Rv

m

ASSESS As the proton moves away from the metal sphere, its potential energy decreases. However, by energy

conservation, its kinetic energy increases.

Electric Potential 22.3

Section 22.3 Potential Difference and the Electric Field

28. For a uniform field, Equation 22.9 can be written as ,V E x where the x axis is in the direction of the field

(V decreases in the direction of the field). Thus, | / | (1 V/2.5 cm) 40 V/m.E V x

29. INTERPRET This problem is about calculating electric field from electric potential.

DEVELOP Given electric potential ( ),V x the x component of the electric field may be obtained as / .xE dV dx

This equation is what we shall use to estimatexE for the seven straight-line segments shown in Fig. 22.21.

EVALUATE Using the equation above, we find 0xE for 0 to 2 m.x Similarly, for 2 m to 4 m,x

( 2 V 2 V)/(4 m 2 m) 2 V/m.xE The field strength in other regions can be calculated in a similar

manner.

The result is sketched below.

ASSESS The field component /xE dV dx is the negative of the rate of change of V with respect to x. The negative

sign means that if we move in the direction of increasing potential, then we’re moving against the electric field.

30. (a) The equipotentials in Fig. 22.23 are most closely spaced along the x axis between 2 m and 5 m.x x (b) The

potential decreases in the direction of the electric field, which, for 2 m 5 m,x is in the negative x direction.

(c) The potential drops by10 V/m, which is the field strength in this region.

31. INTERPRET This problem is about calculating electric field from electric potential.

DEVELOP Given the electric potential V, the corresponding electric field is (see Equation 22.9)

늿늿 늿x y z

V V VE E i E j E k i j k

x y z

r

Thus, taking the partial derivatives of V allows us to get the field components.

EVALUATE (a) Direct substitution gives

2 2( ) 2 3 5 2(1)(1) 3(1)(1) 5(1) 4 VV P xy zx y

(b) Use of Equation 22.9 gives

2 3

2 10

3

x

y

z

VE y z

x

VE x y

y

VE x

z

At ( 1),P x y z we obtain 1 V/m, 12 V/mx yE E and 3 V/m.zE

ASSESS Electric field is strong in the region where the potential changes rapidly. At ( 1),P x y z the potential

changes most rapidly in the direction of the electric field

22.4 Chapter 22

ˆ늿( 12 3 )V/mE i j k r

Section 22.4 Charged Conductors

32. (b) Dielectric breakdown in the air occurs if the field at the surface, 0/ ,E exceeds6

3 10 V/m. Therefore, the

charge (for a uniformly charged sphere) must not be greater than2 2 6 2

04 4 (3 10 V/m)(2.3 m) /q R ER 9 2 2

(9 10 N m /C ) 1.76 mC. (a) From Equation 22.3, / 6.9 MV.V k q R RE

33. INTERPRET This problem is about finding the minimum potential which leads to a dielectric breakdown in air.

DEVELOP We shall treat the field from the central electrode as that from an isolated sphere. Then2

/E k q R

and / ,V k q R so that .V RE

EVALUATE Breakdown of air occurs at a field strength of6

3 10 V/m.E Therefore, dielectric breakdown in air

would occur for potentials exceeding

3 6(1 10 m)(3 10 V/m) 3 kVV RE

ASSESS The result means that if we attempt to raise the potential of the electrode in air above 3 kV, then the

surrounding air would become ionized and conductive; the extra added charge would leak into the air, resulting in

plug sparks.

34. (a) The potential of an isolated metal sphere, with charge Q and radius R, is / ,kQ R so a sphere with charge 3Q and

radius 3R has the same potential. (b) However, the electric field at the surface of the smaller sphere is0/

2/ ,kQ R so tripling Q and R reduces the surface field by a factor of 1

3.

PROBLEMS

35. INTERPRET This problem is about finding the electric field strength, given the potential difference between two

points.

DEVELOP Since the field Er

is uniform, Equation 22.1b, ,ABV E r r r

can be used to relate Er

to the potential

difference .ABV Since the path AB is parallel to ,Er

we write

| |ABV E r

where E is the field strength, and r is the separation between points A and B.

EVALUATE The field strength is

| | 840 V5.60 kV/m

0.15 m

ABVE

r

ASSESS Since dV E dr r r

the potential always decreases in the direction of the electric field.

36. (a) On the line A to B, drv

is antiparallel to ,Er

so Equation 22.1 gives .B B

B A A AV V E dr E dr Ed

r v v(b) The

line B to C makes an angle of 45 with ,Er

so cos 45 / 2.C

C B BV V E dr Ed v

(c) Addition yieldsC AV V

(1 1/ 2) 0.293 .C B B AV V V V Ed Ed

37. INTERPRET This problem is about finding the potential difference between two points, given the electric field

strength.

DEVELOP Since the field Er

is uniform, Equation 22.1b, ,V E r r r

can be used to relate Er

to the potential

difference V across the membrane. For a uniform electric field parallel to the thickness of the membrane, we write

| |V E r

where E is the field strength, and r is the thickness of the membrane.

EVALUATE Using the equation above, the potential difference is

| | (8.0 MV/m)(10 nm) 80 mVV E r

ASSESS Since dV E dr r r


38. The magnitude of the potential difference is 19 19| / | 7.2 10 J/1.6 10 C 4.5 V.W q (The energy imparted per

electron is 4.5 eV.)


39. INTERPRET The problem asks for the speed of the electrons which have been accelerated through a potential

difference.

DEVELOP According to the work-energy theorem, the work done on an electron equals the change in its kinetic

energy:

21| |

2W e V mv

The electron starts from rest. Knowing V allows us to determine the speed of the electron.

EVALUATE Using the equation above, we find the electron speed to be

19 37

31

2 | | 2(1.6 10 C)(25 10 V)9.37 10 m/s

9.11 10 kg

e Vv

m

ASSESS The electron speed is about 1

3of the speed of light. This value is typical of electrons in the cathode ray

tube (CRT) TV.

40. The energy gained is15 41.6 10 J 10 eV, so the charge (really its magnitude) on the ion is | / |q W V

410 eV/2500 V 4 .e

41. INTERPRET This problem is about the potential difference between two conducting plates separated by a distance

d and having opposite charge densities.

DEVELOP We first calculate the electric field between the plates. Using the result obtained in Example 21.6 for one

sheet of charge, and applying the superposition principle, the electric field strength between the plates is0/ ,E and is

directed from the positive to the negative plates. Once E is known, we can use Equation 22.1b to calculate V.

EVALUATE Equation 22.1b gives

0 0

( )d

V V V E r d

r r

ASSESS The displacement from the negative to the positive plate is opposite to the field direction. In other words,


42. The work-energy theorem (for an electron under the influence of just an electric force) gives AB ABW q V

,AK K whereABW is the work done by the electric field (also equal to ), and 0.AB BU K Thus,

2 31 6 2

19

v (9.11 10 kg)(6.5 10 m/s)120 V

2( ) 2( 1.6 10 C)

e AA

AB

mKV

q e

(To stop an electron, a negative potential difference must be applied.)

43. INTERPRET The problem asks for the charge of the particle which has been accelerated through a potential

difference.

DEVELOP The speed acquired by a charge q, starting from rest at point A and moving through a potential

difference of V, is given by

21 2( )

2A B

qVmv q V V qV v

m

This is the work-energy theorem for the electric force. A positive charge is accelerated in the direction of decreasing

potential. If we have two masses moving through the same potential difference, the ratio of their

speed would be

2 2 2 2 1

1 1 1 1 2

2 /

2 /

v q V m q m

v q V m q m

EVALUATE If the second object acquires twice the speed of the first object2 1( / 2),v v moving through the same

potential difference, from the equation above, we find its charge to be

2

22 22 1

1 1

2g(2) (3.8 C) 6.08 C

5g

m vq q

m v

22.6 Chapter 22

ASSESS The speed of the particle moving through a potential difference is proportional to the square root of its

charge, and inversely proportional to the square root of its mass.

44. Since / and / ,A A B BV kQ r V kQ r division yields ( / ) (280/130) 2.15 .B A B A A Ar V V r r r But 20 cm, soB Ar r 1 9 2 2(20 cm)(2.15 1) 17.3 cm. Then / (280 V)(17.3 cm)/(9 10 N m /C ) 5.39 nC.A A Ar Q V r k

45. INTERPRET The positively charged proton is attracted to the negatively charged sphere via Coulomb interaction.

Work must be done to pull the proton away from the sphere.

DEVELOP The work done by the electric field, when a proton escapes from the surface to an infinite distance,

equals the change in kinetic energy, or

2

surf surf surf

1( )

2e V V eV K K mv

where we have assumed zero kinetic energy for the proton at infinity, and that the sphere is stationary.

EVALUATE For a uniformly negatively charged sphere,surf /V kQ R (see Example 22.3), Thus, the escape speed is

surf2 2eV keQ

vm mR

ASSESS The escape speed of a proton from the electric field of the charged sphere in this problem is analogous to

the escape speed of a rocket from the Earth’s gravitational field.

46. The potential at the surface of the shell is /kQ R (as in Example 22.3). The electric field inside a uniformly charged

shell is zero, so the potential anywhere inside is a constant, equal, therefore, to its value at the surface.

47. INTERPRET The problem is about the potential difference between the center of a charged sphere and a point on

its surface.

DEVELOP From Equation 22.1a, we see that the potential difference from point A to point B is given by

B

AB B AA

V V V E dr r r

As shown in Example 21.1, the electric field inside a uniformly charged sphere is3

/E kQr R and is radially

symmetric.

EVALUATE The integration above gives

2

3 300

( ) (0)22

RR kQr kQr kQ

V R V drRR R

ASSESS The potential is higher at the center if Q is positive.

48. Since21

0 0 0 2(0) 0, ( ) .

r r x

V V r E dr axn dr ax dx ax rr r r r

49. INTERPRET The problem is about the potential difference between two conductors that make up the coaxial cable.

DEVELOP If the length of the cable is long compared to its outer diameter, the electric field inside, away from the

ends, will be due to the inner conductor only (apply cylindrical symmetry and Gauss’s law). The potential difference is

given by Equation 22.4:

0

ln 2 ln2

A AAB B A

B B

r rV V V k

r r

EVALUATE Substituting the values given in the problem statement, we find the potential difference to be

9 2 2 9 1.6 cm2 ln 2(9 10 N m /C )(0.56 10 C/m) ln 21.0 V

0.2 cm

AAB

B

rV k

r

ASSESS In this problem, we have chosen B to be a point on the inner conductor, and A to be a point on the outer

conductor. The inner conductor has a positive charge density while the outer one has a negative charge density.

We expectB AV V since the electric field always points in the direction of decreasing potential.

50. If we approximate the potential from the line by that from an infinitely long charged wire, Equation 22.4 can be

used to find 9 2 2 1

0: 2 /ln( / ) (3.9 kV)[2(9 10 N m /C ) ln(100/1.5)]AB A BV r r

51.6 nC/m. ( ABV Note: B AV V so B is at the surface of the wire and A is100 cm distant.)


51. INTERPRET This problem is about the electric potential at a point due to a system of charges.

DEVELOP The electric potential at a point P due to a collection of charges is given by Equation 22.5:

i

P

i i

kqV

r

This is the equation we shall use to compute the electric potential at the center of a triangle.

EVALUATE The center is equidistant from each vertex, with

2cos30 3

a ar

Since each charge contributes equally to the potential, the potential at the center is

3 3 3kq kqV

r a

ASSESS Electric potential is a scalar, so there is no need to consider angles, vector components, or unit vectors.

52. The potential on the x axis, / /| | ( 3 )/| |i ikq r kQ x k Q x a (from Equation 22.5), is zero when 3| | | |.x x a

For 0,x this implies 3 , or /2.x a x x a Note that the absolute value of a negative number is minus the

number, and we assume that a is a positive constant. For 0 ,x a the condition is 3 , or /4.x a x x a

For ,x a there are no solutions. (The same results follow from the quadratic2 2

8 2 0,x ax a which results

from

the square of the above condition.)

53. INTERPRET This problem is about the electric potential at a point due to a system of charges.

DEVELOP The electric potential at a point P due to a collection of charges is given by Equation 22.5:

i

P

i i

kqV

r

Consider a point ( , ).P x y The distance from P to (0, a) is2 2

( ) .r x a y Similarly, the distance from P to

(0,a) is2 2

( ) .r x a y

EVALUATE (a) The equations above give

2 2 2 2

1 1

( ) ( )P

kq kqV kq

r r x a y x a y

(b) If2 2

,r x y a then a can be neglected relative to x or y, so

2 2 2 2 2 2 2 2

1 1 1 1 2

( ) ( )P

kqV kq kq

rx a y x a y x y x y

which is the potential of a point charge of magnitude 2q.

ASSESS At a distance much greater than the separation of two charges1q and

2 ,q the electric potential is like that

due to one single point charge1 2 .q q Note that electric potential is a scalar, so there is no need to consider angles,

vector components, or unit vectors.

54. Equation 22.6 gives the potential from a point dipole as a function of distance and angle from the dipole axis.

For the dipole moment and distance given,2 2 2 2

( , ) cos / (9 GN m /C )(2.9 nC m)cos /(10 cm)V r k p r

(2.61 kV)cos . For the three given angles,

(a) (2.61 kV)cos 0 2.61 kV;V

22.8 Chapter 22

(b) (2.61 kV)cos 45 1.85 kV; andV

(c) (2.61 kV)cos90 0.V

55. INTERPRET The problem is about finding the electric potential due to a continuous charge distribution.

DEVELOP From Example 22.6, we see that the electric potential at the center of a charged ring of radius a is

kQV

a

EVALUATE (a) The radius of the circle is /2 ,a L where L is the length of the rod. Therefore, the potential at the

center of a uniformly charged ring is

9 2 2(9 10 N m /C )(3.2 nC)905 V

0.20 m/2

kQV

a

(b) Since each charge element, ,dq of a circular arc, in Fig. 22.12, is the same distance from the center of the arc,

Example 22.6 also gives the potential at the center of a uniformly charged semicircular ring. However, the radius is

now 20 cm/ ,a or twice the value in part (a). Thus, the potential is

1 1(905 V) 452 V

2 2V V

ASSESS Electric potential is a scalar, so there is no need to consider angles, vector components, or unit vectors.

56. The result in Example 22.6 did not depend on the ring being uniformly charged. For a point on the axis of the ring, the

geometrical factors are the same, and ring totdq Q for any arbitrary charge distribution, so2 2 1/2

tot ( )V kQ x a still

holds. Thus, at the center ( 0)x of a ring of total chargetot 3 2 ,Q Q Q Q and radius ,a R the potential is

2 / .V kQ R

57. INTERPRET This problem is about the electric potential due to a charged ring, which is a continuous charge

distribution.

DEVELOP From Example 22.6, we see that the electric potential at the center of a charged ring of radius a is

(0)kQ

Va

At a distance x along the ring axis from the center of the ring, the potential is

2 2( )

kQV x

x a

These two equations allow us to determine the radius a and the total charge Q.

EVALUATE Substituting the values given in the problem statement yields

2 2(0) 45 kV, and (0.15 m) 33 kV

(0.15 m)

kQ kQV V

a a

Thus, we find

2 2 2

33 kV 33 1(0.15 m) 0.162 m

45 kV 45(0.15 m) 1 (33/45)

aa

a

The charge is9 2 2

(0) (45 kV)(0.162 m)

9 10 N m / C809 nC.

V a

kQ

ASSESS In this problem, we are given two conditions, which allow us to solve for two unknowns—the radius and the

charge of the ring. Note that the electric potential is the greatest at the center of the ring and falls off as x increases.

When ,x a the potential resembles that of a point charge: ( ) / .V x kQ x

58. The annulus can be considered to be composed of thin rings of radius ( )r a r b and charge 2dq rdr

(see Example 22.7 and Figs. 22.12 and 12.13). The element of potential from a ring on its axis, a distance x from the

center, is2 2

/dV k dq x r (see Example 22.6) so the potential from the whole annulus is:


2 2 2 2 2 2

2 2

2 2 2

bb

a a

r drV dV k k x r k x b x a

x r

(Note: This reduces to the potential on the axis of a uniformly charged disk if 0.)a

59. INTERPRET This problem is about calculating the electric potential from the electric field.

DEVELOP Equation 22.2b gives the potential for a uniform field. We take the zero of potential at the origin

(point A in Equation 22.2b) and let ˆ늿r xi yj zk r

be the vector from the origin to the field point (point B in

Equation 22.2b).

EVALUATE (a) The potential difference between A and B is

0 0늿( ) 0 ( ) ( )AB B AV V V V r E i j r E x y

r r

Since the potential is independent of z, it can be written as ( ) ( , ).V r V x yr

(b) Using the result obtained in (a), we see that the potential difference between2 2( , )x y and

1 1( , )x y is

12 0 2 2 0 1 1 0 2 2 1 1( ) ( ) ( )V E x y E x y E x y x y

Thus, we obtain

(3.5 m, 1.5 m) (2.0 m, 1.0 m) (150 V/m)(3.5 m 1.5 m 2.0 m 1.0 m) 150 VV V

ASSESS An alternative approach is to note that the displacement vector from (2.0 m, 1.0 m) to (3.5 m, 1.5 m) is

늿 늿(3.5 m 2.0 m) ( 1.5 m 1.0 m) (1.5 m) (2.5 m)r i j i j r

Using Equation 22.1b, the potential difference is then equal to

늿 늿(150 V/m)( ) [(1.5 m) (2.5 m) ] (150 V/m)(1.5 m 2.5 m) 150 VV E r i j i j r r

Thus, the same result is obtained.

60. (a) The x and y components of the electric field can be found from Equation 22.9:

/ / ( ) ,xE V x x axy ay and 늿/ . Thus ( ).yE V y ax E a yi xj r

(The field has no z

component.)

(b) See sketch below. The field lines (dashed) are perpendicular to the equipotentials (solid) in the direction of

decreasing potential (arrows for 0,a in this case). These equipotentials and field lines are confocal hyperbolas,

proportional to xy and 2 21

2( )x y respectively, and are sketched only for x and y in the first quadrant.

61. INTERPRET In this problem we want to use the expression for the electric dipole potential to find the electric field

at a point on the perpendicular bisector of the dipole.

DEVELOP The dipole potential is given by Equation 22.6:

2

cos( , )

k pV r

r

Using Equation 22.9, the general expressions for the r and components of the electric fields are

22.10 Chapter 22

3

3

2 cos

1 sin

r

V k pE

r r

V k pE

r r

EVALUATE On the bisecting plane, 90 , which yields 0rE and3

/ ,E k p r or3늿 / ) .E E k p r

rTo

compare with Equation 20.6a, we take the origin at the center of the dipole, the dipole moment along the x

axis ˆ( ),p pir

and the y axis up in Fig. 22.10, so ˆ 늿 ?sin 90 cos90i j i and2 20r y y on the

bisecting plane. This leads to3 ˆ( / ) .E kp y i

r

ASSESS Instead of using polar coordinates, one could first express V in terms of x and y (using cosx r and

sin ) :y r

2 2 3/ 2( , )

( )

k pxV x y

x y

and then differentiate, /xE V x and / .yE V y The result is the same.

62. On the axis of a uniformly charged ring (the x axis),2 2

/V kQ x a (Equation 22.8), and the electric field only

has an x component (by symmetry). Then2 2 3 /2늿( / ) ( ) ,E dV dx i kQx x a i in accord with 20.6. (In general, one

needs to know the potential in a 3-dimensional region in order to calculate the field from its partial derivatives.)

63. INTERPRET We are given the electric potential and asked to find the corresponding electric field.

DEVELOP We first note that the potential0( ) ( / )V r V R r depends only on r. This implies that the electric field is

spherically symmetric and points in the radial direction. The field can be calculated using Equation 22.9.

EVALUATE Using Equation 22.9, we obtain 0 .VdV

r dr RE In vectorial form, the field can be written as 0 ˆ ,

V

RE rr

where r is a unit vector that points radially outward.

ASSESS The electric field is uniform, but the potential is linear in r. The difference of one power in r is because

the potential is an integral of the field over distance.

64. (a) Since the spheres are far apart (approximately isolated), we can use Equation 22.3 to find their potentials: 2 2

1 1 1 2 2 2/ (9 GN m /C )(38 nC)/(1 cm) 34.2 kV and / 9 kV.V kQ R V kQ R (b) When connected by a thin

wire, the spheres reach electrostatic equilibrium with the same potential, so1 1 2 2/ / .V kQ R kQ R Since the radii are

equal, so must be the charges,1 2 .Q Q The total charge is

1 2 138 nC 10 nC 28 nC 2Q Q Q (if we assume

that the wire is so thin that it has a negligible charge), so1 2 14 nC. Then (14 nC)/(1 cm) 12.6 kV.Q Q V k

(c) In this process, the first sphere loses 38 14 24 nC to the second.

65. INTERPRET We are given two charge-carrying conducting spheres, and we want to find the electric potential and

electric field at various points.

DEVELOP Since the spheres are small and widely separated, they behave like isolated spheres, and their charge

distributions are essentially spherical.

EVALUATE (a) Using the result obtained in Example 22.3, at the surface of either sphere, the potential is

9 2 2 7(9 10 N m /C )(1.2 10 C)( ) 43.2 kV

0.025 m

kqV R

R

(b) The electric field at the surface of each sphere is

6

2

( ) 43.2 kV( ) 1.73 10 V/m

0.025 m

kq V RE R

RR

(c) Midway between the spheres, the potential from each one is the same, so

9 7

mid-pt.

2 2(9 10 V m/C)(1.2 10 C)540 V

4 m

kqV

r

(d) Since the spheres are at the same potential, the difference is zero.


ASSESS In this problem, the two conducting spheres can be treated as being isolated and the superposition

principle applies because they are far apart ( ).r R If they were brought close to each other, then the charge

distribution would no longer be spherical.

66. (a) The electric field outside the sphere1(radius 2 cm),R but inside the shell

2(inner radius 10 cm),R is only

due to the charge1q on the sphere (recall Gauss’s law), and equals

2

1/kq r radially outward. The potential difference

is

1 2V V 1

2

2 1 1 2 2 1 1

1 1 1 2/ ( ) (9 GN m /C )(75 nC)[(2 cm) (10 cm) ] 27.0 kV.R

Rk q dr r kq R R (We used

Equation 22.2a, with A at R2 and B at R1. Note that2

1 /E dr kq dr r r r

regardless of the choice of points A and B.)

(b) Adding more charge to the conducting shell does not affect the field inside, nor the potential difference between

points inside, like 1 2V V in part (a). The field outside the shell and the potential relative to infinity would change.

67. INTERPRET Our coaxial cable consists of an inner conductor and an outer conductor, each carrying different charges.

DEVELOP For a long straight cable, the field between the two conductors is approximately cylindrically

symmetric and depends only on the charge density of the wire. In fact, the field strength in the region between the

conductors is wire2,

k

rE

and its direction is radially outward. The potential difference can be calculated using

Equation 22.4:

wire2 lnB

A

rA

ABr

B

rV E dr k

r

r r

whereAr and

Br are the radius of the inner and outer conductors, respectively.

EVALUATE (a) Substituting the values given in the problem, we find the potential difference to be

9 2 2 9

wire

2 mm2 ln 2(9 10 N m /C )(75 10 C/m) ln 2.17 kV

10 mm

AAB

B

rV k

r

(b) The potential difference in the region between the conductors does not depend on the charge density of the outer

conductor. Therefore, the result does not change.

ASSESS The negative sign inABV means that the outer conductor is at a lower potential than the inner one. This is

precisely what we expect because the electric field points radially outward (in the direction of decreasing potential).

68. In this case, Equation 22.2a gives2 2

0 0 0 0(0) ( ) ( / ) /3.

R R

rV V R E dr E R r dr E R

69. INTERPRET This problem is about an electric potential function and its corresponding electric field.

DEVELOP We first note that the potential can be rewritten as

( ) ( 3)(1 )V x x x x

Since the potential V is independent of y and z, the electric field has only an x component, which can be computed

using Equation 22.9: / .xE V x

EVALUATE (a) From the equation above, we see that ( ) 0V x at 0,x 1 m and 3 m.

(b) Differentiating V with respect to x, we find23 4 3.dV

x dxE x x

(c) Solving the quadratic equation, the solutions for 0xE are

4 16 4(3)( 3)0.535 m and 1.87 m

2(3)x

ASSESS The component of the electric field is the negative of the rate of change of the potential. So the points

where 0xE are where the slope of ( )V x vanishes (see the figure).

22.12 Chapter 22

70. The potential difference between the sphere and the shell depends only on the electric field inside the shell (which is

due to sphereq only). For a spherically symmetric configuration, this is

sphere shell sphereV V kq 1

shell ).R

(Here,shellR is the

inner radius of the shell.) Outside the shell, the potential depends on the total charge, which, at the surface of a

spherical distribution, was found to be shell sphere shell shell( )/V V k q q R in Example 22.3. (Here,

shellR is the outer

radius of the shell, the conducting material of which is assumed to be an equipotential region in electrostatic

equilibrium, and1 1

sphere sphere shell shell sphere sphere shell sphere shell0.) Thus, ( ) ( )V V V V V V V kq R R k q q

1 1 1

shell sphere sphere shell shell.R kq R kq R (For a thin shell, the inner and outer radii are approximately equal.) (a) In this

problem, 1sphere shell sphere shell3

60 nC, and 5 cm,q q R R so 6 1sphere 5 3

(9 )(1 ) kV 7.2 kV.V (b) If shellq is

changed to equal 6 1sphere sphere 5 3

, (9 )(1 ) kV 14.4 kV.q V

71. INTERPRET This problem is about the electric field of a charged disk. We want to show that the expression found

in Example 22.8 has the correct asymptotic behavior.

DEVELOP From Example 22.8, the electric field on the axis 0x of a charged disk of radius a is

2 2 2

21x

kQ xE

a x a

In the limit where ,x a the quantity2 2/x x a may be approximated as

22 1/ 2

22 2

1[1 ( / ) ] 1

2

x aa x

xx a

L

EVALUATE Substituting the above expression into the first one, we obtain

2

2 2 2 22 2

2 2 11 1 1

2x

kQ x kQ a kQE

a a x xx a

which is like that of a point charge.

ASSESS As always, a finite-size charge distribution looks like a point charge at a large distance.

72. Combining the given data with the potential in Example 22.7, we find2 2 2150 V 2 / ( (5 cm) 5 cm)kQ a a

and2 2 2110 V 2 ( (10 cm) 10 cm).kQ a a The charge can be eliminated by division,

2

2

1 ( /5 cm) 1150

110 4 ( /5 cm) 2

a

a

Several lines of algebra to remove the square roots finally yields (5 cm) 105 209/52 14.2 cm.a We can

now solve for Q from either of the first two equations,2 2 2 1( /2 )[ ] 1.67 nC.Q Va k x a x

73. INTERPRET This problem is about the work done by the electric field in separating the thorium nucleus and the

alpha particle.

DEVELOP The work done by the Coulomb repulsion as the thorium nucleus and the alpha particle separate is

ThTh

Th,RR

kq qkqW U q V q

R R


where 7.4 fmR is the initial separation. (The final separation is essentially infinite.) On the other hand, with the

neglect of other interactions, the work-energy theorem requires that this equal the change in the kinetic energy of

the two particles:

2 2

Th Th

1 1

2 2W m v m v

Note that the two particles start from rest. The conservation of momentum (under the same assumptions) requires

also that Th Th0 m v m v (since the total momentum is zero initially), so that Thandv v can be determined.

EVALUATE Combining the two expressions, noting that Th| | (4/234) | |v v gives

2 2 2

Th

(2 )(90 ) 1 1 16234 u 4 u u 4 u

2 2 234

k e ev v v

R

The speeds are74.07 10 m/sv and

5

Th 6.96 10 m/s ,v where we have used271 u 1.66 10 kg.

ASSESS SinceTh ,m m we expect

Th .v v This indeed is what we have found.

74. (a) As in Example 22.7,2 2

0( ) / ,a

V x k dq x r where 2 ,dq rdr but now0 .r a Reference to standard

integral tables gives

2 2 2 22 20 0

0 2 2

2 2 2

0

2 2 ( ) ln

2 2

1 ( / ) ( / ) ln( / 1 ( / ) )

ak kr dr a x a x aV x x a

a a xx r

k a x a x a a x a x

(b) As in Example 22.8, /xE dV dx results in

2 2 2 2 2

0 22 2 2 2 2 2

2 2 1/ 2

0

2 1ln

(2 / ) ln( / 1 ( / ) ) ( / )(1 ( / ) )

x

x a x a x x x a x aE k

a x a xx a a x a x a

k x a a x a x a x a x

(c) The logarithm has to be expanded carefully, up to order3( / ) ,a x to evaluate

xE for .x a Thus,

2 32 2 2 2 2 3

2 2 2 2 3

1 1ln 1 ln 1

2 32 2 2 2 6

a a a a a a a a a a a a

x x x x x x xx x x x x

L L

Also,

1/22 2 3

2 2 31 1

2 2

a a a a a a

x x xx x x

Then,

23 3

0 0

3 3 2

2 2

6 2 3x

k x k aa a a aE

a x xx x x

This is like a point charge field for2 2

0 0 0(2 / ) 2 /3.

a

Q a r dr a

75. INTERPRET The cylinder is a continuous charge distribution, and we want to find the potential at its center on

the axis.

DEVELOP Following the hint given in the problem, we consider the cylinder to be composed of rings of radius a,

width dx, and charge ( /2 ) .dq q a dx The potential at the center of the cylinder (which we take as the origin, with x

axis along the cylinder axis) due to a ring at ( )x a x a is (see Example 22.6)

cyl2 2 2 2

=2

kdq kqdxdV

x a a x a

EVALUATE The whole potential at the center follows from integration:

cyl2 2

2ln ln(1 2) 0.881

2 2 2

a

a

kq dx kq a a kq kqV

a a a aa ax a

22.14 Chapter 22

ASSESS The result can be compared with that at the center of a charged ring of radius a:ring / .V kq a In the

cylinder case, the charge elements generally are farther away from the center compared to the ring. Hence, we

expectcyl ring.V V

76. INTERPRET We use a given equation for electric field in a region, and the potential at one point in that region, to

calculate the potential at another point in the region.

DEVELOP The potential difference between two points is given by .B

AV E dr

r rIn this case, the electric field

is 4 ,a

xE ir

where3

55 Vm .a At1 0,x the potential is 0.V We want to find the potential at

2 2.8 m,x so we

will integrate from 1x to 2,x and find .V V

EVALUATE

2

1

2.8 m

4 3

1.3 m

7.5 V3

xx

xx

a aV dx

x x

ASSESS This is a direct application of the definition for potential.

77. INTERPRET We need to find the potential due to a line of charge with a non-constant charge density. The problem

is one-dimensional: the line and the point of interest are all on the x axis. We will use the integral expression for

potential.

DEVELOP We start with ,k dq

rV where

2

0( )x

Ldq dx dx and .r x x We will integrate from

2

Lx to

2.Lx

We must also check to see that our expression reduces to the expected result for .x L

EVALUATE

2 2/2 / 2

0 0

2/2 / 2

/22

20

2

/2

20

2

ln2

2ln

2

L L

L L

L

L

k kk dq x xV dx dx

r x x L x xL

k xV x x x x x

L

k x LV Lx x

x LL

For ,x L we rearrange slightly and use the approximation ln(1 ) for small :

20

2

2 3 2 3

20

2

2 220

2 2 2

ln 1 ln 12 2

1 1 1 1

2 2 2 3 2 2 2 2 3 2

8 8

k L L LV x

x x xL

k L L L L L L LV x

x x x x x x xL

k L L L LV x

x xL x x

3 3 3

20 0

3 3 2 3 1224 24 12

k k LL L Lx

xx x L x

The total charge on the rod is

/22 3 3/2 /2

20 0 0

0 2 2 2/2 /2/2

0

2

3 83

12

LL L

L LL

x x Lq dq dx x dx

L L L L

Lq

so in the limit of ,x L .kq

xV

ASSESS The answer we obtain appears to be a point charge from a large distance, as we would hope.


78. INTERPRET We need to find the potential due to a line of charge with a non-constant charge density. The problem

is one-dimensional: the line and the point of interest are all on the x axis. We will use the integral expression for

potential.


rV where 0( )x

Ldq dx dx and .r x x We will integrate from

2

Lx to

2.Lx

We must also check to see that our expression reduces to the expected result for .x L Since the total charge is

obviously zero for this arrangement, we would expect that the potential at large x would go as the potential for

a dipole: 2

1 .x

V

EVALUATE

/2 /20 0

/2 /2

/20

/2

0

ln

2ln

2

L L

L L

L

L

k kk dq x xV dx dx

r x x L L x x

kV x x x x

L

k x LV L x

L x L

For ,x L we rearrange slightly and use the approximation ln(1 ) for small :

0

2 3 2 3

0

2 2

0

2 2

ln 1 ln 12 2

1 1 1 1

2 2 2 3 2 2 2 2 3 2

8 8

k L L LV x

L x x x

k L L L L L L LV x

L x x x x x x x

k L L L LV x

L x x x x

23 3 3

0 0

3 3 3 224 24 12 12

k k LL L Lx

Lx x x x

ASSESS The potential at large distances goes as 2

1 ,x

as predicted.

79. INTERPRET Behind all the details of this problem, the actual question is to determine the potential a distance r

from the center of a uniform line of charge. We will use the integral form for potential, and verify whether the given

equation is correct.


rV where

Q

Ldq dz and

2 2.r r z We will integrate from 0z to

2,Lz

and multiply by 2.

Hopefully, the result of our calculation will be 2

2

2

2 4( ) ln 1 .

kQ L L

L r rV r

EVALUATE

/2 /2

0 0 2 2

/2

2 2 2 2

0

2 2 2

2

22

2 2 1ln( ) ln 4 ln( )

2 2

2 4 2ln ln 1

2 2 4

L L

L

k dq k Q kQ dzV dz

r r L L r z

kQ kQ LV z r z L r r

L L

kQ L L r kQ L LV

L r L r r

ASSESS The equation is correct.

80. INTERPRET We find the potential, and then the field, a distance x from the base of a line charge ( )y by which

extends from 0y to .y h We find the potential using the integral form, and then find the field from the derivative

of the potential.


rV where ( )dq y dy by dy and

2 2.r x y We integrate from 0y to

.y h

22.16 Chapter 22

Once we have ( ),V x we use( )

( )dV x

dxE x to find the x component of the electric field.

EVALUATE

2 2

0 2 2 0

2 2( 1)

hhk dq y dyV kb kb x y

r x y

V x kb x h

Now take the derivative:

2 2

( )( )

dV x kbxE x

dx x h

ASSESS The units look somewhat odd on that answer: but remember that the units of b are charge per length

squared, so it works out.

23.1

ELECTROSTATIC ENERGY

AND CAPACITORS

EXERCISES

Section 23.1 Electrostatic Energy

14. Number the charges 50 C, 1, 2, 3, 4,iq i as they are spaced along the line at 2 cma intervals. There are

six pairs, so2 1 1

pairs 1 2 1 3 1 4 2 3 2 4 3 4 2 3/ ( / /2 /3 / /2 / ) ( / )(1 1i j ijW kq q r k q q a q q a q q a q q a q q a q q a kq a

2 9 21

21) 13 /3 13(9 10 m/F)(50 C) /(3 2 cm) 4.88 kJ.kq a

15. INTERPRET We find the electrostatic energy of a collection of point charges, using the method outlined in

Section 23.1.

DEVELOP The point charges are arranged on the corners of a square of side length a. Three of them are the

same:1 2 3 .q q q q The fourth has charge 1

4 2.q q The first charge takes no work to bring into position,

since there is no field present initially.

1 0W

The work to bring in charge 2 takes the potential energy the charge will have due to charge 1:

1 2

2

q qW k

a

The work to bring in charge 3 takes the potential energy due to the presence of charges 1 and 2:

1 3 2 3

32

q q q qW k k

a a

We’ve assumed that charges 2 and 3 are diagonally opposite on the square, thus the 2 term. The work to bring in

charge 4 is the potential energy due to the presence of charges 1, 2, and 3:

3 41 4 2 44

2

q qq q q qW k k k

a aa

Again, charges 1 and 4 are diagonally opposite each other, while charges 2 and 3 are separated from charge 4 by

only the side length a. The energy of the configuration is the sum of these works.

EVALUATE

2 3 1 4

1 2 3 4 1 2 1 3 2 4 3 4

2 2

2 2

1 1 1 1 11 1 1

2 22 2 2 2 2

q q q qkE W W W W q q q q q q q q

a

kq kqE

a a

ASSESS This is not a particularly convenient method of calculating energies. Fortunately, the charge on a single

electron is small enough that we can usually approximate real charge distributions as continuous and integrate.

16. INTERPRET We find the electrostatic energy of a collection of point charges, using the method outlined in

Section 23.1.

23

23.2 Chapter 23

DEVELOP The point charges are arranged on the corners of a square of side length a. Three of them are the

same: 1 2 3 .q q q q The fourth has charge 4 .q q The first charge takes no work to bring into position, since

there is no field present initially.

1 0W

The work to bring in charge 2 takes the potential energy the charge will have due to charge 1:

1 2

2

q qW k

a

The work to bring in charge 3 takes the potential energy due to the presence of charges 1 and 2:

1 3 2 3

32

q q q qW k k

a a

We’ve assumed that charges 2 and 3 are diagonally opposite on the square, thus the 2 term. The work to bring in

charge 4 is the potential energy due to the presence of charges 1, 2, and 3:

3 41 4 2 4

42

q qq q q qW k k k

a aa

Again, charges 1 and 4 are diagonally opposite each other, while charges 2 and 3 are separated from charge 4 by

only the side length a. The energy of the configuration is the sum of these works.

EVALUATE

2 3 1 4

1 2 3 4 1 2 1 3 2 4 3 4

2

2 2

1 11 1 1 1 0

2 2

q q q qkE W W W W q q q q q q q q

a

kqE

a

ASSESS Although it takes work to bring the second and third charges into place, that energy is regained by the

negative work done in bringing in the fourth.

17. INTERPRET We find the speed of the charges in an “electrostatic explosion.” We use conservation of energy to

solve this problem.

DEVELOP The three charges are initially in a symmetric arrangement, as shown in Figure 23.1. They have the same

charge q, and as shown in Section 23.1, the work done to assemble the configuration is 1 31 2 q qq q

a aW k k

22 3 3

.q q kq

a ak This is the initial energy of the configuration. From the symmetry of the initial configuration, we can see

that they will each gain 1

3of that energy when they fly apart, and we can find the speed v from their kinetic energy.

EVALUATE The initial energy is23.

kq

i aE The final energy is

21

2(3) .fE mv These energies must be equal, so

2323 2

2.

kq k

a ammv v q

ASSESS Conservation of energy applies to more than just mechanical systems. We use this technique throughout

all areas of physics.

18. The electrostatic potential energy of the water molecule (in this approximation) ispairs / .i j ijU W kq q r The two

oxygen-hydrogen pairs have separation1010 m,a while the hydrogen-hydrogen pair has separation

2 cos37.5 1.59 .a a Therefore,2 2 9 19 22 ( )( 2 )/ /1.59 3.37 / 3.37(9 10 )(1.6 10 )U k e e a ke a ke a

10 18J/10 7.76 10 J 48.5 eV. ( U is called the ionic separation energy.)

Section 23.2 Capacitors

19. INTERPRET This problem is about a parallel-plate capacitor. We are given the plate separation and the charges on

the plates, and asked to find the electric field between the plates, the potential difference, and the energy stored.

Electrostatic Energy and Capacitors 23.3

DEVELOP The electric field between two closely spaced, oppositely charged, parallel conducting plates is

approximately uniform (directed from the positive to the negative plate), with strength (See the last paragraph of

Section 22.6.)

0 0

qE

A

Since the electric field is uniform, the potential difference between the plates is given by Equation 22.1b,

,V Ed where d is the plate separation. Finally, the energy stored in the capacitor can be calculated using

Equation 23.3:2/2 /2,U CV qV where .q CV

EVALUATE (a) Using the equation above, we find the electric field to be

212 2 2

0

(1.1 C)1.99 MV/m

(8.85 10 C /N m )(0.25 m)

qE

A

(b) The potential difference is

(1.99 MV/m)(5 mm) 9.94 kVV Ed

(c) The energy stored is

21 1 1(1.1 C)(9.94 kV) 5.47 mJ

2 2 2U CV qV

ASSESS For completeness, the capacitance of the capacitor is

212 2 2100

3

(8.85 10 C /N m )(0.25 m)1.1 10 F 11 nF

5 10 m

AC

d

The value is typical of a capacitor.

20. The separation is much smaller than the linear dimensions of the plates, so the discussion in Section 23.2 applies.

(a) From Equations 23.1 and 23.2, we have 2 2 2 21 1

02 2/2 (7.2 C) (1.2 mm)/2W CV Q C Q d A

12 2(8.85 10 F/m)(5 cm) 1.41 J.

(b) The additional work required to double the charge on each plate

is2

0(2 ) /2W Q d A W 3 4.22 J.W

21. INTERPRET This problem is about the energy stored in a parallel-plate capacitor.

DEVELOP Using Equations 23.1–3, the energy stored in a parallel-plate capacitor is related to the charges on the

plates as

2 2 2 2 22 0

0 0

21 1 1 1 1

2 2 2 2 2 / 2

AUQ Q Q Q Q dU CV C Q

C C C A d A d

Once Q is found, the potential difference V between the plates is simply given by

21 1 1 2( )

2 2 2

UU CV CV V QV V

Q

EVALUATE (a) Using the values given in the problem statement, we find the charge to be

212 2 2 3

0

3

2 2(8.85 10 C /N m )(0.05 m) (15 10 J)0.744 C

1.2 10 m

AUQ

d

(b) The potential difference is3

6

2(15 10 J)2

0.744 10 C40.3 kV.U

QV

ASSESS Since the electric field between the plates is0/ ,E the potential can also be calculated as

0 0

40.3 kVE Qd

Vd d A

22. From Equation 23.1, /V 1.3 C/60 V 0.0217 F.C Q

23. INTERPRET This problem deals with the electrostatic energy stored in a capacitor.

DEVELOP Equation 23.1, / ,C Q V provides the connection between capacitance C, charge Q, and potential

difference V. This is the equation we shall use to solve for Q.

23.4 Chapter 23

EVALUATE The magnitude of the charge on either plate is

(4.0 F)(3.5 V) 14.0 CQ CV

ASSESS This is a very large capacitor, since the capacitance of most capacitors falls in the range of1 pF to1 F.

24. Equation 23.1 gives / 1.6 mC/100 F 16 V.V Q C

25. INTERPRET This problem is about the units of0, the permittivity constant.

DEVELOP The value of0 is

212 28.85 10 C /N m ,

with units2 2

C /N m . On the other hand, since / ,C Q V 1 F

(farad) is equivalent to1 C/V.

EVALUATE From the above, we see that the units of0 are

2 2C /N m (C/N m)(C/m) (C/J)(C/m) C/V m F/m

ASSESS To see that the result makes sense, we can use Equation 23.2 and write

0

Cd

A

The units of C are F, while the units of /d A are1/m.

26. For a (closely spaced) parallel plate capacitor, with circular plates, Example 23.1 shows that2

0 /C r d 2

(8.85 pF/m) (20 cm) /(1.5 mm) 741 pF.

27. INTERPRET The configuration is a parallel-plate capacitor, and the object of interest is the area of each plate.

DEVELOP Using Equations 23.1 and 23.2 for capacitance, we have

0

0

AQ QdC A

V d V

This is the equation we shall use to solve for the area A.

EVALUATE Substituting the values given, we find the area to be

6 3 2

212 2

0

(2.3 10 C)(1.1 10 m)1.91 m

(8.85 10 C /N m )(150 V)

QdA

V

ASSESS If we take the plate to be square, then each side has a length of 1.38 m, which indeed is much greater than

the distance of 1.1 mm between the plates.

28. From Equation 23.3,2 21 1

2 2CV (2500 F)(35 V) 1.53 J.CU

29. INTERPRET The configuration is a parallel-plate capacitor, and the object of interest is the capacitance, given the

energy stored in the capacitor.

DEVELOP Equation 23.3,2/2,U CV provides the connection between the stored energy U, the capacitance C,

and the potential V. This is the equation we shall use to solve for C.

EVALUATE Substituting the values given, we find the capacitance to be

2 2

2 2(350 J)70 nF

(100 V)

UC

V

ASSESS The value is within the typical range of capacitance12 6

(10 F to 10 F).

Section 23.3 Using Capacitors

30. There are only two ways to connect the two capacitors, in parallel,1 2 1 F 2 F 3 F,C C C and in

series,1 2 1 2/( ) 1 2 F/(1 2) 2/3 F.C C C C C (See Equations 23.5 and 23.6a.)

31. INTERPRET This problem is about connecting two capacitors in series.

DEVELOP For two capacitors connected in series, the total voltage is the sum of the voltages across each

one,1 2 ,V V V whereas the charge on each capacitor is the same,

1 2 1 1 2 2Q Q C V C V

Given1 2, , and ,V V V we can use the above two equations to solve for

1 2and .C C

EVALUATE The above equations can be combined to give1 1 2 1( / ) ,V V C C V or


21

1 2

CV V

C C

Similarly, we find 1

1 22 .

VC

C CV For

1 2 /2 50 VV V V as in this problem, either equation implies1 2 .C C

ASSESS The equivalent capacitance of two capacitors connected in series is

1 2

1 2 1 2

1 1 1 C CC

C C C C C

In our case, 1/2C C since 1 2,C C and the voltage across C is

1 1

1

1 1

22

/2

Q QQV V

C C C

This is precisely the condition given in the problem statement.

32. (a) C1 is in series with the parallel combination of C2 and C3. Thus,1 2 3 1 2 3( )/( ) (0.02 F)C C C C C C C

(1 2)/(2 1 2) 0.012 F. (b) The net charge on the entire combination is (0.012 F)(100 V)Q CV

1.2 C. Since C1 is in series with the capacitors in parallel,1 2 31.2 C .Q Q Q Q Moreover, for the parallel

capacitors,2 2 2 3 3 3/ / ,V Q C V Q C so

3 2 3 2/ / 2.Q Q C C Thus,2 (1/3) 0.4 CQ Q and

3 (2/3) 0.8 C.Q Q (In general, for two capacitors in parallel,2 2 2 3/( )Q C Q C C etc.) (c) Equation 23.1,

applied to each capacitor, gives1 1 1/V Q C 1.2 C/0.02 F 60 V, and

2 3 40 V.V V (Alternatively, one can

first use the general result in the solution to Exercise 31 (with C2 replaced by2 3 )C C to obtain the

voltages,1 2 3 1 2 3( ) /( )V C C V C C C (3/5)(100 V), 2 3V V 1 1 2 3/( ) (2/5)(100 V),C V C C C and then

use Equation 23.1 to find the charges.)

33. INTERPRET This problem is about all possible values of equivalent capacitance that could be obtained by

connecting three capacitors.

DEVELOP The equivalent capacitance has a maximum value when all three capacitors are connected in parallel, with

1 2 3C C C C

On the other hand, C is a minimum when the capacitors are connected in series:

1 2 3

1 2 3 1 2 1 3 2 3

1 1 1 1 C C CC

C C C C C C C C C C

Intermediate values are obtained when one is in parallel with the other two in series, or one in series with the other

two in parallel.

EVALUATE (a) When all capacitors are in parallel, we have

1 2 3 1 2 3 6 FC C C C

(b) When all are in series, the equivalent capacitance is

1 2 3

1 2 1 3 2 3

(1 F)(2 F)(3 F) 6F 0.545 F

(1 F)(2 F) (1 F)(3 F) (2 F)(3 F) 11

C C CC

C C C C C C

(c) When one is in parallel with the other two in series, the possible values are

23.6 Chapter 23

2 3

1

2 3

1 3

2

1 3

1 2

3

1 2

(2 F)(3 F) 111 F F 2.20 F

2 F 3 F 5

(1 F)(3 F) 112 F F 2.75 F

1 F 3 F 4

(1 F)(2 F) 113 F F 3.67 F

1 F 2 F 3

C CC C

C C

C CC C

C C

C CC C

C C

Similarly, when one is in series with the other two in parallel, the equivalent capacitance is

( )1 1 1 i j k

i j k i j k

C C CC

C C C C C C C

Therefore, the possible values are

1 2 3

1 2 3

2 1 3

2 1 3

3 1 2

3 1 2

( ) (1 F)(2 F 3 F) 5F 0.933 F

1 F 2 F 3 F 6

( ) (2 F)(1 F 3 F) 4F 1.33 F

2 F 1 F 3 F 3

( ) (3 F)(1 F 2 F) 3F 1.50 F

3 F 1 F 2 F 2

C C CC

C C C

C C CC

C C C

C C CC

C C C

ASSESS With three capacitors, each having two options (parallel or series), there are eight possible outcomes.

Section 23.4 Energy in the Electric Field

34. Equation 23.7 relates the field strength and the electric energy density,

35

0 12

2(3 J/m )2 / 8.23 10 V/m

(8.85 10 F/m)E u

(Note: The manipulation of units is facilitated by the relations V J/C and F C/V. Thus,3 3

(J/m )/(F/m) (VC/m )/ 2

(C/V m) (V/m) .)

35. INTERPRET This problem is about the volume required for storing a given amount of electrostatic energy.

DEVELOP In a uniform field, Equation 23.8 can be written as

2

0

1(volume)

2U E

Knowing U and E allows us to deduce the volume of the region.

EVALUATE Substituting the values given in the problem statement, the volume is

9 3 3

2 2

0

2 2(4 MJ)volume 1.00 10 m 1 km

(8.85 pF/m)(30 kV/m)

U

E

ASSESS This is a very big volume occupied by a car battery. In reality, not all the energy stored goes to creating

the field.

36. The energy content of gasoline is644 10 J/kg (Appendix C), and the density of gasoline is

3670 kg/m , so the

equivalent energy density is6 3 10 3(44 10 J/kg)(670 kg/m ) 2.95 10 J/m .u The field strength giving the same

electrostatic energy density is (Equation 23.7; see also the solution to Exercise 34)

10 310

0 12

2(2.95 10 J/m )2 / 8.16 10 V/m,

(8.85 10 F/m)E u

which greatly exceeds the breakdown field in air.

37. INTERPRET In this problem we are asked to find the electric energy stored in a proton by assuming it to be a

uniformly charged sphere.

DEVELOP For this model of the proton, the field strength at the surface is2

/E ke R (from spherical symmetry and

Gauss’s law). Thus, the energy density in the surface electric field is


2 22

0 2 4

1 1 1

2 2 4 8

ke keu E

k R R

EVALUATE With151 fm 1 10 m,R the energy density is

2 9 2 2 19 230 3 3

4 15 4

(9 10 N m /C )(1.6 10 C)9.17 10 J/m 57.3 keV/fm

8 8 (1 10 m)

keu

R

ASSESS The energy density is enormous, given the small size of the proton.

PROBLEMS

38. The work necessary to bring up Qx is2

0 0/ 2 / ,x xW kQ Q a kQ a while the work necessary to subsequently bring up

Qy is 0 0/ / 2 (1 2)/ .y y x y yW kQ Q a kQ Q a kQ Q a If 2 ,y xW W then 0(1 2) 4 ,yQ Q 0or 4 /( 2 1)yQ Q

01.66 .Q (Note:1/( 2 1) 2 1.)

39. INTERPRET This problem is about the work done to create a certain charge distribution. The work is equal to the

energy stored in the system.

DEVELOP When a charge q (assumed positive) is on the inner sphere, the potential difference between the spheres is

2

1 1a a

b b

kqdrV Edr kq

a br

To transfer an additional charge dq from the outer sphere requires work .dW V dq

EVALUATE The total work required to transfer charge Q (leaving the spheres oppositely charged) is

2

0 0

1 1 1 1 1

2

Q Q

W Vdq kq dq kQa b a b

ASSESS Since2/2 ,U Q C this shows that the capacitance of this spherical capacitor is

1 1

1

( )( )

abC

k b ak a b

Note that capacitance depends only on the geometry of the system, and is independent of V and Q.

40. Combining Equation 23.7 with Equation 21.8, one finds 0 0/ 2 / ,E u or 02q A A u

2 3(10 cm) 2(4.5 kJ/m )(8.85 pF/m) 2.82 C.

41. INTERPRET We find the capacitance of a pair of coaxial cylinders, using the definition of capacitance.

DEVELOP From Example 22.4, we can see that the potential difference between two points distances a and b from

a cylinder of charge per length is02ln( ).a

bV

The outer cylinder contributes nothing to the potential between the

two surfaces, so we may set the potential at b to zero and use02ln( )b

aV

as the magnitude of the voltage

difference between the two conductors. The total charge on the inner cylinder is ,Q L and the definition of

capacitance is .Q

VC

EVALUATE

0

0

2

2

ln lnb ba a

LQ LC

V

ASSESS This capacitance depends only on the geometry, as we would expect.

42. This result, mentioned in the solution to Problem 39, also follows from Equation 23.1 and the potential difference

between two concentric conducting spheres,1 1

0( ) ( )/4 / .V kQ a b Q b a ab Q C

43. INTERPRET This problem is about the capacitance of a charge configuration.

DEVELOP When the third (middle) plate is positively charged, the electric field (not near an edge) is

approximately uniform and away from the plate, with magnitude0/ .E Since half of the total charge Q is on

either side (by symmetry), the charge density is /2 .Q A The potential difference between the third plate and the

outer two plates (which are both at the same potential and carry charges of /2Q on their inner surfaces) is

23.8 Chapter 23

0 02

d QdV Ed

A

EVALUATE Using Equation 23.1, the capacitance is

02.

AQC

V d

ASSESS As expected, capacitance only depends on the geometric factors (e.g., area, distance, . . .) of the

configuration. Note that the above arrangement is like two capacitors in parallel.

44. (a) Equation 23.3, expressed as a ratio for the same capacitor charged to two different voltages, gives 2

2 1 2 1/ ( / ) .U U V V Therefore,2

2 (25/100) (0.04 J) 2.5 mJ.U (b) From the same Equation 23.3,2

1 12 /C U V 2

2(0.04 J)/(100 V) 8 F. 2

2 2( 2 / ,C U V of course.)

45. INTERPRET In this problem, we are asked to compare the amount of energy stored in two different capacitors.

DEVELOP The energy stored in a capacitor can be calculated using Equation 23.3:2/2.U CV

EVALUATE The energies stored in the two capacitors are

2 2

1 1 1

2 2

2 2 2

1 1(1 F)(250 V) 31.3 mJ

2 2

1 1(470 pF)(3 kV) 2.12 mJ

2 2

U C V

U C V

Thus, the energy stored in the second is about 14.8 times less than the first one.

ASSESS The general expression for the ratio of the energies stored in two capacitors is

22

2 2 2 2 2

2

1 1 11 1

/2

/2

U C V C V

U C VC V

46. The capacitor must withstand a potential difference of 2 / 2(12 mJ)/10 F 49.0 V,CV U C so one rated at

50 V would barely suffice.

47. INTERPRET In this problem we are asked to compare the amount of charge and energy stored in three different

capacitors.

DEVELOP The charge stored in a capacitor is Q CV (Equation 23.1), and the energy stored is2/2U CV

(Equation 23.3).

EVALUATE (a) The stored charges are

1 1 1 (0.01 F)(300 V) 3 CQ C V for the first capacitor

2 2 2 (0.1 F)(100 V) 10 CQ C V for the second

3 3 3 (30 F)(5 V) 150 CQ C V for the third

(b) The stored energies for the three capacitors are

2

1 1 1

2 2

2 2 2

2 2

3 3 3

1 1(3 C)(300 V) 450 J

2 2

1 1(0.1 F)(100 V) 500 J

2 2

1 1(30 F)(5 V) 375 J

2 2

U C V

U C V

U C V

(c) The cost effectiveness, measured in J/? is 18.0, 14.3, and 4.26 for these capacitors, respectively.

ASSESS The first capacitor is the most cost effective of the three.

48. (a) From Equation 23.3, 2 / 2 950 J/100 F 4.36 kV.V U C (b)av / 300 J/2.5 ms 120 kW.P U t

49. INTERPRET This problem is about the power consumption of a camera flashtube and the capacitor required for its

operation.

DEVELOP Power, / ,P W t as defined in Equation 6.15, is the rate at which energy is used. This is the equation we

shall use for our calculations.


EVALUATE (a) The power delivered to the tube during the flashing process is

flash

5.0 J5 kW

1 ms

WP

t

(b) The energy stored in a capacitor is21

2.U CV Thus, the capacitance to be used to supply the flash energy is

2 2

2 2(5 J)250 F

(200 V)

UC

V

(c) The average power consumption during the 10-second interval is 5.0 Jav 10 s

0.5 W.P

ASSESS The average poweravP is only about

4

flash10 times .PThis makes sense because the camera only flashes

for 1 ms every 10 s.

50. (a)14 9 5

(10 W)(10 s) 10 J.W Pt (b)

3 21

2(0.17%) 1.7 10 ( ),CW U CV

so5 32 10 J/(1.7 10 )(0.26 F) 21.3 kV.V

51. INTERPRET This problem is about the change in the capacitance and the stored energy due to an insertion of a

conducting slab between the capacitor plates.

DEVELOP For part (a), we first note that the charge on the plates remains the same, and so does the electric

field0( / )E in the gaps between the plates and the slab. However, the separation (i.e., the thickness of the field

region) between the plates has been changed to .d Therefore, the capacitance becomes0 / .C A d For part (b),

when the charge is constant (no connections to anything isolates the system), the energy stored is inversely

proportional to the capacitance,2/2 .U Q C

EVALUATE (a) With the plate separation reduced to 40% of its original value1 2 0.4 ,d d d d the capacitance

is increased to

0 0 2.5 0.4

A AC C

d d

(b) The stored energy becomes

2 2

0.402 2(2.5 )

Q QU U

C C

or the energy decreases to 40% of its original value.

ASSESS With the slab inserted, there is less field region and less energy stored. While the slab is being inserted,

work is done by electrical force to conserve energy. Note that the configuration behaves like a series combination of

two parallel plate capacitors:

1 2 1 2

1 2 0 0 0 0

1 1 1 0.4 12.5

2.5

d d d d dC C

C C C A A A A C

52. Relative to points A and B, the combination of capacitors 2, 3, and 4 is in parallel with 1 (see numbering added to

Fig. 23.15), sotot 1 234 .C C C However, C234 consists of 2 in series with the parallel combination of 3 and 4,

so234 2 34 2 34 2 3 4 2 3 4/( ) ( )/( ).C C C C C C C C C C C Since each individual capacitance is equal to

C, 2234 3

C C and 5tot 3

.C C

23.10 Chapter 23

53. INTERPRET In this problem we want to connect capacitors of known capacitance and voltage rating to obtain the

desired equivalent capacitance and voltage rating.

DEVELOP In parallel, the voltage across each element is the same, so to increase the voltage rating of a

combination of equal capacitors, series connections must be considered. The general result of Problem 31 shows

that for two equal capacitors in series, the voltage across each is one half the total, so the voltage rating of a series

combination is doubled.

EVALUATE (a) To obtain the desired voltage rating, we must use two capacitors in series so that the voltage

becomes 50 V 50 V 100 V. However, the capacitance is now (2 F)(2 F)/(2 F 2 F) 1 F,C and we

need to increase the total capacitance to twice that of just two in series, without altering the voltage rating. This can

be accomplished with a parallel combination of two pairs in series, i.e., a parallel combination of two1 F, 100 V

series pairs. (Note that for equal capacitor elements, a parallel combination of two pairs in series has the same

properties as a series combination of two pairs in parallel.)

(b) In this situation, four capacitors in series are required (rating 200 V, and capacitance1 1

4(2 F)C or 1

2F).C

ASSESS Schematically the connections described look like the following.

One may use Equations 23.5 and 23.6 to verify that the configurations indeed have the desired capacitance and

voltage rating as specified in the problem statement.

54. Number the capacitors as shown. Relative to points A and B, C1, C4, and the combination of C2 and C3 are in series,

so the capacitance is given by1 1 1 1

1 4 23.ABC C C C C23 is a parallel combination, hence

23 2 3 ,C C C therefore 1

ABC 1 1 1 6

7(3 F) (2 F) (2 F 1 F) , or F 0.857 F.ABC

55. INTERPRET This problem involves an assemblage of capacitors in a circuit, and we are interested in the energy

stored in one of the capacitors.

DEVELOP Number the capacitors as shown. Relative to points A and B, C1, C4, and the combination of C2 and C3

are in series. The energy of the3 1 FC capacitor is

213 3 32

,U C V where the voltage drop across C3 is 3 23 ,V V

since C2 and C3 are in parallel. On the other hand, since C1, C4, and C23 are in series, we have1 4 23V V V V and

1 1 4 4 23 23.Q C V C V C V Once we know23 ,V 3U can be calculated.

EVALUATE From the equation above, we find

23 23 1 23 4 23 23 23

3 3 7 2(1 / / ) 1

3 2 2 7V V C C C C V V V V

This gives2 21 1

3 3 32 2(1 F)(2 50 V/7) 102 J.U C V

ASSESS Since the combination of C2 and C3 are in series with C1 and C4, C3 does not “feel” the full 50 V applied

across AB, so its working voltage is lower. In fact we find it to be3 (2/7) .V V

56. This is shown in the solution to Exercise 31.

57. INTERPRET This problem is about connecting two capacitors with different voltage ratings in series.

DEVELOP In Problem 56, we have verified that the voltages across the individual capacitors are


2 11 2

1 2 1 2

and C V C V

V VC C C C

where V is the voltage across the combination. The equations allow us to determine the maximum voltage that could

be applied.

EVALUATE The separate ratings require

21

1 2

0.2 F 250 V

0.1 F 0.2 F 3

CV V V V

C C

Similarly, we find

1

2

1 2

0.1 F 1200 V

0.1 F 0.2 F 3

CV V V V

C C

The more stringent limit is 3

2(50 V) 75 V.V

ASSESS We use the more stringent condition for V. If the less stringent one were used, then 3(200 V)V

600 V and the voltage across1C would be 400 V, which exceeds the voltage rating of 50 V.

58. (a) From Equations 23.2, 23.4, and Table 23.1, one obtains2

0 / (2.3)(8.85 pF/m)(50 cm )/(25 m)C A d

4.07 nF. (b) Dielectric breakdown in polyethylene occurs at a field strength of 50 kV/mm, corresponding to a

maximum voltage, for this capacitor, of (50 kV/mm)(25 m) 1.25 kV.V Ed (See ASSESS in the solution for

Problem 59.)

59. INTERPRET This problem involves inserting a piece of dielectric material in a capacitor.

DEVELOP From Equation 23.4, one finds that the capacitance in the presence of dielectric is 0

0 .A

dC C

As for part (b), the maximum voltage for this capacitor ismax max .V E d

EVALUATE (a) With reference to Table 23.1, the thickness is

2

0 (2.6)(8.85 pF/m) (0.15 m)3.46 mm

470 pF

Ad

C

Since this is much less than the radius of the plates, the parallel plate approximation (plane symmetry) is a good one.

(b) The dielectric breakdown field for polystyrene ismax 25 kV/mm,E so the maximum voltage is

max maxV E d

(25 kV/mm)(3.46 mm) 86.5 kV.

ASSESS In practice, the working voltage would be less than this by a comfortable safety margin.

60. If we assume that the inner and outer surfaces of the membrane act like a parallel plate capacitor, with the space

between the plates filled with material of dielectric constant 3, then the capacitance per unit area is0/ / .C A d

Thus,23(8.85 pF/m)/(1 F/cm ) 2.7 nm.d . .

61. INTERPRET This problem is about connecting two capacitors in parallel.

DEVELOP The equivalent capacitance of two capacitors connected in parallel is (Equation 23.5)

1 2C C C

In this problem,1 0.001 F 1000 pF,C and

2C has a range of 10–30 pF.

EVALUATE The combination covers a range from 1010 to 1030 pF, or about

100.0098 0.98%

1020

from the central value.

ASSESS We expect the variation to be very small since1 2 .C C

62. Since there is no y or z dependence, the volume element of the cube can be written as2 ,dV L dx where 1 mL is

the cube’s edge. Then22 2 2 51 1

0 0 0 0 0 0 02 2( / ) ( / ) /3.

LU udV E x x L dx E x L Numerically, 1

6(8.85 pF/m)U

2 5 2(24 kV/m) (1 m) /(6 m) 23.6 J.

63. INTERPRET This problem involves electrostatic energy contained within a charged sphere. Our system has radial

symmetry.

23.12 Chapter 23

DEVELOP From Example 21.1, we see that the radially symmetric field inside the sphere is3/ .rE kQr R so the

energy density is

2 2 22

0 3 6

1 1( )

2 2(4 ) 8r

kQr kQ ru r E

k R R

The energy within the sphere can be found by integrating over the volume.

EVALUATE With thin spherical shells of radius r for volume elements,24 ,dV r dr the integral for the energy is

2 2 2 22 4

6 60 0sphere

4108 2

R RkQ r kQ kQU udV r dr r dr

RR R

(This is just the energy stored inside the sphere. For the energy outside the sphere, and the total energy, see

Problems 64 and 65.)

ASSESS The result shows that U is inversely proportional to R. This means that the stored energy decreases if the

same amount of charge Q is distributed over a greater volume. Our result can be compared to the situation (Problem

64) where the total charge Q is distributed over its surface. In that case, the total energy stored in its electric field

is2/2 .U kQ R

64. The calculation of the electrostatic energy for a sphere with uniform surface charge density is, in fact, given in

Example 23.5. We simply set2 ,R R the radius of the sphere, and

1R (so the integral covers all the space

where the field is non-zero).

65. INTERPRET This problem is about the electrostatic energy of a charged sphere, both within and outside the sphere.

Our system has radial symmetry.

DEVELOP The field outside a spherically symmetric distribution of radius R is the same for the charge Q uniformly

spread over the volume or the surface (thanks to Gauss’s law). Thus, Problem 64 gives the energy in the electric field

outside a uniformly charged spherical volume, while Problem 63 gives the energy inside. The total energy is

2 2 23

2 10 5

kQ kQ kQU

R R R

EVALUATE Applied to a U235-nucleus, the result gives

2 9 2 2 2103 3(9 10 N m /C )(92 )

3.55 10 J 2.22 GeV5 5(3.3 fm)

kQ eU

R

ASSESS This Coulomb energy is about 1% of the mass energy of the U235-nucleus, mc2.

66. The initial electrostatic energy of two isolated spherical drops, with charge Q on their surfaces and radii R, is

iU 21

22( / )kQ R (see Problem 64 and Example 23.5). Together, a drop of charge 2Q, radius

1/32 ,R and energy

2 1/3 2/3 21

2(2 ) /(2 ) 2 / ,fU k Q R kQ R is created. The work required is the difference in energy,

f iW U U 2/3 2 9 8 2 3 4

(2 1) / (0.587)(9 10 m/F)(1.5 10 C) /(2 10 m) 5.95 10 J.kQ R

67. INTERPRET This problem involves electrostatic energy contained within a volume around a charged wire. Our

system has line symmetry.

DEVELOP The electric field outside the wire (assumed to have line symmetry) is radially away from the axis with

magnitude 2 /rE k r (see Equation 21.6). The energy density in a cylindrical shell of radius r, length ,L and

volume 2 ,dV rLdr is

2 22

0 2

1 1 2

2 2(4 ) 2

k ku E

k r r

EVALUATE Thus, the energy in the space mentioned in this problem is

23

2

2

9 2 2 2

2 ln 32

(9 10 N m /C )(28 C/m) (1 m) ln 3 7.75 J

R

R

kU udV rLdr k L

r


ASSESS The energy density (energy/volume) decreases as21/ .r This means that more energy is concentrated in the

region of space closer to the wire.

68. The energy in the thunderstorm of Example 23.4 was about111.4 10 J, while the energy in a lightning flash

is8(30 C)(30 MV) 9 10 J.qV Thus, there is energy for about

11 81.4 10 /9 10 156 flashes, which at a rate

of one flash in 5 s, would last for156 5 s 13 min.

69. INTERPRET Our object of interest is a spherical capacitor, and we want to explore the limit where .b a a

DEVELOP The capacitance for a spherical capacitor can be derived by noting that the potential difference between

two concentric conducting spheres is

1 1 ( )

( )

kQ b a Q abV kQ C

a b ab V k b a

The limit d b a a can be taken to show that C reduces to that of a parallel-plate capacitor.

EVALUATE With ,d b a a we find

2

0 0 0 04 4 ( ) 4

( )

ab a a d a AabC

k b a d d d d

which is the result of Equation 23.2, with2

4A a being the area of the spherical plates.

ASSESS The limit d b a a means that the radius of the sphere is much greater than the separation between

the two spheres. So locally we have effectively two plates that appear to be parallel.

70. The results of Problems 63 and 65 (with the aid of the argument in the solution to Problem 64) show that the

fraction is just2 2 1

6( /10 )/(3 /5 ) .kQ R kQ R

71. INTERPRET This problem is about the size an electron has that makes its electrostatic energy equal to its mass energy.

DEVELOP The electrostatic energy stored in the field of a classical electron, whose charge e&

is distributed

uniformly over the surface of a sphere of radius R, is2/2U ke R (see Problem 64). The radius of the electron can

then be obtained by setting this equal to the electron’s mass energy,2.eU m c

EVALUATE Equating the two energies gives2 2/2 ,em c ke R or

2

21.41 fm

2 e

keR

m c

where constants from the inside front cover were used.

ASSESS The “classical radius of the electron,” based on a consideration of the scattering of electromagnetic waves

from free electrons, called Thomson scattering, is actually equal to2 22 / .e er R ke m c

72. (a) In so far as fringing fields can be neglected, the electric field between the plates is uniform, E V d (but

when the dielectric is inserted,0V V and E depends on x). In fact, on the left side, where the slab has penetrated,

0(1/ )( / ),LE and on the right,0/ ,rE where

land

r are the charge densities on the left and right sides.

Thus,0 E

land

0 ,r E and the charge can be written (in terms of geometrical variables superposed on

Fig. 23.19) as q wx l 0 0( ) ( ) ( / ) ( ).r w L x Ew x L x V d w x L x From Equation 23.1,

/C q V 0 ( )/ ,C x L x L where0 0 /C A d and .A Lw Although the question specifies 1

2,x L for which

value the capacitance is 102( 1),C we give C as a function of x, because we will need to differentiate with respect

to x in part (c). (b) When the battery is disconnected, the capacitor is isolated and the charge on it is a

constant,0 .q q The stored energy is

2 21

2( ) /2U CV U q C 2

0 0 0/2 ( ) /( ),q L C x L x U L x L x

where2 21 1

0 0 0 0 02 2/ .U q C C V For 1

2,x L the energy is

2

0 0 /( 1).C V (c) The force on a part of an isolated system is

related to the potential energy of the system by Equation 8.9. The force on the slab is therefore

0 0

2

( 1),

( )x

U L U LdU dF

dx dx x L x x L x

in the direction of increasing x (so as to pull the slab into the capacitor). For 1

2,x L the magnitude of the force is

2 2

0 02 ( 1)/ ( 1) .C V L It turns out that if we rewrite the force, for any value of x, in terms of the voltage for that x,

using0 0 0 0 ( )/ ,q C V CV C V x L x L the expression can be used in the succeeding problem. Thus,

23.14 Chapter 23

22 2

0 0 0 0

2

( 1) ( 1)( 1)

2 22( )x

C V L C C VVF L

L Lx L x

73. INTERPRET This problem is about the effect of inserting a dielectric material into a parallel-plate capacitor, which

is connected to a battery.

DEVELOP We first note that the capacitance depends on the configuration and electrical properties of the plates

and insulating materials, not on the external connections, so

0

( )x L xC C

L

as in the preceding problem. If the capacitor remains connected to a battery, the voltage is constant,0 .V V

EVALUATE (b) The energy is( )2 21 1

0 0 02 2.

x L x

LU CV C V

For /2,x L we get2

0 0 ( 1)/4.U C V Note that this

is different from the preceding problem, because the battery does work.

(c) When the capacitor is connected to a battery, Equation 6.8 ( / )xF dU dx for the force does not apply. However,

for particular values of charge and voltage on the capacitor, the force on the slab considered here is the same,

regardless of the external connections. In the preceding problem we found that21

02( 1)/ ,xF C V L where V was

the particular voltage (and, because of the special form of the capacitance, ( ),C x the particular charge q did not

appear). Since 0V V in this problem,

2

0 0 ( 1)1

2x

C VF

L

ASSESS The force on the slab is to the right, drawing the dielectric between the plates.

74. The potential difference between the plates, via path A B is 0,AB A BV E d L r

since the field is non-zero

and parallel to .d L If there were no fringing field, then the integral of Er

over path C D would vanish,

0,CD C DV E d L s

in contradiction to the path-independence of a conservative field. Since the plates are

equipotentials,ABV must equal .CDV If we choose a path along an electric field line, we can define an average field

strength by .avE d L Ed L E L r

It is clear that the average field strength is weaker along the longer field line

between the same two points.

75. INTERPRET We find the energy per length stored in the electric field of a uniformly charged rod. We will use

Gauss’s law to determine the electric field within the rod, and then integrate the energy density.

DEVELOP We use Gauss’s law, with cylindrical symmetry, to find the electric field using enclosed

0.

qE da r r

Once

we have this field, we integrate the energy density21

02Eu E over the cylinder to find the total

energy .EU u dV

EVALUATE

2

enclosed

0 0 02

22

3

00 0 0 0

0 0

2 2 44

0 0

(2 )2

12

2 2 4

1

2 4 8

L R R

q r L rE da E rL E

rU rdrd dz L r dr

U RR

L

r r

ASSESS Increasing the radius increases the energy per length dramatically.


76. INTERPRET We are given the electric field in a region, and are asked to find the total electrostatic energy

contained in that cylinder. We will integrate the energy density to find the answer.

DEVELOP The electric field is given as 0

3ˆE

xE ir

on the region between x L and 2 .x L The cylinder has radius R,

and lies along the x axis.

EVALUATE

2 2 22 2

2 0 0 0102 3 6

22 2 2 2

0 0 0 0

5 5 5

2 2

0 0

5

1

2

1 1 1 1

2 5 10 32

31

320

L L

EL L

L

L

E R EU u dV R dx dx

x x

R E R EU

x L L

R EU

L

ASSESS These units look wrong, because the0E given does not have units of electric field. Instead it has units

of3electric field length , which makes things work out correctly.

77. INTERPRET Find the electrostatic energy stored between two parallel plates of a parallel-plate capacitor, and then

differentiate to find the force between the plates.

DEVELOP We find the electric field between the plates using0 0

.Q

AE

This is constant, so total energy stored

in this field is then .EU u V We can find the force by using .dUx dx

F

EVALUATE

(a)

22

0

0

1( )

2 2E

QU u V E Ax x

A

(b)

2

02x

dU QF

dx A

This is half the value you would obtain by multiplying the charge on one plate by the field between the plates.

ASSESS The answer we get for (b) is half the field times the charge on one plate: but we must remember that the

field between the plates is created by both charged plates. A charge is not affected by the field it creates. Only the

field created by the other plate causes a force on each plate, and the other plate creates half the field.

78. INTERPRET We use the energy stored in a capacitor network at a given voltage to find the capacitance of an

unknown capacitor in that network. We use the equation for the energy stored in a capacitor, and the rules for

adding capacitors in series and parallel.

DEVELOP We first find the capacitance of the entire networknetworkC in terms of the given values and C the

unknown. Then we use21

network2U C V with 5.8 mJU and 100 V,V and work backward through the network

to solve for C.

EVALUATE

(a) See the figure.

(b)

2

network network 2

left

network left

left top top

top

1 21.16 F

2

1 1 12.76 F

2.0 F

1.0 F 1.76 F

1 1 14.27 F

3.0 F

UU C V C

V

CC C

C C C

CC C

23.16 Chapter 23

ASSESS This trick of “unraveling” the network is often useful. We will use the same trick in dealing with resistor

networks later in the course.

79. INTERPRET Given the electric field as a function of radius inside a cylindrical capacitor, we find the energy

density and the total energy stored, and we show that this total energy is consistent with the energy stored in a

cylindrical capacitor.

DEVELOP The electric field is given in the problem as

0

ˆ2

E rr

r

The energy density is21

02,Eu E and the capacitance for a given length L (from Problem 23.41) is

02

ln ba

LC

EVALUATE First we find the energy density as a function of radius:

2

2 2

0

1

2 4Eu

r

Next we integrate this over the volume of the capacitor to find the total energy U:

2 2 22

2 2 20 000 0

2 1ln

48 8

L b b

a a

L L bU rdrd dz dr

r ar

From Problem 23.41, we have

22

2 0

0 0 0

21 1 1ln so ln ln

2 2 2 ln 2 2 2ba

Lb b L bV CV U

a a a

ASSESS It is always a relief, in a complicated problem like this, to discover that things are consistent with what we

already know. Checking that we get the same value for U by multiple methods is a good method of verification.

80. INTERPRET We find capacitance per length for a polyethylene-filled coaxial cable.

DEVELOP From Problem 23.41, we see that the capacitance per length for a vacuum-filled coaxial cylinder is

02

ln ba

C

L

This cable is filled with polyethylene, which has a dielectric constant 2.3.

EVALUATE

02

116 pF/mln b

a

C

L

ASSESS This is a typical value for coaxial cable capacitance. It’s not much, but it can have a significant effect,

especially with high-frequency signals as we will see in later chapters.

81. INTERPRET We find the capacitance of a “twin lead” antenna with given dimensions, using the potential given

and the definition of capacitance

DEVELOP We are given that

0

lnb a

Va

The characteristics of the cable are 0.5 mm,a 12 mm,b and 30 m.L The capacitance is .Q

VC

EVALUATE


0

0 266 pFlnln

b ab aaa

LQ LC

V

ASSESS This is less than the maximum allowable capacitance, so our design is ok.

24.1

ELECTRIC CURRENT

EXERCISES

Section 24.1 Electric Current

14. The current is the amount of charge passing a given point in the wire, per unit time, so in one second,

(1.5 A)(1s) 1.5 C.q I t The number of electrons in this amount of charge is 191.5 C/1.6 10 18C 9.38 10 .

15. INTERPRET This problem is about the number of charges involved in setting up an electric current.

DEVELOP For a steady current, the amount of charge crossing a given area in time t can be found by using

Equation 24.1a, / .I Q t

EVALUATE A battery rated at 80 A h can supply a net charge of

5(80 C/s)(3600 s) 2.88 10 CQ I t

ASSESS By definition,1 A 1C/s, or1 C 1 A s. So the result makes sense.

16. The charge moving through the membrane each second is 30 nC. Since singly-charged ions carry one elementary

charge (about160 zC), this corresponds to 11 2330 nC/(160 zC/ion) (1.88 10 ion)/(6.02 10 ion/mol) 0.311

pmol. (See Example 24.2(a), Appendix B, and Table 1.1.) Chemical drug-testing instrumentation can detect

amounts of substances this low.

17. INTERPRET In this problem we are asked to find the current density, given the electric current and the cross section.

DEVELOP The current density J, is defined as the current per unit area, or / .J I A The area of the cross section is 2 2/4.A R d

EVALUATE The cross section of a wire is uniform, so the density is

6 2

2 3 2

10 A7.65 10 A/m

/4 (1.29 10 m) /4

I IJ

A d

ASSESS If the current I is kept fixed, the smaller the cross-sectional area A, the greater the current density J.

Section 24.2 Conduction Mechanisms

18. Aluminum obeys Ohm’s law, so 8 2/ (0.085 V/m)/(2.65 10 m) 3.21 MA/mJ E E

(see Table 24.1).

19. INTERPRET In this problem we are asked to calculate the electric field in a current-carrying conductor.

DEVELOP To find the electric field, we make use of Ohm’s law (which applies to silver), / ,J E and the

definition of current density, which is assumed to be uniform in the wire.

EVALUATE Using Table 24.1, we find the electric field to be

8

2 4 2

(1.59 10 m)(7.5 A)0.168 V/m

/4 (9.5 10 m) /4

IE J

d

ASSESS The value is a lot smaller than the electric field we discussed in electrostatic situations. Since silver is

such a good conductor, a small field can drive a substantial current.

20. Assuming a uniform current density obeying Ohm’s law, we find 214

/ / , or 4 /J E I d d I E 1/22[(0.22 m)(350 mA)/ (21 V/m)] 6.83 cm (see Table 24.1).

24

24.2 Chapter 24

21. INTERPRET This problem is about applying Ohm’s law to find the resistivity of a rod.

DEVELOP If the rod has a uniform current density and obeys Ohm’s law (Equations 24.4a and 4b), then its

resistivity is

2( /4)

/

E E E d

J I A I

EVALUATE Substituting the values given in the problem, we find the resistivity of the rod to be

2 2 26( /4) (1.4 V/m) (10 m) /4

2.20 10 m50 A

E d

I

ASSESS With reference to Table 24.1, we see that the value is within the range of resistivity of a metallic conductor.

22. Equation 24.4a and 24.4b show that the conductivity and the resistivity are reciprocals of one another. Thus,

(a) 1 8 1 7 1(1.68 10 m) 5.95 10 ( m) for copper, and (b)

1 1(0.22 m) 4.55 ( m)

for typical seawater. (The salinity of open-ocean water varies between 33 and 37 parts per thousand, but can vary

from 1 to 80 parts per thousand in shallow coastal waters.)

Section 24.3 Resistance and Ohm’s Law

23. INTERPRET This problem involves using Ohm’s law to calculate the resistance of a heating coil.

DEVELOP The macroscopic form the Ohm’s law is probably applicable to the heating coil, which is typically a

coil of wire. Equation 24.5 gives / .R V I

EVALUATE Substituting the values given in the problem statement, we find the resistance to be

120 V25

4.8 A

VR

I

ASSESS This is a fairly large resistance. The resistance of the coil used for heating is usually quite high.

24. For an Ohmic resistor, (300 mA)(1.2 k ) 360 V.V IR

25. INTERPRET In this problem we want to apply Ohm’s law to calculate the current across a resistor.

DEVELOP Ohm’s law, / ,I V R given in Equation 24.5 provides the connection between current, resistance, and

voltage. This equation allows us to compute the current I.

EVALUATE Substituting the values given in the problem statement, we find the current to be

110 V2.34 mA

47 k

VI

R

ASSESS From Ohm’s law, we see that current is inversely proportional to resistance when V is kept fixed. In our

case, a large resistance yields a small current.

26. If the third rail behaves like a uniform ohmic material, its resistance is given by Equation 24.6 (with resistivity from

Table 24.1),8 2/ (9.71 10 m)(5 km)/(10 15 cm ) 32.4 m .R L A

27. INTERPRET This problem is about applying Ohm’s law to calculate the current across a resistor.

DEVELOP Ohm’s law, / ,I V R given in Equation 24.5 provides the connection between current, resistance, and

voltage. This equation allows us to compute the current I.


45 V25 mA

1.8 k

VI

R

ASSESS From Ohm’s law, we see that current is inversely proportional to resistance when V is kept fixed. In our

case, a large resistance yields a small current.

28. If the density (and mass) of the wire is constant, then so is its volume, constant.LA Thus,2

/ /R L A L LA 2

constant ,L and doubling L increases the resistance four times.

Section 24.4 Electric Power

Electric Current 24.3

29. INTERPRET This problem is about electric power of a motor, given the current drawn and the voltage across the

terminals.

DEVELOP Equation 24.7, ,P IV provides the connection between electric current, voltage and electric power.

This equation is what we shall use to compute the power P.

EVALUATE Substituting the values given in the problem, we find the power consumption to be

(125 A)(11 V) 1.38 kWP IV

ASSESS A large electric power is needed to start a car. This is why a very high current is required in the starter

motor to crank the engine.

30. (a) From Equation 24.7, / 4.5 W/750 mA 6.0 V.V P I (b) From Equation 24.5, / 6.0 V/0.75R V I

A 8.0 .

31. INTERPRET This problem is about the current drawn from a battery, given its voltage and power rating.

DEVELOP Equation 24.7, ,P IV provides the connection between electric current, voltage, and electric power.

This equation is what we shall use to compute the current I.

EVALUATE Rearranging Equation 24.7, we find

240 W160 A

1.5 V

PI

V

ASSESS We expect the current drawn from the battery to be very small since little power is needed to operate a watch.

32. Equation 24.8b gives (1.5 kW)(35 ) 229 V.V PR

33. INTERPRET In this problem we are asked to calculate the resistance of a light bulb.

DEVELOP Since voltage is fixed, we use Equation 24.8b,2/ ,P V R to find the bulb’s resistance.

EVALUATE Substituting the values given in the problem, we obtain

2 2(120 V)240

60 W

VR

P

ASSESS This is the resistance of the light bulb at its operating temperature. Light bulbs actually are non-Ohmic

because their resistance varies with temperature.

Section 24.5 Electrical Safety

34. INTERPRET We use Ohm’s law to find the resistance necessary for a given voltage to drive a specified current.

DEVELOP Ohm’s law is .V IR We are given 30 VV and 100 mA,I and need to find R.

EVALUATE

30 V300

100 mA

VR

I

ASSESS The resistance of dry skin is much higher than this, so in dry conditions this voltage does not pose a threat.

35. INTERPRET We use Ohm’s law to find the resistance necessary for a given voltage to drive a specified current.

DEVELOP Ohm’s law is .V IR We are given 120 VV and 2.5 mA,I and need to find R.

EVALUATE

30 V12 k

2.5 mA

VR

I

ASSESS If your skin was wet, your resistance would be much lower and this voltage would pose a serious threat.

36. INTERPRET Given a resistance and voltage, what current would flow? We use Ohm’s law, and also estimate

whether the resulting current could be felt.

DEVELOP Ohm’s law is .V IR The resistance is 100 k ,R and the voltage is 12 V.V The threshold for

sensation is listed in Table 24.3 as 0.5–2 mA.

EVALUATE

(a)

24.4 Chapter 24

3

12 V0.12 mA

100 10

VI

R

(b) This current is below the threshold for sensation, and would not be felt.

ASSESS The resistance of human skin varies considerably with moisture. If your hand was wet, the resistance

would be lower and the current would be high enough to deliver a noticeable shock.

PROBLEMS

37. INTERPRET This problem is about the average current in a given time interval and the number of ions in the channel.

DEVELOP The average current is a time-weighted average over sub-intervals

av = i i i i

i

I t I tQI

t t t

The number of singly-charged ions passing through the channel while current is flowing is / .N Q e

EVALUATE (a) The average current is

av (20%)(2.4 pA) (80%)(0) 0.48 pAi i

i

I tI

t

(b) The number of ions is

4

19

(2.4 pA)(1.0 ms)1.5 10

(1.60 10 C)

Q I tN

e e

ASSESS This is a very small current. Nevertheless, there exists instrumentation that can measure current down to

the pico-ampere range.

38. The current density in the filament (with numbers from Example 24.5) is2(0.833 A)/ (0.025 mm)J

8 24.24 10 A/m , in the direction of the current. A 12-gauge wire, carrying the same current, would have a current

density smaller by a factor of the square of the ratio of the diameters,2 4(0.05 mm/2.1 mm) 5.67 10 , or

5 22.41 10 A/m .J

39. INTERPRET This problem is about finding the total current in a film, given the current density and the linear

dimensions of the film.

DEVELOP Since current density J is current per unit area, assuming the current density is perpendicular to the

cross-sectional area of the film, the current is .I JA


5 2A (6.8 10 A/m )(2.5 m 0.18 mm) 306 AI J

ASSESS The current density 0.68 MA/m2 is approximately the maximum safe current density in typical household

wiring.

40. (a) Since the current is the same in both wires, but the cross-sectional area of the aluminum wire is four times that

of the copper wire, Equation 24.3 gives29

Cu Al Cu ,Cu Al ,Al ,Cu ,Al/ 1 /4 , or / 4(2.1 10 )/d d d dI I n ev n ev v v 29(1.1 10 ) 7.64. (b) The same equation also gives

Cu Cu Al Al Cu Al Al Cu/ 1, or / / 4.J A J A J J A A

41. INTERPRET In this problem we are given the microscopic parameters, and asked to calculate the drift speed of

charge carriers in different regions.

DEVELOP The drift speed of a charge can be calculated using either Equation 24.2 or Equation 24.3:

d

I Jv

nAq nq

EVALUATE In the copper wire, the drift speed is

1

29 3 19 8 214

10 A0.723 mm/s

(1.1 10 m )(1.6 10 C)( 10 m )d

Iv

neA

In the solution, the positive and negative ions have equal charge magnitudes, number, and current densities, by

hypothesis, so (2 ) ( 2 )( ) 4 .d d dJ n e v n e v nev Thus, with uniform current density, / ,J I A the drift speed is


1

23 3 19 4 214

10 A= = 3.26 mm/s

4 4 4(6.1 10 m )(1.6 10 C)( 10 m )d

J Iv

ne neA

Similarly, in the vacuum tube, we have

17

16 3 19 6 214

10 A3.62 10 m/s

(2.2 10 m )(1.6 10 C)( 10 m )d

Iv

neA

ASSESS Since the drift speeddv is inversely proportional to the number density n, it is greatest in the vacuum tube,

but smallest in the copper wire.

42. (a) If we use Equation 24.1b for the charge, then5s 2 2 3 3 4 5s1 1 10 02 3 4

|(60 nA/s) (200 nA/s ) (4 nA/s ) |q Idt t t t 13

(30 1000 25)(25 nC) 9.71 C.

43. INTERPRET This problem involves using Ohm’s law to find the current density and current, given the potential

difference between two ends of a wire.

DEVELOP Equation 24.4b, / ,J E can be used to find the current density. The corresponding current is simply

equal to .I JA

EVALUATE (a) Substituting the values given in the problem statement, we find the current density to be

6 2

8

/ (2.5 V)/(6 m)4.29 10 A/m

9.71 10 m

E V LJ

(b) The current is2 6 2 6 21

4( /4) (4.29 10 A/m )( 10 m ) 3.37 A.I JA J d

ASSESS Since the potential difference between the ends of the wire is only 2.5 V, we expect the resistance to be

small as well. The resistance of the wire is2.5 V

3.37 A0.74 .V

IR We can also find R by using Equation 24.6:

/ .R L A

44. (a) For a wire carrying uniform current density,2 3 2 21 1

4 4/ 20 A / (2.1 10 m) 5.77 MA/m .J I d (b) In an

Ohmic material like copper,8 2(1.68 10 m)(5.77 MA/m ) 97.0 mV/mE J (see Table 24.1).

45. INTERPRET In this problem we are asked to compare the voltages across wires made up of silver and iron.

DEVELOP The resistance of a uniform wire of Ohmic material is given by Equation 24.6: / .R L A Therefore, for

equal , , and ,L A I the ratio of the voltages is

Fe Fe Fe

Ag Ag Ag

V IR

V IR

EVALUATE Using Table 24.1 for resistivities, we obtain

8

Fe Fe

8

Ag Ag

9.71 10 m6.11

1.59 10 m

V

V

ASSESS Since iron has higher resistivity than silver, for the same length, area, and current, the voltage across the

iron wire would be higher.

46. For a uniform piece of material, Equations 24.5 and 24.6 imply 214

/ / ( )/RA L VA IL V d IL 21

4(9 V)( )(2 mm) /(2.6 mA)(2.4 cm) 0.453 m. This is closest to the resistivity of germanium in Table 24.1.

47. INTERPRET This problem compares resistances of wires made up of different materials. Ohm’s law is involved.

DEVELOP The resistance of a uniform wire of Ohmic material is given by Equation 24.6: / .R L A Therefore, the

resistance per unit length is

2/4

R

L A d

EVALUATE Equal values of /R L for copper and aluminum wires imply that

8

Cu Al Al Al2 2 8

Cu CuCu Al

2.65 10 m = 1.26

1.68 10 m

d

dd d

where we have used Table 24.1 for resistivities.

24.6 Chapter 24

ASSESS Since1/ 2,d z the higher the resistivity, the greater the diameter of the wire.

48. (a) From Equation 24.6 and Table 24.1,8 6 21

4/ / (1.68 10 m)/( 10 m ) 21.4 /m.R L A (b) From

Equation 24.5, (7 A)(21.4 m /m)(8 m) 1.20 V.V IR

49. INTERPRET We need to find the diameter of copper and of aluminum wire that would provide the requested

resistance. We will use the resistivity of the two materials to find the resistance of the wires. We will also find the

mass and cost for each wire, using the price per kilogram and the density.

DEVELOP The resistance of a cylinder is .L

AR

The resistivity of copper, from Table 24.1, is 81.68 10 mc and the resistivity of aluminum is 82.65 10 m.a

For parts (a) and (b), we will use

1.0 km,L and solve for the radius of each wire such that 50 m .R For part (c), we will find the mass of each

1-km wire, multiply by the cost per kg for each type, and determine which is cheaper. $4.65/kg,cc

$2.30/kg,ac 38900 kg/m ,cdensity and 32700 kg/m .adensity

EVALUATE

2

L L LR r

A Rr

(a) 2 21 mmc

c c

Ld r

R

(b) 2 26 mma

a a

Ld r

R

(c) The cost per meter of each wire is2(cost/kg) ( ) .density r For copper, the cost/meter is $13.90, and for

aluminum the cost/meter is $3.29.

ASSESS Although copper is a better conductor, and less is required, in this case aluminum is more economical due

to its lower density and lower cost per kg.

50. If opposite faces are equipotentials, the current density is uniform over any parallel cross-section. Then Equation

24.6 gives 1 1 2 3 1( ), where R L L L L is the length in the direction of the potential drop, and 2 3L L is the cross-

sectional area of an equipotential face. For 1 2 320 cm, 1 cm, and 0.5 cm,L L L and permutations thereof, we

find 1 (9.71R 8 210 m)(20 cm)/(1 0.5 cm ) 388 , and 2 30.971 , 0.243 .R R

51. INTERPRET In this problem we are asked to calculate resistance, given the voltage drop and the current. Ohm’s

law is involved.

DEVELOP The resistance at the battery terminal can be found by using Ohm’s law, / .I V R

EVALUATE From Ohm’s law, we have

4.2 V33.6 m

125 A

VR

I

ASSESS The resistance of the battery cable in Example 24.4 was 0.60 m ,R so most of the resistance just

calculated was in the connection.

52. Since the resistance of the heating element is assumed constant, its power output is proportional to the square of the

line voltage (Equation 24.8b). Thus, 2 2

0 0/ ( / ) (105/120) 76.6%,PP V V a drop of 23.4%. (The burner’s output

during a brownout is about1150 W.)

53. INTERPRET This problem is about the dependence of resistance and power dissipation on the geometry of the

material.

DEVELOP Equation 24.6, / ,R L A relates the resistance of a material to its geometry (length and cross-sectional

area). When voltage is fixed, the power dissipated is given by Equation 24.8b,2/ .P V R Thus, we see that at the

same voltage, the ratio of the power dissipated is the inverse of the ratio of the resistances, which in turn, goes as

the inverse of the square of the ratio of the diameters.

22 2

1 1 2 2 1 1 1

2 2

2 1 2 1 22 2

/ / /4

// /4

P V R R L A A d d

P R L A A dV R d


EVALUATE From the equation above, we see that if1 22 ,P P then 1 22 .d d

ASSESS Our result shows that power dissipated increases with the area, or the square of the diameter,2.P A dz z

54. (a) If we neglect possible energy losses in the locomotive’s engine,in out, or 2000 hp.P P VI Thus,

4(2000 hp)(746 W/hp)/(10 V) 149 A.I (b) The power loss in getting current to and from the locomotive is 2

loss ,P I R where I is the current from part (a) and R is the resistance of the feed/return circuit. The high voltage

wire has resistance (0.2 /km) ,L where its length, ,L is also the distance from the power station, and the resistance

of the return rail is negligible (see Exercise 26). Then2

loss in1%, implies 0.01 , or P P I R VI L 4(0.01)(10 V)/(149A)(0.2 /km) 3.35 km.

55. INTERPRET This problem is about using electric power to do mechanical work.

DEVELOP If there are no losses, the electrical power supplied to the motor,in ,P IV equals the mechanical power

expended lifting the weight,out .P Fv

EVALUATE Within out ,P P we find the current to be

(15 N)(0.25 m/s)0.625 A

6 V

FvI

V

ASSESS In reality, the motor will not be 100% efficient. So the current drawn will be somewhat higher than 0.625 A.

56. (a) If the power supplied by the plant is1 GW at 115 kV, then the current supplied is9

out / 10 W/115 kVI P V

8.70 kA. (b) The power loss in the transmission line is2 2

loss line (8.70 kA) (0.050 /km 40 km)P I R

151 MW, or 15.1% of the plants,1 GW output. (Note: The voltage drop along the transmission line is

line 17.4 kV, not 115 kV.)IR

57. INTERPRET This problem is about doing mechanical work with electric power.

DEVELOP The electrical power supplied to the motor isin .P IV With 90% efficiency, the mechanical power

expended lifting the weight isout in0.90 0.90 .P Fv P IV

EVALUATE Thus, the current drawn is

(200 N)(3.1 m/s)= 2.87 A

0.90 0.90(240 V)

FvI

V

ASSESS The less efficient the motor operates, the more the current that must be drawn to supply the required power.

58. The current density is 2 ˆ(0.1 A/cm )( /10 cm) ,J x k

rin the coordinate system superposed on Fig. 24.19. The cross

section can be divided into strips of area ˆ(5 cm) dA dxkr

(over which Jr

is constant), so the total current in the

bar is:

10 cm2 3

-sect 0

2 2

(10 A/cm )(5 cm)

1(0.05 A/cm ) (10 cm) 2.5 A

2

xI J dA xdx

rr

59. INTERPRET This problem is about using electric power to boil water. Electric energy is converted to thermal energy.

DEVELOP The power required to bring the water to its boiling point is

w wm c TQP

t t

EVALUATE (a) Substituting the values given in the problem, we find the power to be

24.8 Chapter 24

(250 ml)(1 g/ml)(4.184 J/g k)(100 C 10 C)1.11 kW

85 s

w wm c TP

t

(b) With2/P V R (appropriate to an Ohmic device), one finds the heater’s resistance to be

2 2(120 V)13.0

1.11 kW

VR

P

ASSESS Besides the power supplied to heat up water, additional power is also needed to compensate for any power

lost to the surroundings or cup, plus that used by the heater itself. All these factors have been neglected in the

problem.

60. Taking room temperature to be 20C, we find that the resistivity doubles when02 1 ( ), or T T T

3

0 1/ 20 C 1 C/4.3 10 253 C.T

61. INTERPRET In this problem we are asked to calculate the drift speed of electrons in aluminum.

DEVELOP Using Equation 24.2, the drift speed of electrons in the wire is

2( /4)d

I Iv

neA ne d

where n is the number density of conduction electrons.

EVALUATE For aluminum, the number density is

329 3

27

(3.5 electrons/ion)(2702 kg/m )2.11 10 electrons/m

(26.98 u/ion 1.66 10 kg/u)n

Thus, the drift speed is

2 29 3 19 2

20 A0.171 mm/s

( /4) (2.11 10 m )(1.6 10 C) (0.21 cm) /4d

Iv

ne d

ASSESS The drift speed is very small. With this speed, it would take about 100 minutes for an electron to travel

1 m. However, as explained in Example 24.1, electrons inside the conducting wire all get their “marching orders”

from the electric field that’s established almost instantaneously. Consequently, when you flip the switch, electrons

throughout the wire start to move almost instantaneously and light comes on immediately.

62. Consider a concentric cylindrical surface S, of radius r and height h, between the two metal electrodes. S completely

surrounds the disk, so the current flowing (assumed from the center to the sides) is ( / ) ,s sI J dA E dA r rr r

for

an ohmic solution. If the electrodes behave like perfect conductorsmetal( 0) . they are essentially equipotentials,

and the electric field in the solution has cylindrical (line) symmetry. (There are no fringing fields because outside

the solution, the current density is zero.) Thus, without changing the integral, circular, flat surfaces above and below

may be added to close S, so that .sI E dA rr

r This integral is the same as the one appearing in Gauss’s law

for two conductors in an identical configuration, as in Example 21.5. Thus,0/ 2 /ln( / ) ,s E dA q hV b a I

rrr

wherea bV V V is the potential difference and we used h instead of L for the length. A comparison of this with

Ohm’s law, ,V IR shows that the resistance between the electrodes is ln( / )/2 .R b a h In terms of the

capacitance of the same configuration of electrodes in air,0 0 0/ / ( / ), or / .q CV I V R R C This relation

holds for electrodes of arbitrary shape.

63. INTERPRET We find the radius of a conical wire. We do this by integrating over each thin disk that makes up the

wire, as shown in Figure 24.21. We also show that this resistance is the same as for a wire with elliptical cross section.

DEVELOP The resistance of each disk dx that makes up this cone is

2

dL dxdR

A r

The radius of each disk changes linearly from a to b as we move down the wire, so


( )b ar a x

L

We find the total resistance by integrating dR from 0x to .x L For comparison at the end of the problem, we will

use the area of an ellipse,

, andL

A ab RA

EVALUATE

2 ( )0

0( )

L

L

b a xb aLL

L LR dx

aba b aa x

ASSESS This is the same result as for a wire with elliptical cross-section and area ,A ab since .L L

A abR

64. INTERPRET We find the current in a particle beam. The beam is circular, has a current density0J at the center, and

the current density falls to half that value at the edge. We will integrate current density over the beam to find the

total current.

DEVELOP We will assume that the current density decreases linearly. The current density as a function of radius is

then

00

2

J rJ J

a

To find the total current, we will integrate over circular rings of radius r, each of which has area 2 .dA rdr

EVALUATE

0 00 0 0

2

0 00

( ) 1 2 2 12 2

22 1

2 3

a a a

a

r rdI JdA I J r dA J rdr J rdr

a a

rI J rdr J a

a

ASSESS The beam current is 23

as much as it would be if the current density was a constant value.

65. INTERPRET We find the resistance of a cylinder which has varying resistivity. We integrate over the length of the

cylinder, using the equation for resistance

( )x dxdR

A

DEVELOP We are given an equation for the resistivity:

/

0( ) 1 x Lxx e

L

where3

0 2.41 10 m. The radius of the cylinder is 0.0025 m,r and the length is 0.015 m.L We integrate

( )x dxdR

A

from 0x to .x L

EVALUATE

/

0 /0

2 20 0

/0 0

02 2

1( )1

[ ]

x LxL L

L x L

Lx L

ex dx xdR R dx e dx

A Lr r

LeR xe

r r

ASSESS The resistivity increases with length, and this value of R is larger than the constant-resistivity value 0

2 .L

r

That makes sense.

66. INTERPRET We find the capacitance of a homemade capacitor, as well as the energy stored and the discharge time.

We will approximate the capacitor as a parallel-plate capacitor.

24.10 Chapter 24

DEVELOP The capacitance of a parallel-plate capacitor with dielectric constant is 0 ,A

Cd

and the energy

stored in a capacitor is21

2.E CV The capacitor has area

2 210 cm 0.001 m ,A plate spacing 0.001 m,d and

dielectric constant 5.6. The voltage on the capacitor is 100 V,V and it self-discharges through the dielectric

which has resistance ,d

RA

where

131.2 10 m. ,dQ

Idt

and .Q CV

EVALUATE First, we calculate the capacitance:0 / 49.6 pF.C A d The energy stored in this capacitor at

100 V is2 71

22.48 10 J.E CV The resistance of the dielectric is

131.2 10 .d

AR

To find something about the time it takes to discharge this capacitor, we use

0[ ] ( )tRC

dQ d V dV dt dVI CV C V t V e

dt dt R dt RC V

This never completely discharges, but it has a time constant 595.2 s 9.92 minutes.RC

ASSESS This time constant is probably sufficient: the capacitor will still be at 60% of its initial value after 5 minutes.

67. INTERPRET We use the dimensions of a resistor, and the desired resistance, to find the necessary resistivity of the

material. We use the relationship between resistivity and resistance to solve the problem.

DEVELOP .L

RA

We are told that the length of the resistor is 10 m,L the width is 1.4 m,w and the

thickness is 0.85 m.t The desired resistance is 470 .R

EVALUATE

55.59 10 mL Rwt

Rwt L

ASSESS With the initial resistivity, the resistance would have 71 ,L

wtR

which is different enough that it

would probably cause trouble.

68. INTERPRET We use work, energy, and electric power to determine whether a given electric motor can lift an

elevator in the desired time.

DEVELOP The height the elevator must move in 20 st is 60 ft 18.3 m.h The 85%-efficient motor draws

24 AI at 480 V,V and .P IV The weight of the elevator is 8.5 kN,F so the work done isW Fh and the

power should be at least .W Fh

Pt t

EVALUATE The desired power is 7.77 kW.Fh

Pt

The power of the motor is 85% 9.8 kW.P IV

ASSESS Whether the customer will be happy or not is more the realm of philosophy than physics, but the motor is

certainly sufficient for the task.

69. INTERPRET We find the speed at which an electric vehicle can climb a given slope at a required speed, given the

voltage, the amp-hour rating of the batteries, the mass of the vehicle, and the efficiency of the motor. We will use

the equation for electrical power.

DEVELOP We will first use energy methods to find the power required, then the current needed to supply that

power at a voltage of 312 V,V using .P IV Only 85% of the electrical power is converted to mechanical power,

so we will have to take that into consideration as well. The batteries can supply 100 amp-hours, so once we know

the current we can find the time.

The slope is 10 , the mass of the car is 1500 kg,m and the desired speed is 45 km/h 12.5 m/s.v

EVALUATE The power necessary is ( sin ) .P Fv mg v The electrical power converted to mechanical power is

(85%) ,P IV sos

sinsin 0.85 120 A

0.85

mgvmgv IV I

V

The batteries can supply current such that


100 Amp hour,It so100 A hour

50 minutes120 A

t

ASSESS Unfortunately, this number is about right. Improving battery technology is one of the research areas

critical for further EV and hybrid vehicle development.

25.1

ELECTRIC CIRCUITS

EXERCISES

Section 25.1 Circuits, Symbols, and Electromotive Force

14. A literal reading of the circuit specifications results in connections like those in sketch (a). Because the connecting

wires are assumed to have no resistance (a real wire is represented by a separate resistor), a topologically

equivalent circuit diagram is shown in sketch (b).

15. INTERPRET This problem is about how various circuit elements can be connected to form a closed series circuit.

DEVELOP In a series circuit, the same current must flow through all elements.

EVALUATE One possibility is shown below. The order of elements and the polarity of the battery connections are

not specified.

ASSESS An important feature about a series circuit is that the current through all the components must be the

same. With two batteries, the direction of the current flow is determined by the polarity of the larger of the two

voltage ratings.

16. The circuit has three parallel branches: one with 1R and 2R in series; one with just 3;R and one with the battery

(an ideal emf in series with the internal resistance).

17. INTERPRET This problem explores the connection between the emf of a battery and the energy it delivers.

DEVELOP Electromotive force, or emf, is defined as work per unit charge, / .W q

EVALUATE Substituting the values given in the problem statement, we find the emf to be

27 J9 V

3 C

W

q

25

25.2 Chapter 25

ASSESS For an ideal battery with zero internal resistance, the emf is equal to the terminal voltage (potential

difference across the battery terminals).

18. The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the

battery. For an ideal battery, ,P I therefore 4.5 kJ,It or34.5 kJ/(1.5 V)(0.60 A) 5 10 s 1.39 h.t

19. INTERPRET This problem is about the chemical energy used up in the battery for the work done.

DEVELOP The power delivered by an emf is .P I Therefore, if the voltage and current remain constant, then

the energy converted would be .W Pt I t

EVALUATE Substituting the values given, the energy used in

(5 A)(12 V)(3600 s) 216 kJW I t

ASSESS The result makes sense; the energy used up is proportional to the current drawn, the emf, and the

duration the headlights were left on.

Section 25.2 Series and Parallel Circuits

20. From Equations 25.1 and 25.3b, 22 k 47(39 k )/(47 39) 43.3 k .R

21. INTERPRET This problem is about connecting two resistors in parallel.

DEVELOP The equivalent resistance of two resistors connected in parallel can be found by Equation 25.3a:

parallel 1 2

1 1 1

R R R

The equation allows us to determine2R when

parallelR and1R are known.

EVALUATE The solution for R2 in Equation 25.3a is

1 parallel

2

1 parallel

(56 k )(45 k )229 k

56 k 45 k

R RR

R R

ASSESS Our result shows that2 parallel .R R This is consistent with the fact that the equivalent resistance

parallelR is

smaller than1R and

2 .R

22. The starter circuit contains all the resistances in series, as in Figure 25.10. (We assumeLR includes the resistance

of the cables, connections, etc., as well as that of the motor.) With the defective

starter,int 6 V 12 VTV IR

int(300 A) ,R soint 0.02 .R With a good

starter, 12 V (100 A)(0.02 ) 10 V.TV

23. INTERPRET This problem is about the internal resistance of the battery in Exercise 22.

DEVELOP The starter circuit contains all the resistances in series, as in Figure 25.9. (We assume LR includes the

resistance of the cables, connections, etc., as well as that of the motor.) With the defective starter, the terminal

voltage is

int int6 V 12 V (300 A)TV IR R

EVALUATE From the equation above, we find the internal resistance to be

int

(12 V 6 V)0.02

300 A

TVR

I

ASSESS The terminal voltage 6.0 VTV is substantially less than the battery’s emf 12 V. The two are equal

only in the ideal case where the internal resistance vanishes.

24. From the equation for a battery short-circuited, in the subsection “Real Batteries,”int / 9 V/0.2 A 45 .R I

25. INTERPRET In this problem we are asked to find all possible values of equivalent resistance that could be

obtained with three resistors.

DEVELOP Since each resistor can be placed either in parallel or in series, there are eight combinations using all

three resistors.

EVALUATE Let1 21.0 , 2.0 ,R R and

3 3.0 .R The possible results are (a) one in series with two in

parallel:

Electric Circuits 25.3

1 2

3

1 2

1 3

2

1 3

2 3

1

2 3

(1 )(2 ) 113

1 2 3

(1 )(3 ) 112

1 3 4

(2 )(3 ) 111

2 3 5

R RR

R R

R RR

R R

R RR

R R

(b) one in parallel with two in series:

3 1 2

1 2 3

2 1 3

1 2 3

1 2 3

1 2 3

( ) (3 )(1 2 ) 9 3

1 2 3 6 2

( ) (2 )(1 3 ) 8 4

1 2 3 6 3

( ) (1 )(2 3 ) 5

1 2 3 6

R R R

R R R

R R R

R R R

R R R

R R R

(c) three in series:1 2 3 1 2 3 6 .R R R

(d) three in parallel:

11 1 1 1 2 3

1 2 3

1 2 1 3 2 3

(1 )(2 )(3 ) 6

(1 )(2 ) (1 )(3 ) (2 )(3 ) 11

R R RR R R

R R R R R R

ASSESS The equivalent resistance is a maximum when all three are connected in series, as in (c), and a minimum

when all are connected in parallel, as in (d).

Section 25.3 Kirchoff’s Laws and Multiloop Circuits

26. If the switch is irrelevant, then there is no current through its branch of the circuit. Thus, points A and B must be at

the same potential, and the same current flows through1R and

2 .R Kirchhoff’s voltage law applied to the outer loop,

and to the left-hand loop, gives1 1 2 2 0,IR IR and

1 1 3 0,IR respectively. Therefore,

1 2 2 1 1 23 1 1 1 1

1 2 1 2

R RIR R

R R R R

27. INTERPRET This problem asks for the currents in a multi-loop circuit.

DEVELOP The general solution of the two loop equations and one node equation given in Example 25.4 can be

found using determinants (or1I and

2I can be found in terms of3,I as in Example 25.4). The equations and the

solution are:

1 1 3 3 1

2 2 3 3 2

1 2 3

0 (loop 1)

0 (loop 2)

0 (node A)

I R I R

I R I R

I I I

25.4 Chapter 25

1 3 1 3

1 2 3 2 3

2 3 1 2 2 3 3 1 1 2 2 3

1 1 3 1 1

1 3 2 1 3 2 1 1 22 2 3 3 2 2

0 0( )1

0 ,

1 1 1 0 1 1

0( )1 1

0 , 0

1 0 1 1 1 0

R R RR R R

R R R R R R R R I R R

R R RR R R R R

I R I R

EVALUATE With the particular values of emf’s and resistors in this problem, we have

2

1 2 2 3 3 1 1(2 )(4 ) (4 )(1 ) (1 )(2 ) 14 R R R R R R R

and the currents are

1 2

1 2 3 1 3 2

1 2

2 3 1 1 3 2

1 2

3 2 1 1 2

[( ) ] [(4 1 )(6 V) (1 )(1 V)]/(14 ) 2.07 A

[ ( ) ] [(1 )(6 V) (2 1 )(1 V)]/(14 ) 0.214 A

( ) [(4 )(6 V) (2 )(1 V)]/(14 ) 1.86 A

I R R R

I R R R

I R R

ASSESS The same results could be obtained by retracing the reasoning of Example 25.4, with2 1.0 V

replacing the original value in loop 2. Then, everything is the same until the equation for loop 2:2 31.0 4 0.I I

28. The current is3 3 3/ 6 V/3 2 A,I V R from Ohm’s law. The answer is trivial because the potential

difference across the 3 resistor is evident from the circuit diagram. (However, if the 6 V battery had internal

resistance, an argument like that in Example 25.4 must be used.)

29. INTERPRET We find the current through a resisitor in a given circuit, using Kirchhoff’s laws. We will use the

loops and nodes drawn in Example 25.4.

DEVELOP The circuit is given to us in Figure 25.14, with one change:2 2.0V. We will use node A and loops 1

and 2. These will give us three equations, which we will use to solve the three unknown currents. At node A,

1 2 3 0.I I I For loop 1,1 1 1 3 3 0.I R I R For loop 2,

2 2 2 3 3 0.I R I R

EVALUATE We want current2 ,I so eliminate the other two currents. The node equation gives us

1 2 3.I I I

Substitute this into the equation for loop 1 and solve for 3:I

1 2 11 2 3 1 3 3 3

1 3

( ) 0I R

I I R I R IR R

Now substitute this value into the equation for loop 2 and solve for2 .I

1 2 12 2 2 3 2 1 3 2 2 1 3 1 3 2 1 3

1 3

2 1 3 2 1 2 2 3 1 3 1 3

1 3 2 1 3

2 2 2 2

1 2 2 3 1 3

0 ( ) ( ) 0

( ) ( ) 0

( ) (6V)(1 ) (2V)(3 )0 V/ 0 A

8 4 2

I RI R R R R I R R R R I R R

R R

R R I R R R R R R R

R R RI

R R R R R R

ASSESS The current through resistor2R is zero! Looking back at the original diagram, we can see that this would

mean that battery 2 is supplying no current and the voltage drops through resistors 1 and 3 equal the voltage

supplied by battery 1. This is a somewhat unexpected solution, but it is consistent.

Section 25.4 Electrical Measurements

30. The voltage across the10 k resistor in Fig. 25.29 is (150 V)(10)/(10 5) 100 V (the circuit is just a voltage

divider as described by Equations 25.2a and 25.2b), as would be measured by an ideal voltmeter with infinite

resistance. With the real voltmeter connected in parallel across the10 k resistor, its effective resistance is

changed to|| (10 k )(200 k )/(210 k ) 9.52 k ,R and the voltage reading is only

(150 V)(9.52)/(9.52 5) 98.4 V, or about 1.64% lower.

31. INTERPRET This problem is about the measurement error caused by the non-zero resistance of an ammeter.

DEVELOP The current in the circuit of Fig. 25.29 is

1 2

150 V10 mA

5 k 10 k

VI

R R


With the ammeter inserted (in series with the resistors), the resistance is increased by 100 .AR

EVALUATE The resulting current after including AR is

1 2

150 V9.93 mA

5 k 10 k 0.10 kA

VI

R R R

The value is about 0.662% lower than I.

ASSESS The current reading by the ammeter is lower due to its internal resistance.

32. The current through the misconnected ammeter isint/( ),mI R R so the power dissipated in it is

2

mP I R 2 2 2

int/( ) (12 V/0.11 ) (0.1 ) 1.19 kWm mR R R (comparable to a small toaster-oven).

Section 25.5 Capacitors in Circuits

33. INTERPRET In this problem we are asked to show that the quantity RC, the product of resistance and capacitance,

has the units of time.

DEVELOP The SI units for R and C are W and F, respectively. The units can be rewritten as

V V V s C1 1 1 1 , 1 F 1

A C/s C V

EVALUATE From the expressions above, the SI units for the time constant, RC, are

V s C1 F 1 1 s

C V

as stated.

ASSESS The quantity RC is the characteristic time for changes to occur in an RC circuit.

34. (a) ( )( F) s (see Exercise 33), (b)3 6(k )( F) 10 10 s ms, (c) (M )( F) s.

35. INTERPRET This problem is about the time dependence of the capacitor voltage in a charging RC circuit.

DEVELOP The capacitor voltage as a function of time is given by Equation 25.6:

/(1 )

t RC

CV e

EVALUATE After five time constants, 5 ,t RC the equation above gives a voltage of

5 31 1 6.74 10 99.3%CV

e

of the applied voltage.

ASSESS As time goes on and after many more time constants, we find essentially no current flowing to the

capacitor, and the capacitor could be considered as being fully charged for all practical purposes.

36. For the RC circuit described, Equation 25.6 gives the voltage across the capacitor, as a function of time. Thus,

CV /(1 )

t RCe or ln[ /( )] (10 F)(470k ) ln(250/50) 7.56 s.Ct RC V

37. INTERPRET We find the voltage across a capacitor in a circuit, given that the capacitor is “fully charged.” We

will take this to mean that the current through the capacitor is zero, and use the results of Example 25.7b, with

Ohm’s law, to find the voltage required.

DEVELOP In Figure 25.24a, we see the circuit in question. If the capacitor is fully charged, then no current flows

through it and the circuit is equivalent to the circuit shown in 25.24c. So we find the current through resistor2R in

Figure 25.24c and then determine the voltage across resistor2 ,R which is the same as the voltage across the

capacitor.

EVALUATE The current through resistor2R is given in Example 25.7 as

1 2.

R RI

The voltage is given by Ohm’s

law as 2

1 2 1 22 .

R

R R R RV IR R

ASSESS In the limit of long charging times, this circuit behaves like a voltage divider.

PROBLEMS

38. (a) The resistance between A and B is equivalent to two resistors of value R in series with the parallel combination

of resistors of values R and 2R. Thus, (2 )/( 2 ) 8 /3.ABR R R R R R R R (b)ACR is equivalent to just one

25.6 Chapter 25

resistor of value R in series with the parallel combination of R and 2R (since the resistor at point B carries no

current, i.e., its branch is an open circuit). Thus (2 )/3 5 /3.ACR R R R R R

39. INTERPRET The problem asks for the current in a resistor which is part of a more complex circuit.

DEVELOP The circuit in Fig. 25.30, with a battery connected across points A and B, is similar to the circuit

analyzed in Example 25.3. In this case, we have one 1.0 W resistor in parallel with two 1.0 W resistors in series:

||

||

1 1 1 3 2

1 1 1 2 3R

R

and the total resistance is 82tot 3 3

1 1 .R The total current (that through the battery) is

tot

6V 9tot 8 / 3 4

A 2.25 A.R

I

EVALUATE The voltage across the parallel combination is

|| tot ||

9 2 3A V

4 3 2V I R

which is the voltage across the vertical v 1 R resistor. Thus, the current through this resistor is then

v

3V/2v 1

1.5 A.V

RI �

ASSESS We have a total of 2.25 A of current flowing around the circuit. At the vertex of the triangular loop, it is

split intov 1.5 AI and

tot v 0.75A.I I I The voltage drop across the vertical resistor ( 1.5 V)V �

is the same

as that going through point C and the two 1.0-W resistors: (0.75 A)(1 1 ) 1.5V.V

40. The circuit diagram is like Fig. 25.9, and the voltage across the load (from Kirchhoff’s voltage law) is

int .LV IR Sinceint int/( ), /( )L L L LI R R V R R R (as for a voltage divider). With the given numerical

values,int int(1.5 V)(1 )/(1 ) 1.49 V, 1.36 V, and 0.750 V, for 0.01 , 0.1 , and 1 ,LV R R

respectively.

41. INTERPRET The circuit has two batteries connected in series. We apply Kirchhoff’s law to solve for the current

that flows through the discharged battery.

DEVELOP Terminals of like polarity are connected with jumpers of negligible resistance. Kirchhoff’s voltage

law gives

1 2 1 2 0IR IR

EVALUATE Solving the equation above, we obtain

1 2

1 2

12 V 9V 30 A

0.02 0.08 I

R R

ASSESS When you try to jump start a car, you connect positive to positive and negative to negative terminals. The

connection is what was illustrated in the figure.

42. The 50 combination must be capable of dissipating2 2/ (12 V) /50 2.8 WP R of power (when connected

across an ideal 12V battery), so combinations with at least six resistors (capable of dissipating 0.5 W each) must

be considered. In order to get the same total resistance as each individual resistor, n parallel branches of n resistors

in series are needed (or n branches in series of n resistors in parallel), making a total of 2

n resistors. The

smallest2

n greater than 6 is for 3.n (Alternatively, one could argue that the total current

is12 V/50 0.24 A, while the maximum current in each resistor element is 0.5 W/50 0.1 A, so at least

three equal branches in the circuit are needed.)

43. INTERPRET This problem is about rate of energy dissipation in the resistor.

DEVELOP For a short-circuited battery, the current isint/ ,I R so the dissipated power is

2

int

2

int .R

P I R


EVALUATE Substituting the values given in the problem, we find the rate of energy dissipation to be

2 2

int

(6 V)14.4 W

2.5 P

R

ASSESS With held fixed at 6 V, we see that the power dissipated is inversely proportional to the internal

resistanceint .R

44. The circuit breaker is activated if min120 V/ 20 A,I R or if min 6 .R The resistance of each light bulb is 2 2/ (120V) /100 W 144 ,R V P and n bulbs in parallel have resistance || / .R R n Therefore || minR R implies

144/6 24,n so more than 24 bulbs would blow the circuit.

45. INTERPRET The circuit in this problem contains a battery – the emf source, and two resistors in series.

DEVELOP Kirchhoff’s loop law gives 1 2 0,IR IR or 1 2/( ).I R R Therefore, the voltage across 2R is

2

2 2

1 2

RV IR

R R

This is the equation we shall use to solve for2 .R Once

2R is known, the power dissipated is simply equal

to2

2 2 2/ .P V R

EVALUATE (a) The equation above gives

1 22

2

(270 )(4.5 V)162

12 V 4.5 V

RVR

V

(b) The power dissipated is2 22

2

(4.5 V)

2 162125 mW.

V

RP

ASSESS For completeness, let’s calculate the power dissipated in1R and the total power. The voltage across

1R is

1 2 12V 4.5V 7.5V,V V and the power dissipated in1R is

2 22

1

(7.5 V)

1 270 208 mW.

V

RP The total power

dissipated in the circuit is

2 2

tot

1 2

(12V)333 mW

270 162 P

R R

which indeed is equal to the sum of1P and

2 .P

46. Resistors of1 k and10 k connected in parallel have a combined resistance of (10/11) k (Equation 25.3b), so

the altered circuit is a voltage divider (Figure 25.4) with1 1 kR and

2 (10/11) k .R Equation 25.2b gives

2 2 1 2/( ) (160V)(10/11)/(1 10/11) 76.2V.V R R R

47. INTERPRET The circuit in this problem contains a battery—the emf source, and three resistors. We want to

analyze the voltage across the one which is a variable resistor.

DEVELOP The resistors in parallel have an equivalent resistance of|| 1 1/( ).R RR R R The other R, and

|| ,R is a

voltage divider in series with voltage .

EVALUATE (a) Using Equation 25.2, we find the voltage across1R to be

|| 1

||

|| 12

R RV

R R R R

(b) The sketch of ||V as a function of

1R is shown on the right.

(c) If1 0,R then

|| 0.V On the other hand, if1 ,R then

|| /2V (the value when1R is removed). If

1 10 , R R

|| (10/21) .V

25.8 Chapter 25

ASSESS The limit 1 0R corresponds to the situation where the second resistor is shorted out. The limit 1R is

an open circuit with no current going through it.

48. (a) With reference to Problem 49’s solution, the resistance of the three parallel resistors is (12/11) , so the current

supplied by the battery is1 ||/( ) (6V)/(23/11) 2.87A.I R R (b) The voltage drop across the resistors in

parallel is || 1 || ,V IR IR and the current through the 6 resistor is 6 ||/6 .I V Thus, 6 (2.87A)(2/11)I

522 mA.

49. INTERPRET This problem asks for the power dissipated in a resistor which is part of a more complex circuit.

DEVELOP The three resistors in parallel have an effective resistance of

||

||

1 1 1 1 11 12

2 4 6 12 11R

R

The equivalent resistance of the circuit is 2312tot 1 11 11

1 .R R R �

Equation 25.2 gives the voltage across

them as

||

tot

(6V)(12 /11) 72V

23 /11 23

RV

R

�

EVALUATE Thus, the power dissipated in the 4 resistor is

2 2||

4

4

(72V/23)2.45 W

4

VP

R

ASSESS With held fixed at 6 V, we see that the power dissipated is inversely proportional to the resistance.

50. If the ammeter has zero resistance, the potential difference across it is zero, or nodes C and D are at equal

potentials. If I is the current through the battery, 1

2 I must go through each of the 2 -resistors connected at node A

(because 1

2(2 ) ).A C A DV V I V V At node B, the 2 -resistor inputs twice the current of the 4 -resistor ,

or 2

3I and 1

3 I respectively (because 2 1

3 3(2 ) (4 ) ).C B D BV V I I V V Therefore 1

6 I must go through the

ammeter from D to C, as required by Kirchhoff’s current law. To find the value of I, note that the upper pair of

resistors are effectively in parallel ( )C DV V as is the lower pair. The effective resistance between A and B is

74eff 3 3

2 2 /(2 2) 2 4 /(2 4) 1 ( ) ( ) .R Thuseff/ ,I V R and the ammeter current is

7 31 1

6 6 3 7 (6 V)/( ) ( ) A 0.429 A.I

51. INTERPRET The problem asks for the equivalent resistance between two points in a complex circuit.

DEVELOP The equivalent resistance is determined by the current which would flow through a pure emf if it were

connected between A and B: / .ABR I Since I is but one of six branch currents, the direct solution of Kirchhoff’s

circuit laws is tedious (6 6 determinants). However, because of the special values of the resistors in Fig. 25.34,

a symmetry argument greatly simplifies the calculation.

The equality of the resistors on opposite sides of the square implies that the potential difference between A and C

equals that between D and B, i.e.,

A C D BV V V V


Equivalently, .A D C BV V V V Since 1 ,A CV V I R 2 (2 ),A DV V I R etc., the symmetry argument requires that

both R-resistors on the perimeter carry the same current, 1,I and both 2R-resistors carry current 2 .I Then

Kirchhoff’s current law implies that the current through E is 1 2 ,I I and the current through the central resistor

is 1 2I I (as added to Figure 25.34). Now there are only two independent branch currents, which can be found

from Kirchhoff’s voltage law, applied, for example,

1 2

1 1 2 2

(2 ) 0 (loop )

( ) (2 ) 0 (loop )

I R I R ACBA

I R I I R I R ACDA

These equations may be rewritten as

1 2

1 2

2

2 3 0

I IR

I I

with solution1 3 /7I R and

2 2 /7 .I R

EVALUATE The sum of the two currents gives1 2 5 /7 ,I I I R which leads to

7

5AB

RR

I

ASSESS The configuration of resistors in Figure 25.34 is called a Wheatstone bridge.

52. The general solution of the two loop equations and one node equation given in Example 25.4 can be found using

determinants (or1I and

2I can be found in terms of3 ,I as in Example 25.4). The equations and the solution are:

1 1 3 3 1

2 2 3 3 2

1 2 3

0 (loop 1),

0 (loop 2),

0 (node A);

I R I R

I R I R

I I I

1 3

2 3 1 2 2 3 3 1

1 3

1 2 3 2 3

1 2 2 3

1 1 3

1 3 2 1 3

2 2 3

1 1

2 1 1 23 2 2

0

0 ,

1 1 1

0( )1

,

0 1 1

( )10 ,

1 0 1

01

0 .

1 1 0

R R

R R R R R R R R

RR R R

I R R

R RR R R

I R

RR R

I R

With the particular values of emf’s and resistors in this problem, we find currents of1 [(4 1)6 1( 9)] A/14I

2.79A, 2 [1 6 (2 1)( 9)] A/14 2.36A,I and3 [4 6 2( 9)] A/14 0.429A.I Or, one could retrace the

reasoning of Example 25.4, with2 9 V replacing the original value in loop 2. Then, everything is the same until

the equation3 39 2(6 3 ) 0,I I or 3 31

3 27 2 7( ) A, (6 3 ) A (33/14) A,I I and

1 2 3 (39/14) A.I I I

53. INTERPRET This problem asks for the current in a resistor which is part of a more complex circuit. The solution

requires analyzing a circuit with series and parallel components.

25.10 Chapter 25

DEVELOP Let us choose the positive sense for each of the three branch currents in Figure 25.35 as upward

through their respective emf’s (at least one must be negative, of course), and consider the two smaller loops shown.

Kirchhoff’s circuit laws give:

1 2 3 1 2

3 4 3 2

0 (top node)

( ) (left loop)

(right loop)

a b c

a b

b c

I I I

R R I R I

R I R I

Solve for aI and cI from the loop equations and substitute into the node equation:

1 2 3 3 2 3

1 2 4

( ) ( )0b b

b

R I R II

R R R

The current in3R is .bI

EVALUATE Solving for ,bI we find

1 2 4 3 2 1 2

3 4 1 2 3 4

( ) ( )( ) (6 V 1.5 V)(820 ) (4.5 V 1.5 V)(420 )

( )( ) (560 )(820 ) (420 )(1380 )

4.77 mA

b

R R RI

R R R R R R

A negative current is downward through2 in Figure 25.35.

ASSESS Substituting 4.77 mAbI into the equations above, we find 4.35 mAaI and 0.42 mA.cI One can

readily verify that the solutions satisfy all the equations.

54. Let us choose the positive sense for each of the three branch currents in Figure 25.35 as upward through their

respective emf’s (at least one must be negative, of course), and consider the two smaller loops shown. Kirchhoff’s

circuit laws give:

1 2 3 1 2

3 4 3 2

0 (top node)

( ) (left loop)

(right loop)

a b c

a b

b c

I I I

R R I R I

R I R I

Solve foraI and

cI from the loop equations and substitute into the node equation:

1 2 3 3 2 3

1 2 4

( ) ( )0b b

b

R I R II

R R R

Then

4 2 1 1 2 2 3

3 4 1 2 3 4

( ) ( )( )

( )( )b

R R RI

R R R R R R

with similar expressions foraI and .cI One can see that

bI is positive if4 2 1 1 2 2 3( ) ( )( ) 0,R R R or

4 1 1 2 3

2

1 2 4

( ) (820 )(6 V) (420 )(4.5V)5.49 V

(1240 )

R R R

R R R

55. INTERPRET In this problem the voltage across a given resistor is measured using a voltmeter which behaves like

a resistance.

DEVELOP With a meter of resistancemR connected as indicated, the circuit reduces to two pairs of parallel

resistors in series. The total resistance is

tot

(30 k ) 40 k

30 k 2

m

m

RR

R

The voltage reading is


tot(30 k )

30 k

m

m m m

m

R IV R I

R

wheretot tot(100V)/I R (the expression for

mV follows from Equation 25.2, with1R and

2R as the above pairs, or

frommI as a fraction of

tot ).I

EVALUATE For the three voltmeters specified,tot 2.58 mA,I 2.14 mA, and 2.00 mA,

while 48.4 V, 57.3 V,mV and 59.9 V, respectively. (After checking the calculations, round off to two figures.)

ASSESS Of course, 60 V is the ideal voltmeter reading. This reading corresponds to an ideal voltmeter that has

infinite resistance.

56. (a) An ideal voltmeter has infinite resistance, so AB is still an open circuit (as shown on Figure 25.37) when such a

voltmeter is connected. The meter reads the voltage across the 20 k resistor (part of a voltage divider), or

(30V)20/(20 10) 20V (see Equation 25.2a or Equation 25.2b). (b) An ideal ammeter has zero resistance, and

thus measures the current through the points A and B when short-circuited (i.e., no current flows through the

20 k resistor). In Figure 25.37, this would be 30V/10 3 mA.ABI (Such a connection does not measure the

current in the original circuit, since an ammeter should be connected in series with the current to be measured.)

57. INTERPRET In this problem an ammeter is used to measure the current in a circuit. The ammeter is connected in

series with the resistor.

DEVELOP The internal resistance of an ideal battery is zero, so the resistor has a value of

12V)/(1A) 12 ./ (R I With the ammeter in place, the current would be

m

VI

R R

EVALUATE (a) Substituting the values given in the problem statement, we find the current to be

12 V0.992 A

12.1 m

VI

R R

(b) If the resistance of the ammeter were neglected in the calculation, one would obtain

12 V12.1

0.992 AR

ASSESS This value R differs from the actual value of 12 R by 0.83%. However, subtraction ofmR is a

correction that could be included easily.

58. (a) Equation 25.6 and the given circuit characteristics imply that the time constant is 5.0 ms.RC Therefore,

in three time constants, or 15 ms, the capacitor is charged to3

1 e of the battery voltage. (b) Evidently, /C R

5 ms/22k 0.227 F.

59. INTERPRET This problem is about a discharging capacitor in an RC circuit, and we want to find the time to reach

a given voltage.

DEVELOP A capacitor discharging through a resistor is described by exponential decay, with time constant RC

(Equation 25.8):

/( ) (0)

t RCV t V e

The energy in the capacitor is

2 2 2 / 2 /1 1( ) ( ) (0) (0)

2 2

t RC t RC

C CU t CV t CV e U e

EVALUATE (a) The time it takes to reach ( ) 5 VV t is

25.12 Chapter 25

(0)ln (500 k )(1 F) ln 2 347 ms

( )

Vt RC

V t

(b) Similarly, the time it takes for the energy to decrease to half its initial value is

(0)1 1ln (500 k )(1 F) ln 2 173 ms

2 ( ) 2

C

C

Ut RC

U t

ASSESS The time constant in this problem is (500 k )(1 F) 500 ms.RC When ,t RC the voltage is

reduced by a factor of ,e or1( )/ (0) 0.368.V t V e Therefore, it takes less than one time constant for the value

of ( )V t to be halved from its initial value.

60. (a) The effective resistance of a circuit that draws 1.2 A from a constant 35 V supply is 35 V/1.2 A 29.2 .

(b) To keep the voltage within the prescribed range for the discharging capacitor (Equation 25.8), the time constant

must satisfy/

0/ 34/35,t RCV V e or /ln(35/34).RC t For 1/60 st and 29.2 ,R one finds 19.7 mF.C

61. INTERPRET This problem involves energy dissipation in an RC circuit. The object of interest is the capacitance.

DEVELOP A capacitor discharging through a resistor is described by exponential decay, with time constant RC

(Equation 25.8):

/( ) (0) t RCV t V e

The energy in the capacitor is

2 2 2 / 2 /1 1( ) ( ) (0) (0)

2 2

t RC t RC

C CU t CV t CV e U e

EVALUATE If 2 J is dissipated in time t, the energy stored in the capacitor drops from (0) 5 JCU to ( ) 3 JCU t

(assuming there are no losses due to radiation, etc.). From the equation above, we find the capacitance to be

2 2(8.6 ms)3.37 F

ln( (0)/ ( )) (10 k ) ln(5 J/3 J)C C

tC

R U U t

ASSESS In this problem the time constant is (10 k )(3.37 F) 33.7 ms.RC Therefore, at 8.6 ms (about

0.255 ),RC the energy decreases by a factor2(0.255)

0.6.e This is precisely what we found.

62. When current stops flowing (at ),t the potential difference across the capacitors is equal, but the total charge is

just the initial charge.

Thus,1 2( ) ( ) ( ),V V V and

2 1 2(0) ( ) ( ).Q Q Q Since1 1 2 2, ( ) ( )Q CV C V C V

2 2 (0)C V or 22 2 1 2 23

( ) (0) /( ) (0).V V C C C V The energy stored in the capacitors is

initially21

2 22(0) (0)U C V

21

2(2 F)(150 V) 22.5 mJ, and finally

211 22

( ) ( ) ( )U C C V 21

2(3 F)(100 V) 15.0 mJ. The difference, | | 7.50 mJ,U is dissipated in the resistor (except for a negligible

amount of radiated energy).

63. INTERPRET This problem is about the long-term and short-term values of current and voltage of an RC circuit.

DEVELOP In addition to the explanation in Example 25.7, we note that when the switch is in the closed position,

Kirchhoff’s voltage law applied to the loop containing both resistors yields1 1 2 2 ,I R I R and to the loop

containing just R2 and C,2 2 .CV I R

EVALUATE (a) If the switch is closed at 0,t Example 25.7 shows that (0) 0,CV 2 (0) 0,I and

1

1

100 V(0) 25 mA

4 kI

R

(b) After a long time, ,t Example 25.7 also shows that

1 2

1 2

100 V( ) ( ) 10 mA

10 kI I

R R

and2 2( ) ( ) (10mA)(6k ) 60V.CV I R

(c) Under the conditions stated, the fully charged capacitor ( 60 V)CV simply discharges through R2. (R1 is in

an open-circuit branch, so1 0I for the entire discharging process.) The initial discharging current is


2

2

60 V10 mA

6 k

CVI

R

(d) After a very long time,2I and VC decay exponentially to zero.

ASSESS We deduced the short-term and long-term behavior of the RC circuit without having to solve a

complicated differential equation. A long time after the circuit has been closed, the capacitor becomes fully

charged with no current flowing to it. When the circuit is reopened, the capacitor starts to discharge and eventually

loses all its stored energy.

64. (a) An uncharged capacitor acts instantaneously like a short circuit (see Example 25.7), so initially ( 0)t all of the

current from the battery goes through R1 and C1, and none goes through R2 and R3. Thus,1(0) / ,I R and

2 3(0) (0) 0.I I (b) A fully charged capacitor acts like an open circuit (when responding to a constant applied

emf), so after a long time ( ),t all of the current goes through R1 and R2 in series, and none goes through R3.

Thus1 2( ) ( ) /2 ,I I R and

3 ( ) 0.I (c) One can easily guess that I1 and I2 respectively decrease and

increase monotonically from their initial to their final values, and that3I first increases from, and then decreases to

zero. (One can use the loop and node equations to solve for the currents. They turn out to be linear combinations of

two decaying exponentials with different time constants.)

65. INTERPRET In this problem we are asked to find the voltage and internal resistance of a battery using the

measured voltage values of two voltmeters.

DEVELOP The internal resistance of the battery ( )iR and the resistance of the voltmeter ( )mR are in series with the

battery’s emf, so the current is /( ).i mI R R The potential drop across the meter (its reading) is

m

m m

i m

RV IR

R R

From the given data, we can write

(1 k ) (1.5 k )4.36 V and 4.41 V

1 k 1.5 ki iR R

or 1k (1k /4.36V)iR and 1.5k (1.5 k /4.41 V).iR

EVALUATE Solving the simultaneous equations for and Ri gives

11.5 k 1 k

(1.5 k 1 k ) 4.51 V4.41 V 4.36 V

and (4.51V)(1 k /4.36V) 1 k 35.2 .iR

ASSESS An ideal voltmeter has infinite resistance. Thus, when we let ,mR its reading approaches the battery

voltage .

66. Using// 45%CV t RCe

gives /ln[1/(1 45%)] 140 ms/ln(1/0.55) 234 ms,RC t so

234 ms/20 FR 11.7 k .

67. INTERPRET The electric field at the node increases due to charge accumulation and eventually reaches the

breakdown field strength.

DEVELOP The charge on the node (whether positive or negative) accumulates at a rate

of / 1 A 1 C/s,I dq dt so | ( )| (1 A)q t t (where we assume that (0) 0).q If the node is treated

approximately as an isolated sphere, the electric field strength at its surface is

2 2

| |k q kItE

r r

Electric breakdown occurs when 3 MV/m .bE E

25.14 Chapter 25

EVALUATE The time when the breakdown happens is

2 2

9

(3 MV/m)(0.5 mm)= 83.3 s

(9 10 m/F)(1 A)

bE rt

kI

ASSESS This problem shows that Kirchhoff’s node law must hold, or else there would be a charge buildup at the

node which quickly leads to an electric breakdown.

68. The current for the connection described is int/( ),I R R so the power dissipated in the resistor is2P I R

2 2

int/( ) .R R R This inherently positive function of R is zero for 0R and ,R so it must have a maximum

at some intermediate value of R. The condition for the maximum is / 0,dP dR which implies 2

int int( ) 2 ( ) 0,R R R R R or int .R R

69. INTERPRET This problem is about energy stored in the capacitor that’s part of the RC circuit. We are asked to

show that it only stores half the energy the battery supplies.

DEVELOP The power supplied by the battery charging a capacitor (initially uncharged) in an RC circuit is

2/t RCP I e

R

where the current is given by Equation 25.5:/( / .) t RCI R e The total energy supplied is

2/ 2 0 2

battery 00

( )t RCU Pdt e dt C e e C

R

EVALUATE The energy stored in the fully charged capacitor is

2 2

battery

1 1 1( ) ( )

2 2 2CU CV C U

Thus, we see that the energy stored in the capacitor is only half of that supplied by the battery.

ASSESS The other half of the energy supplied by the battery is dissipated in the resistor:

22 2 / 2

0 0

1

2

t RC

RU I R dt e dt CR

70. (a) There are two parallel pairs 112

( )R in series, so 1 11 1 12 2

.ABR R R R (b) Here, there are two series pairs1(2 )R

in parallel, so1 1 1 1 1(2 )(2 )/(2 2 ) .ABR R R R R R (c) Symmetry requires that the current divides equally on the

right and left sides, so points C and D are at the same potential. Thus, no current flows through2 ,R and the circuit

is equivalent to (b). (Note that the reasoning in parts (a) and (b) is easily generalized to resistances of different

values; the generalization in part (c) requires the equality of ratios of resistances which are mirror images in the

plane of symmetry.)

71. INTERPRET This problem is about finding the voltage and internal resistance of a battery. We are given the

current values when the battery is connected to resistors of known resistance.

DEVELOP The circuit diagram is like Fig. 25.10, and Kirchhoff’s voltage law is

int 0LIR IR

For the two cases given, this may be written as


int

int

(26 mA) (26 mA)(50 ) 1.3 V

(43 mA) (43 mA)(22 ) 0.946 V

R

R

EVALUATE Solving for and int,R we find

int

1.3 V 0.946 V20.8 and (26 mA)(50 20.8 ) 1.84 V

43 mA 26 mAR

ASSESS The terminal voltage of the battery is int 1.84 V (20.8 ),V R II which is lower than . When

the battery is connected to a resistor of resistance R, the current in the circuit is int/( ).I R R

72. The loop law for a battery charging a capacitor through a resistor is 0.CIR V Differentiate this and use

Equation 25.4 to obtain ( / ) ( )/ ( / ) ( / ) / .CdV dt d IR dt R dI dt R I RC I C ( ( )I t is given in Equation 25.5.)

For an initially uncharged capacitor,0 /I R (an uncharged capacitor acts like a short circuit), so the initial rate of

increase of the capacitor’s voltage is 0( / ) / / .CdV dt RC If the capacitor were to charge steadily at this rate

(i.e., if the voltage were a linear function of time,0( / ) / ),CdV dt t t the voltage would reach its final value

( ( ) )CV E in just one time constant (i.e., ( ) /CV t for ).t

73. INTERPRET Our circuit consists of an array of resistors of infinite extent, and we’re asked to find the equivalent

resistance.

DEVELOP Since the circuit line is infinite, the addition or deletion of one more element leaves the equivalent

resistance unchanged. This can be represented diagrammatically as

The right-hand picture represents R in series with the parallel combination R and Req. Thus,

eq

eq

eq

RRR R

R R

EVALUATE Solving foreq ,R one finds

2 2

eq eq 0,R RR R or

eq (1 5) /2 1.62R R R

Note that only the positive root is physically meaningful for a resistance.

ASSESS Let’s see how this limiting value is reached. With only two resistors, the equivalent resistance is

1 2 .R R R R Next, consider four resistors (the four on the left of Fig. 25.41). The equivalent resistance is

2

1 21.67

1/ 1/2 3

RR R R R

R R

Continuing the same line of reasoning leads to the quadratic equation which we solved to obtain

eq (1 5) /2 1.62 .R R R

74. (a). For the circuit considered, the voltage across the capacitor asymptotically approaches the battery voltage after

a long time (compared to the time constant). In Fig. 25.42, this is about 9 V. (b) The time constant is the time it

takes the capacitor voltage to reach1

1 63.2%e of its asymptotic value, or 5.69 V in this case. From the graph,

1.5 ms. . (c) The time constant is RC, so 1.5 ms/4700 0.319 F.C

75. INTERPRET This problem asks for the current through an emf source which is part of a more complex circuit.

The solution requires analyzing a circuit with series and parallel components.

25.16 Chapter 25

DEVELOP Let us choose the positive sense for each of the three branch currents in Fig. 25.43 as upward through

their respective emf’s (at least one must be negative, of course), and consider the right loop and the big loop.

Kirchhoff’s circuit laws give:

1 3

3 2

0 (top node)

(2 ) (2 ) (big loop)

(2 ) (right loop)

a b c

a b

b c

I I I

I R I R

I R I R

Solve for aI and cI from the loop equations and substitute into the node equation:

1 3 3 2( ) 2 2 ( )0

2

b b

b

RI RII

R R

The current in3 is .bI

EVALUATE Solving For ,bI we find

3 2 1(3 2 ) (60 mV 90 mV 75 mV) 8.75 nA

8 8(1.5 M )bI

R

The negative sign means that the direction ofbI is opposite of what was shown in the diagram.

ASSESS The negative sign inbI can be easily understood by noting that

3 is smaller than1 and

2 .

76. The relation betweenbI and the circuit emf’s and resistances, given in the solution to Problem 75, can be solved for

3b in Fig. 25.43, resulting in 1

3(8 2 ).b b c aRI For 40 nAbI and the rest of the circuit elements the

same, 13 3

(8 1.5 M 40 nA 90 mV 75 mV) 215 mV.

77. INTERPRET We represent a “leaky” capacitor with an equivalent circuit diagram, and determine the time constant

for this circuit. We will also show that the time constant does not depend on the geometry of the capacitor, but only

on its material properties.

DEVELOP For part (a), see the figure later. For part (b), we will use the resistance of the insulation material,

,d

AR where d is the thickness of the material and A is the area of the plates. We will also use the equation for

parallel-plate capacitance, 0 ,A

dC where 5.6 is the dielectric constant of glass. The time constant we are

seeking is .RC

EVALUATE

(b) d

ARC 0

A

d 0 . This is independent of the geometrical terms d and A, and depends only on the

material properties.0 595 s.

ASSESS This is actually pretty good for a capacitor. Materials with high resistivity and high dielectric constant

will make capacitors with longer leakage time constants.

78. INTERPRET We use Kirchhoff’s laws to write a system of equations for the circuit shown in Figure 25.24a, and

from the resulting equations we will determine the time constant of the circuit.

DEVELOP We first sketch our loops and nodes, as shown in the figure below. We have 3 unknowns, so we will

need 3 equations. Node A and node B give us duplicate information, so we will use only one of the two: our

equations must then come from loops 1 and 2, and node A.

Node A gives us1 2 3 0.I I I Loop 1 gives us

1 1 2 2 0,I R I R and loop 2 gives us2 2 0.CI R V


The voltage across the capacitor is given by ,Q

C CV and

3 .dQ

dtI We will eliminate

1I and2I in our system of

equations, then rearrange the results into the form of Equation 25.4, from which we can easily identify the time

constant.

EVALUATE From node A,1 2 3.I I I Substitute this into the equation for loop 1:

3 1

2 3 1 2 2 2

1 2

( ) 0I R

I I R I R IR R

Now we substitute into the equation for loop 2:

3 1

2 2 3 1 2 1 2

1 2

0 ( )Q

C C

I RR V R I R R R R

R R

We take the time derivative of this last equation:

3 31 2 1 21 2 1 2 3

( ) ( )dI dIR R R RdQR R R R I

dt dt C dt C

Rearrange this slightly to obtain

1 2

1 2

3 3

R R C

R R

dI I

dt

Now here’s a trick: rather than solve this equation, we note that it’s the same equation as 25.4, with a different

cluster of constants in the position of the square brackets. In the solution to 25.4, we found that ,RC so here the

time constant must be

1 2

1 2

R R C

R R

ASSESS This trick of putting the equation in a previously solved form can save us a lot of effort. Note that we can

only do it because all the terms in the square brackets are constants: if there was a term involving 3I in those

brackets, then it would be a different equation and we couldn’t use the same solution.

79. INTERPRET We use Kirchhoff’s laws to write a system of equations for the circuit shown in Figure 25.39, and

from the resulting equations we will determine the current through resistor2 .R We will need 4 equations.

DEVELOP First we make a diagram of the circuit, as shown in the figure later. Nodes A and B give us duplicate

information, so we will use only node A, along with the three loops.

Node A: 1 2 3 4 0I I I I

Loop 1: 1 2 0I R I R

Loop 2: 3

2 0Q

CI R

Loop 3: 3 4

4 0Q Q

C CI R

We will solve for2I as a function of time.

EVALUATE

Node A:1 2 3 4.I I I I

Loop 1:

25.18 Chapter 25

2 3 4 2 4 2 3

34 2

( ) 0 2

2

I I I R I R I I IR

dIdI dI

dt dt dt

Loop 2:

2

3 3 32 2

2 2

Q I dIdI d II R R RC

C C dt dt dt

Loop 3:

3 34 4 2 22 3

2

2 2 2 22 2

2

2 222 2 2

10 2 2 0

1 22 0

4 2

( ) ( )

I dII dI dI dIR R I I R

C C dt dt C R dt dt

dI dI dI d II RC

dt RC R RC dt dt dt

d I dII

RC dtdt RC R RC

This is a second-order linear differential equation with constant coefficients. We can solve the homogenous

equation using the characteristic equation:

2

2

2 2

4 20

( )

1 4 16 8 2 22

2 ( ) ( )

2 2(1 2), (1 2)

RC RC

RC RC RCRC RC

RC RC

So the solution to the homogenous equation is

2 2(1 2 ) (1 2 )

2 1 2( )t t

RC RCI t A e A e

and the solution to the inhomogenous equation is

2 2(1 2) (1 2 )

2 1 2( )2

t tRC RCI t A e A e

R

Now we need the initial condition on2I and 2 .

dI

dtSince both capacitors are initially uncharged and essentially short

circuits, 3

01 3(0) (0) |dQ

dt RI I and the initial voltage across the central capacitor is given by

/1( ).t RC

CV e

This voltage creates a current through2R of

/

2 000

(1 )t RCC

ttt

VI e

R R

so

2 (0) 0I and/2

2

0 0

1 t RC

t t

dIe

dt R RC R C

Applying the boundary condition2 (0) 0I to the solution obtained previously gives us

1 20 ,2

A AR

and

applying

2

2

0t

dI

dt R C

gives us 1 22

2 21 2 1 2A A

RC RCR C

from which we can determine that1 2 .

4A A

R

So, our final solution is


2 2(1 2 ) (1 2 )

2 4( ) 2 RC RC

t t

RI t e e

ASSESS This was a difficult problem, but the technique used to set it up is the same as for an easier one:

Kirchhoff’s laws.

80. INTERPRET We must convert a battery energy rating (in watt-hours) at a given voltage to a charge rating of amp-

hours. We will use .P IV

DEVELOP The battery is specified as 50 watt-hours, which means that it can supply 50WP for 1 hour. We will

use P IV to find I, knowing that the voltage is 6V.V

EVALUATE 8.33 AP

VP IV I

ASSESS This is an 8.33 amp-hour battery, which is sufficient for our requirements.

81. INTERPRET We need to create a resistance value using three different resistors. We can combine the resistors

in series and/or parallel. We will find the appropriate combination by making some educated guesses and seeing

what works.

DEVELOP The three resistors are1 23.30 k , 4.70 k ,R R and

3 1.50 k .R Resistors add in series, and in

parallel their reciprocals add.

One guess would be to place1R and

2R in parallel so that the combination was smaller than either, then adding3R in

series to this combination. Try it!

EVALUATE 1 2

1 1 1 1.94k ,p

pR R RR and

3 3.44 kpR R which is just what we wanted.

ASSESS Another reasonable guess might be to put1R and

3R in series, then put2R in parallel across that

combination: but that method gives us only 2.37 k .

82. INTERPRET We use an equation for the time constant of a leaky capacitor to determine if the capacitor will hold

its energy for a required amount of time. We will use the equation for the voltage on a capacitor, and the equation

for energy stored on a capacitor as a function of voltage.

DEVELOP We are given a time constant0, where

1610 m and 2.6. The energy stored in a

capacitor is 21

2,U CV and the voltage is

/

0 .t

V V e We want for the capacitor to still hold half its energy after

425 hours 9 10 s.t

EVALUATE The initial energy stored in the capacitor is2102.U CV After some time t, the energy is

/ 2 2 2 / 2 /

0 0 0

1 1( ) [ ]

2 2

t t tU t C V e CV e U e

so in order to store half (or more) of the initial energy,2 / 1

2.te

40

0

2 1ln ln(2) 7.98 10 s 22.2 hours

2 2

tt

ASSESS The time by which half the energy leaks away is too short—the capacitor does not work.

26.1

MAGNETISM: FORCE AND FIELD

EXERCISES

Section 26.2 Magnetic Force and Field

17. INTERPRET This problem is about the magnetic force exerted on a moving electron.

DEVELOP The magnetic force on a charge q moving with velocity vr

is given by Equation 26.1: .BF qv B r rr

The

magnitude ofBFr

is

| | | | | | sinB BF F qv B q vB r rr

EVALUATE (a) The magnetic field is a minimum when sin 1 (the magnetic field perpendicular to the velocity).

Thus,

153

min 19 7

5.4 10 N1.61 10 T 16.1 G

| | (1.6 10 C)(2.1 10 m/s)

BFB

q v

(b) For 45 , the magnetic field is

15

min19 7

5.4 10 N2 22.7 G

| | sin (1.6 10 C)(2.1 10 m/s)sin 45

BFB B

q v

ASSESS The magnetic force on the electron is very tiny. The magnetic field required to produce this force can be

compared to the Earth’s magnetic field, which is about 1 G.

18. (a) If the magnetic force is the only one of significance acting in this problem, then sin .F ma ev B Thus, 31 15 2 19 5

v / sin (9.11 10 kg)(6 10 m/s )/(1.6 10 C)(0.1 T)sin 90 3.42 10 m/s.ma eB (b) Since

F v Br rrz is perpendicular to ,v

rthe magnetic force on a charged particle changes its direction, but not its speed.

19. INTERPRET In this problem we are asked to find the magnetic force exerted on a moving proton.

DEVELOP The magnetic force on a charge q moving with velocity vr

is given by Equation 26.1: .BF qv B r rr

The magnitude of BFr

is

| | | | | | sinB BF F qv B q vB r rr

The charge of the proton is191.6 10 C.q

EVALUATE (a) When 90 , the magnitude of the magnetic force is

19 5 14sin 90 (1.6 10 C)(2.5 10 m/s)(0.5 T) 2.0 10 NBF qvB

(b) When 30 , the force is

19 5 14sin 30 (1.6 10 C)(2.5 10 m/s)(0.5 T)sin 30 1.0 10 NBF qvB

(c) When 0 , the force is sin 0 0.BF qvB

ASSESS The magnetic force is a maximum,max( | | )BF q vB when 90 and a minimum

,min( 0)BF when 0 .

20. An electron, moving perpendicularly to the Earth’s magnetic field near the surface, experiences a maximum

magnetic force of19 4 3 19 31v 2 / (1.6 10 C)(10 T) 2(10 eV)(1.6 10 J/eV)/(9.11 10 kg)e B eB K m . .

163 10 N.

The weight of an electron near the Earth’s surface is31 2 30(9.11 10 kg)(9.8 m/s ) 9 10 N,mg .

a factor of nearly14

3 10 times smaller.

21. INTERPRET This problem is about the speed of a given charge if it is to pass through the velocity selector

undeflected.

26

26.2 Chapter 26

DEVELOP In the presence of both electric and magnetic fields, the force on a moving charge is (see Equation 26.2):

( )E BF F F q E v B r r r r rr

The condition for a charged particle to pass undeflected through the velocity selector is ,E BF F r r

or / .v E B

EVALUATE Substituting the values given in the problem statement, we obtain

24 kN/C400 km/s

0.06 T

Ev

B

ASSESS Only particles with this speed would pass undeflected through the mutually perpendicular fields; at any

other speed, particles would be deflected.

Section 26.3 Charged Particles in Magnetic Fields

22. From Equation 26.3, the radius of the orbit is27 19 2/ (1.67 10 kg)(15 km/s)/(1.6 10 C)(4 10 T)r mv eB

3.91 mm. (SI units and data for the proton are summarized in the appendices and inside front cover.)

23. INTERPRET This problem is about an electron undergoing circular motion in a uniform magnetic field. We want

to know its period, or the time it takes to complete one revolution.

DEVELOP Using Equation 26.3, the radius of the circular motion is /| | .r mv e B Therefore, the period of the

motion is

2 2 2

| | | |

r mv mT

v v e B e B

EVALUATE Substituting the values given in the problem statement, we find the period to be

31

19 4

2 2 (9.11 10 kg)358 ns

| | (1.6 10 C)(10 T)

mT

e B

ASSESS The period is independent of the electron’s speed and orbital radius. However, it is inversely proportional

to the magnetic field strength.

24. If this is electromagnetic radiation at the electron’s cyclotron frequency, Equation 26.4 implies a field strength of 31 19 32 ( / ) 2 (42 MHz)(9.11 10 kg/1.6 10 C) 1.50 10 T 15.0 G.B f m e

25. INTERPRET In this problem electrons and protons have the same kinetic energy, and they are undergoing circular

motion in the same magnetic field. We want to compare the radii of their orbits.

DEVELOP From Equation 26.3, the radius of the circular motion is / .r mv qB For a non-relativistic particle,

21

2,K mv or 2 / .v K m Therefore,

2 2mv m K Kmr

qB qB m qB

EVALUATE For protons and electrons having the same kinetic energy and in the same magnetic field, the ratio of

their radii is

1836 43p p

e e

r m

r m

ASSESS The fact that1/ 2r mz implies that heavier particles are more difficult to bend.

26. (a) A cyclotron frequency of 2.4 GHz for electrons implies a magnetic field strength of 2 ( / )B f m e 31 19

2 (2.4 GHz)(9.11 10 kg/1.6 10 C) 85.9 mT (see Equation 26.4). (b) The kinetic energy of an electron,

with the maximum orbital radius allowed for this magnetron tube (half the diameter), in the field found in part (a),

is2 19 2 31 16

( ) /2 (1.25 mm 1.6 10 C 85.9 mT) /(2 9.11 10 kg) (1.62 10 J)/K reB m

19(1.6 10 J/eV) 1.01 keV

(see Problem 49). The same calculation in atomic units, explained in the solution to

Exercise 25, is3 3 2 1

(1.25 10 300 85.9 10 ) (2 0.511) MeV. The electron’s kinetic energy could also be

expressed in terms of the cyclotron frequency directly,2 2

(2 ) /2 2 ( ) ,K frm m m rf with the same result.

27. INTERPRET This problem is about two protons undergoing circular motion and colliding head-on.

DEVELOP In an elastic head-on collision between particles of equal mass, the particles exchange velocities

Magnetism: Force and Field 26.3

1 2 2 1( and ).f i f iv v v v Moving in a plane perpendicular to the magnetic field, each proton describes a different

circle of radius / ,r mv eB with period (which is independent of r and v) of 2 / .T m eB

EVALUATE Substituting the values given, we obtain

27

19 2

2 2 (1.67 10 kg)1.31 s

(1.6 10 C)(5 10 T)

mT

eB

After one period, each proton would be back at the site of the collision, and could collide again.

ASSESS In solving the problem, we have ignored Coulomb repulsion between the two protons.

Section 26.4 The Magnetic Force on a Current

28. The force on a straight current-carrying wire in a uniform magnetic field is (Equation 26.5) .F IL B r r r

Thus,

sin (15 A)(0.5 m)(0.05 T)sin 90 0.375 N.F ILB (The direction is given by the right-hand rule.)

29. INTERPRET In this problem we are asked about the magnetic field strength of a straight current-carrying wire.

DEVELOP Equation 26.5 gives the magnetic force on a straight current carrying wire in a uniform magnetic field,

.F IL B r r r

The magnitude of the force is sin .F ILB

EVALUATE (a) From the magnitude of Fr

and the given data, we find the magnetic field strength to be

0.31 N/m48.9 mT

sin (15 A)sin 25

FB

IL

(b) By placing the wire perpendicular to the field (sin 1) a maximum force per unit length of

(15 A)(48.9 mT) 0.734 N/mF

IBL

could be attained.

ASSESS From the definition of cross product between two vectors, we see that the magnetic force Fr

is

perpendicular to both the current direction Lr

and the magnetic field ,Br

and the magnitude of Fr

is a maximum

when .L Br r

30. The magnitude of the force necessary to balance the magnetic force on the bar (Equation 26.5) is | |F IL B r

sin (7.5 kA)(0.3 m)(22 T)sin 60 42.9 kNILB (nearly 5 tons). The direction of this force is perpendicular

to the plane of L Br

in the opposite sense as the magnetic force.

31. INTERPRET Two forces are involved in this problem: the magnetic force and the gravitational force. We want to

find the magnetic field strength such that the two forces are equal in magnitude.

DEVELOP A magnetic force equal in magnitude to the weight of the wire requires that

B gF F ILB mg

since the wire is perpendicular to the field.

EVALUATE The equation above implies that the field strength is

2( / ) (75 g/m)(9.8 m/s )0.119 T

6.2 A

mg m L gB

IL IL

ASSESS This field strength is much greater than the typical value of 0.01 T produced by a bar magnet.

Section 26.5 Origins of the Magnetic Field

32. Equation 26.9 with 0,x gives the magnetic field at the center of a circular loop,0 0 /2B I I a (with direction

along the axis of the loop, consistent with the sense of circulation of the current and the right-hand rule). Thus, the

radius is7 2

0 /2 (2 10 N/A )(15 A)/(80 T) 11.8 cm.a I B

33. INTERPRET This problem is about the magnetic field produced by a current-carrying loop.

DEVELOP As shown in Example 26.4, the magnetic field at a point P on the axis of a circular loop of radius a

carrying current I is (Equation 26.9):

2

0

2 2 3 / 22( )

IaB

x a

26.4 Chapter 26

EVALUATE (a) At the center, 0,x and the field strength is

7 2

0 (4 10 N/A )(650 mA)= 40.8 T

2 2(1 cm)

IB

a

(b) At 20 cmx on the axis, we have

2 7 2 2

0

2 2 3 / 2 2 2 3/2

(4 10 N/A )(650 mA)(1 cm)5.09 nT

2( ) 2[(20 cm) (1 cm) ]

IaB

x a

ASSESS The direction of the field is along the axis. The field strength is greatest at the center of the loop since the

point is closest to the current distribution.

34. The magnetic field at the center of each of the tightly wound turns is the same (Equation 26.9 with 0),x so the

net field is0( /2 ),B N I a where N is the number of turns. If the wire has length L and the coil has diameter D,

then /N L D and2 7 2 2

0 0( / )( / ) / (4 10 N/A )(3.5 A)(2.2 m)/(5 cm) 1.23 mT.B L D I D IL D

35. INTERPRET This problem is about the magnetic field produced by a current-carrying wire.

DEVELOP Equation 26.10 gives the magnetic field strength of an infinitely long straight wire:

0

2

IB

r

The expression is applicable if r is much smaller compared to the length of the wire.

EVALUATE Using the equation above, we find the current to be

7 2

0

2 2 (1.2 cm)(67 T)4.02 A

4 10 N/A

rBI

ASSESS The current is proportional to the magnetic field strength. Note that the magnetic field lines are

concentric circles, as illustrated in Fig. 26.18.

36. Equation 26.11 gives the force per unit length between two long straight parallel current carrying wires: /F L 7 2 2

0 1 2/2 (2 10 N/A )(15 A) /(0.01 m) 4.50 mN/m.I I d (In household wiring, the currents are antiparallel

and the force is repulsive.)

Section 26.6 Magnetic Dipoles

37. INTERPRET This problem is about the magnetic field strength produced by a magnetic dipole.

DEVELOP As discussed in Section 26.6, the magnetic field strength at a distance x along the axis of a magnetic

dipole moment is

0

32B

x

The magnetic poles are two such points on the Earth’s surface.

EVALUATE Substituting the values given, we find the field strength to be

22 27 50

3 2 6 3

N (8 10 A m )2 10 6.20 10 T 0.62 G

2 A (6.37 10 m)E

BR

ASSESS The main component of the Earth’s magnetic field is a dipole. The magnetic field near the surface of the

Earth is about 0.5 G.

38. (a) Equation 26.12 for the magnetic moment of a loop gives2 3 2(1)(450 mA)(5 cm) 1.13 10 A m .NIA

r

(b) Equation 26.14 gives the torque on a magnetic dipole moment in a uniform magnetic field, | |B rr

3 2 3sin (1.13 10 A m )(1.4 T)sin 40 1.01 10 N m.B

39. INTERPRET This problem is about an electric motor. We are asked to find the magnetic field strength, given the

torque and the current.

DEVELOP The maximum torque on a plane circular coil follows from Equations 26.14:

2

max B NI R B


EVALUATE Solving for B using the above equation, we obtain

max max

2 2

1.2 N m482 mT

(250)(3.3 A) (3.1 cm)B

NI R

ASSESS This field strength is rather high, but reasonable for producing the torque needed to rotate the motor.

Section 26.8 Ampère’s Law

40. If the only current encircled by the path is that flowing in the wire, Ampère’s law (Equation 26.16) gives

0 wire8.8 T m ,B dL I rr or

7 2

wire (8.8 T m)/(4 10 N/A ) 7.00 A.I

41. INTERPRET This problem involves application of Ampère’s law since current is encircled by a loop.

DEVELOP Applying Ampère’s law (Equation 26.16) to the loop shown in Fig. 26.42 (going clockwise) gives

7 2

0 encircled encircled(2 ) (75 T)(2 0.2 m) (4 10 N/A )B dr I B r I r r

r

Note that the sides of the loop perpendicular to Br

give no contribution to the line integral.

EVALUATE Thus, the encircled current is

encircled 7 2

(75 T)(2 0.2 m)23.9 A

4 10 N/AI

ASSESS As explained in the text, the current flows along the boundary surface between the regions of oppositely

directed ,Br

positive into the page in Fig. 26.42, for clockwise circulation around the loop.

42. (a) Equation 26.18 gives2 7 2 2

0( /2 ) / (2 10 N/A )(5 A)(0.1 mm)/(0.5 mm) 4 G.B Ir R (b) At the surface

( ),r R Equation 26.18 reduces to Equation 26.17, either of which give7 2

0 /2 (2 10 N/A )(5 A)/B I R

(0.5 mm) 20 G.

43. INTERPRET This problem is about the magnetic field of a long current-carrying wire of radius R. We want to

show that both Equations 26.17 and 26.18 lead to the same result when .r R

DEVELOP Equation 26.17, 0 /2 ,B I r holds for r R while Equation 26.18,2

0 /2 ,B Ir R holds for .r R

EVALUATE When ,r R both equations give 0 /2 .B I R

ASSESS We expect both equations to give the same result for the magnetic field since the encircled current at

r R is encircledI I in both cases.

44. Equation 26.20 for a long thin solenoid (if applicable)

gives7 2

0 (4 10 N/A )(3300/m)(4.1 kA) 17.0 T.B nI

PROBLEMS

45. INTERPRET This problem is about magnetic force exerted on a moving charged particle.

DEVELOP The magnetic force on a moving charge can be calculated using Equation 26.1: .F qv B r rr

The force

is a cross product of vr

and .Br

EVALUATE (a) From Equation 26.1, we find the force to be

6

3

ˆ늿 ?(50 C)[(5 m/s) (3.2 m/s) ] [(9.4 T) (6.7 T) ]

ˆ 늿(50 10 N)(5 6.7 3.2 9.4 3.2 6.7 )

ˆ늿( 1.072 1.504 1.675 ) 10 N

F qv B i k i j

k j i

i j k

r rr

The magnitude and direction can be found from the components, if desired.

(b) The dot products and F v F B r r rr

are proportional to ( 1.072)(5) (1.675)(3.2) 0 and ( 1.072)(9.4)

(1.504)(6.7) 0, respectively, since the cross product of two vectors is perpendicular to each factor. (We did not

round off the components of ,Fr

so that the vanishing of the dot products could be exactly confirmed.)

ASSESS The fact that the product F vr r

vanishes can also be shown as follows:

( ) ( ) 0F v qv B v q v v B r r rr r r r r

where we have used the vector identity: ( ) ( ) .A B C C A B r r r rr r

26.6 Chapter 26

46. The positive ions enter the field region with speed (determined from the work-energy theorem) of21

2,mv qV or

2 ( / ).v V q m They are bent into a semicircle with diameter 2 2 / 2( / ) 2 ( / )x r mv qB m qB V q m

2 2( / ) / ,m q V B as shown in Fig. 26.7 (see Equation 26.3).

47. INTERPRET This problem is about magnetic force exerted on a moving charged particle. We are interested in the

angle between vr

and .Br

DEVELOP The magnetic force on a moving charge can be calculated using Equation 26.1: .F qv B r rr

The magnitude of the force is

| | | | sinF F qv B qvB r rr

where is the angle between vr

and .Br

EVALUATE Equation 26.1 gives

2 2

2 2

(2.5 ) (7.0) Nsin 0.644

(1.4 C)(185 m/s) (42 ) ( 15) mT

F

qvB

Then 40.1 or 140 . Both values are possible since sin sin(180 ).

ASSESS The result is reasonable since |sin | 1.

48. While moving perpendicularly to the magnetic field, the beam is bent into a horizontal circle of radius given by

Equation 26.3,31 6 19/ (9.11 10 kg)(8.7 10 m/s)/(1.6 10 C)(0.018 T) 2.75 mm.r mv eB If the electrons

enter normal to the boundary of the field region, r is also the penetration distance, and the beam will travel 180 in

a circle before exiting the field region in a direction opposite to which it entered. (However, if the beam enters the

field region making an angle with the normal to the boundary, then it penetrates a distance (1 sin )r into the

region, travels180 2 around the circle, and exits at angle on the other side of the normal, as shown,

where 90 90 .)

49. INTERPRET This problem is about a charged particle undergoing circular motion in a magnetic field, and we

want to express the radius of the orbit in terms of its charge, mass, kinetic energy, and the magnetic field strength.

DEVELOP From Equation 26.3, the radius of the circular motion is / .r mv qB For a non-relativistic particle,

21

2,K mv or 2 / .v K m

EVALUATE Therefore, the radius of the orbit is

2 2mv m K Kmr

qB qB m qB


ASSESS Our result indicates that the radius is proportional to ,K or v. Thus, the greater the kinetic energy of the

particle, the larger its radius.

50. (a) The frequency of the accelerating voltage is the cyclotron frequency for deuterons (Equation 26.4), 19 27/2 (1.6 10 C)(2 T)/2 (2 1.67 10 kg) 15.2 MHz.f eB m . (b) We can use the result of Exercise 25

(expressed in atomic units), with the maximum orbital radius equal to the radius of the dees. Thus, 2

max max(300 ) /2K qBr m . 2(300 1 2 0.45) /2(2 938) 19.4 MeV. (c) If the deuterons start with essentially

zero kinetic energy, and gain1500 eV each half-orbit, they will make319.4 MeV/2(1500 eV) 6.48 10 orbits.

(Of course, the same results follow in standard SI units.)

51. INTERPRET In this problem an electron is moving in a magnetic field with a velocity that has both parallel and

perpendicular components to the magnetic field. The path is a spiral.

DEVELOP The radius depends only on the perpendicular velocity component, .mv

eBr On the other hand, the

distance moved parallel to the field is || , where d v T T is the cyclotron period.

EVALUATE (a) The radius of the spiral path is

31 6

19

(9.11 10 kg)(3.1 10 m/s)70.6 m

(1.6 10 C)(0.25 T)

mvr

eB

(b) Since|| ,v v the distance moved parallel to the field is

||

22 2 2 (70.6 m) 444 m

mvmd v T v r

eB eB

ASSESS Since motion parallel to the field is not affected by the magnetic force, with|| ,v v the distance traveled

in t T along the direction of the field is simply 2 .d r

52. The forces on the upper and lower horizontal parts of the circuit are equal in magnitude, but opposite in

direction and thus cancel (see Fig. 26.43), leaving the force on the righthand wire, ( / )ILB R LB

(12 V/3 )(0.1 m)(38 mT) 15.2 mN toward the right, as the net force on the circuit.

53. INTERPRET This problem is about the current that’s needed to produce a magnetic force to balance the current-

carrying rod against gravity.

DEVELOP An upward magnetic force on the rod equal (in magnitude) to its weight is the minimum force

necessary. Since the rod is perpendicular to the magnetic field, the magnetic force is ,F IL B r r r

or

sin .F ILB ILB

EVALUATE (a) The minimum current is obtained by setting ,ILB mg or

2(0.018 kg)(9.8 m/s )5.88 A

(0.2 m)(0.15 T)

mgI

LB

(b) The force is upward for current flowing from A to B, consistent with the right-hand rule for the cross product.

ASSESS A current of 5.88 A sounds reasonable. The weight of the rod is about 0.176 N.gF mg To support the

weight with an upward magnetic force, we need a strong enough magnetic field and big enough current such

that .ILB mg

54. The geometry in this problem is the same as that in the discussion leading to Equation 26.6, which shows that 19 28 3

/ (6.8 A)(2.4 T)/(1.6 10 C)(1.2 V)(1 mm) 8.50 10 mHn IB qV t is the number density if conduction

electrons are the charge-carriers.

26.8 Chapter 26

55. INTERPRET In this problem the magnetic field exerts a torque on a current-carrying loop, causing the normal of

the loop to make an angle with the field.

DEVELOP The magnetic torque exerted on a dipole moment r

by the magnetic field Br

is given by Equation 26.14:

.B rr r

The magnitude ofr

is

| | sinB r

where ˆ,IAn r

with2

A R being the area of the loop, and n the unit vector in the normal direction of the plane

of the loop.

EVALUATE Substituting the values given in the problem, we find the field strength to be

2 2

0.015 N m= 377 mT

sin sin (12 A) (5 cm) sin 5B

I R

ASSESS The torque tends to align the magnetic dipole moment with the magnetic field. It is at a maximum,

max ,B when 90 .

56. (a) From Equation 26.12, the magnetic moment of the coil has magnitude21

4NIA 100(5 A) (0.03 m)

20.353 A m . The direction of , determined from the right-hand rule (see Fig. 26.15), rotates with the coil.

(b) The maximum torque (from Equation 26.14, with sin 1) is2

max (0.353 A m )(0.12 T)B 2

4.24 10 N m.

57. INTERPRET This problem is about the change in potential energy of a magnetic dipole moment.

DEVELOP From Equation 26.15, the potential energy of a magnetic dipole in a magnetic field is .U B rr

Therefore, the energy required to reverse the orientation of a proton’s magnetic moment from parallel to

antiparallel to the applied magnetic field is 2 .U B

EVALUATE Substituting the value given, we find the energy to be

26 2 25 62 2(1.41 10 A m )(7.0 T) 1.97 10 J 1.23 10 eVU B

ASSESS The potential energy of a dipole moment is a minimum ( )U B when it is parallel to the magnetic

field, but a maximum ( )U B when it is antiparallel to the field. Positive work must be done to flip the dipole.

58. INTERPRET Here we find the force on a quarter-circle of current-carrying wire in a magnetic field. We will use

the equation for magnetic force on a wire, which we will express in differential form and then integrate to

determine the net force.

DEVELOP The force on a section of wire dl carrying a current I in a magnetic field B is dF Idl B dF rr r

.IBdl Using the right-hand rule, we can see from Fig. 26.45 that the net force on the wire will be up and to the

right, and by symmetry we can see that the horizontal and vertical components of the force will be equal. We can

then find the horizontal component of force, and knowing that the vertical component has the same magnitude we

can find the total force.

The horizontal component of force on each segment dl of the wire is given by sin ,xdF IBdl where is the angle

between the vertical and a line between the center of curvature and the wire segment dl, as shown in the figure later.

The total horizontal force on the wire is then 2

0 sin .xF IB dl

In terms of , .dl rd The current and field are

constant: 1.5 AI and3

48 10 T.B The radius of curvature is 0.21 m.r

EVALUATE 2 2

0 0sin sin .xF IB rd IBr d IBr

The force in the y direction is the same magnitude,

so the net force is 2 0.0214 NF IBr at an angle of 45° above the horizontal.

ASSESS Finding the force without taking the advantage of the symmetry would be much more difficult, but is

also possible.


59. INTERPRET This problem is about the magnetic field at the center of a current-carrying coil. We are interested in

the number of turns the coil has.

DEVELOP Equation 26.9 (see Example 26.4) can be modified for N turns of wire so at the center of a flat circular

coil, the magnetic field is 0 /2 .B N I a This equation can be used to solve for N.

EVALUATE The number of turns in the coil is

7 2

0

2 (0.2 m)(2.3 mT)732

(4 10 N/A )(0.5 A)

aBN

I

ASSESS This is quite a lot of turns. The same field strength is obtained with one turn of coil carrying a current

(732)(0.5 A) 366 A.NI

60. The field at the center is the superposition of fields due to current in the circular loop and straight sections of wire.

The former is0 /2I a out of the page (Equation 26.9 at 0x for CCW circulation), and the latter is

0 /2I a out

of the page (Equation 26.10 at y a for the very long, straight sections). Their sum is0(1 ) /2B I a out of

the page.

61. INTERPRET This problem asks for the direction of the magnetic compass points when it is placed directly

underneath a current-carrying cable and subject to the influence of the Earth’s magnetic field.

DEVELOP A compass needle (small dipole magnet) is free to rotate in a horizontal plane until it is aligned with

the direction of the total horizontal magnetic field (see Equation 26.14). On the other hand, a long, straight wire

(the power line) carrying a current of 500 A parallel to the ground, in the direction of magnetic north, produces

a magnetic field, at a distance 10 m below, to the west, with magnitude given by Equation 26.17 (see diagram):

7 250 (4 10 N/A )(500 A)

10 T 0.1 G2 2 (10 m)

y

IB

r

The horizontal component of the Earth’s magnetic field is 0.24 G.xB

EVALUATE The compass needle will point

1 1 0.1tan tan 22.6

0.24

y

x

B

B

west of magnetic north.

ASSESS The compass needle points in the direction of the magnetic field, which in this case, is the vector sum of

the Earth’s magnetic field and the field of a current-carrying power line.

62. The Biot-Savart law (Equation 26.7) written in a coordinate system with origin at P, gives ( )B P r

2

0 wireˆ( /4 ) / ,I dL r r where r is a unit vector from an element dL on the wire to the field point P. On the straight

segments to the left and right of the semicircle, dL is parallel to r or ,r respectively, so ˆ 0dL r On the

semicircle, dL is perpendicular to r and the radius is constant, .r a Thus,

0 0 0

2 2semicircle( )

4 4 4

I I IdL aB P

aa a

The direction of ( )B Pr

from the cross product, is into the page.

26.10 Chapter 26

63. INTERPRET Our system consists of three parallel wires carrying current in the same direction. We are interested

in the force on each wire.

DEVELOP Since 4.6 m 3.5 cm ,L d the magnitude of the attractive force between each pair of wires is

approximated by Equation 26.11:2

0 /2 .F I L d

EVALUATE The forces on a given wire, due to the other two, make an angle of 60 , as shown, so the net force is

2 7 2 2

0

1 2

3 3(4 10 N/A )(20 A) (4.6 m)| | 2 cos30 18.2 mN

2 2 (3.5 cm)

I LF F F F

d

r r

ASSESS The net force exerted on one wire by the other two wires is attractive since the currents flow in the same

direction.

64. At any given distance from the long, straight wire, the force on a current element in the top segment cancels that on

a corresponding element in the bottom. The force on the near side (parallel currents) is attractive, and that on the

far side (antiparallel currents) is repulsive. The net force is the sum, which can be found from Equation 26.11

(is attractive):

7 60 1 2

2

1 1 N 1 52 10 (20 A) A (10 cm) 7.14 10 N.

2 2 cm 7 cm 2 14 cmA

I I LF

65. INTERPRET The system is a long conducting rod having a non-uniform current density. We are interested in the

magnetic field strength both inside and outside the rod. The problem involves Ampere’s law.

DEVELOP The magnetic field of a long conducting rod is approximately cylindrically symmetric, as discussed in

Section 26.8. The magnetic field can be found by using Ampere’s law:

0 encircledB dr I r r

r

EVALUATE (a) Inside the rod, Ampère’s law can be used to find the field, as in Example 26.8, by integrating the

current density over a smaller cross-sectional area, corresponding toencircledI for an amperian loop with .r R Then,

3

0

encircled 00

2( / )2

3

r J rI J r R rdr

R


Here, area elements were chosen to be circular rings of radius r, thickness dr, and area 2 .dA r dr Ampère’s law

gives0 encircled2 ,rB I or

2

0 0 /3 ,B J r R for .r R

(b) The field outside ( )r R is given by Equation 26.17, and has direction circling the rod according to the right-

hand rule. The total current can be related to the current density by integrating over the cross-sectional area of

the rod,

2

0

00

2( / )2

3

R J RI J dA J r R rdr

rr

Equation 26.17 can then be written as2

0 0 0

2 3

I J R

r rB

for .r R

ASSESS The magnetic field increases as2r inside the rod, but decreases as1/r outside the rod. At ,r R both

expressions give the same result:0 0( ) /3.B r R J R

66. (b) Since the pipe is assumed to have cylindrical symmetry, the field outside is given by Equation 30.8. (a) For

an amperian loop withencircled, 0,r R I hence 0B inside. (The thickness of the pipe is considered negligible.)

67. INTERPRET The system consists of two large current-carrying plates. The current distributions have plane

symmetry, and we apply the superposition principle.

DEVELOP Equation 26.19,1 0 s /2,B J and Fig. 26.32 give the magnitude and direction of the individual fields.

The total field is the superposition of fields due to two (approximately infinite) flat parallel current sheets.

EVALUATE (a) Between the plates, both fields are in the negative y direction. Thus,

btw 0 ,1 0 ,2 0

1 1늿 ?2 2

s s sB J j J j J j r

(b) Outside the plates, the fields are in opposite directions, and thus cancel, out 0.B r

ASSESS The situation here is analogous to the electric field of two infinite planes carrying opposite charges

(a capacitor). The field is non-zero in between the plates but vanishes outside.

68. See the solution to Problem 73, with , ,a bR a R b and .cR b c

69. INTERPRET This problem deals with the magnetic field of a current-carrying solenoid. We are interested in the

number of turns the solenoid has and the power it dissipates.

DEVELOP A length-diameter ratio of 10 to 1 is large enough for Equation 26.20,0 ,B nI to be a good

approximation to the field near the solenoid’s center. This is the equation we shall use to calculate the number of

turns. On the other hand, the power the solenoid dissipates is given by2

.P I R

EVALUATE (a) Using Equation 26.20, we find the number of turns per unit length to be

13 1

7 2

0

10 T2.27 10 m

(4 10 N/A )(35 A)

Bn

I

This implies that the total number of turns is3

2.27 10 .N nL

(b) A direct current is used in the solenoid, so the power dissipated (Joule heat) is

2 2(35 A) (2.7 ) 3.31 kWP I R

ASSESS That’s a lot of turns in one meter. So the solenoid is very tightly wound to produce such a strong field at

its center.

70. (a) The length of a solenoid, with one layer of N turns of closely spaced wire of diameter d, is ,L Nd so the

number of turns per unit length is1 1 1

/ (0.5 mm) 2000 m .n N L d (Although not needed, the value of

10 m/2 cm 159N turns can be used to check that 159 0.5 mm 7.96 cmL is barely long enough,

compared to 2 cm, to justify the use of Equation 26.20 as an approximation to B at the solenoid’s center.) Then 7 2 1 2

0 (4 10 N/A )(2000 m ) (15 A) 3.77 10 T.B nI (b) The magnetic field at the center of a flat,

circular current loop can be found from Equation 26.9,0 /2 .B I a Here, 2 10 m,a so

0 /10 mB I 2 7 2 6

(4 10 N/A )(15 A)/10 m 5.92 10 T.

71. INTERPRET In this problem we are asked to derive to expression for the magnetic field of a solenoid by treating it

as being made up of a large number of current loops.

DEVELOP Consider a small length of solenoid, dx, to be like a coil of radius R and current .nI dx Using

Equation 26.9, the axial field is

26.12 Chapter 26

2

0

2 2 3/2

( )

2( )

nIdx RdB

x R

with direction along the axis according to the right-hand rule. For a very long solenoid, we can integrate this from

tox x to find the total field.

EVALUATE Integrating over dx from to ,x x we find the magnetic field to be

2 2

0 0

sol 02 2 3/2 2 2 22 2( )

nI R nI Rdx xB nI

x R R x R

This is the expression given in Equation 26.20.

ASSESS For a finite solenoid, a similar integral gives the field at any point on the axis only, for example, at the

center of a solenoid of length L,

0

2 2(0)

4

nILB

L R

72. Supposing that the cylindrical channel of ionized air acts like a long straight wire, we can use Equation 26.10 to

estimate the distance:7 2

0 /2 (2 10 N/A )(250 kA)/(50 T) 1 km.y I B

73. INTERPRET This problem is about the magnetic field of a coaxial cable. Ampère’s law can be applied since the

current distribution has line symmetry.

DEVELOP For a long, straight cable, the magnetic field can be found from Ampère’s law. The field lines are

cylindrically symmetric and form closed loops, hence must be concentric circles, which we also choose as

amperian loops. Take positive circulation counterclockwise so that positive current is out of the page. Then

0 encircled2B dr rB I r r

r

Assume that the current density in each conductor is uniform; i.e., the current is proportional to the cross-sectional

area. We may calculateencircledI in four regions of space.

EVALUATE (a) For ,ar R the encircled current is

2 2

encircled 2 2

a a

r rI I I

R R

so2

0 /2 .aB Ir R Numerically, 0.5 mm,aR 5 mm, 5.2 mm,b cR R and 0.1 A,I so inside the inner

conductor when 0.1 mm ,ar R the magnetic field is

7

0

2 2

(4 10 T m/A)(0.1 A)(0.1 mm)8 T

2 2 (0.5 mm)a

IrB

R

(b) For ,a bR r R the current encircled isencircled ,I I so

0 /2 .B I r Therefore, for 5 mm ,a bR r R the

magnetic field is

7

0 (4 10 T m/A)(0.1 A)4 T

2 2 (5 mm)

IB

r

(c) Forencircled, 0,cr R I so the magnetic field is 0.B Thus, at 2 cm , 0.cr R B

ASSESS For completeness, let’s calculate the magnetic field in the region .b cR r R The current encircled by

the amperian loop is


2 2 2 2

encircled 2 2 2 2

( )

( )

b c

c b c b

r R R rI I I I

R R R R

and the magnetic field is

2

0

2 2( )

2 ( )

c

c b

I RB r r

rR R

The outer radius Rc is the inner radius plus the thickness of the outer conductor. Forbr R and ,cr R the above

equation gives0 /2 bB I R and 0,B respectively, in agreement with the results obtained before.

74. (a) As mentioned in the paragraph following Equation 30.4, the calculation of the magnetic field from a circular

current loop, at points not on its axis, is difficult. Fortunately, at a distance of 1 mm from a loop of radius 15 cm,

the field is approximately that of a long, straight wire, Equation 30.5 gives7 2

0 /2 (2 10 N/A )(2A)/B I y 3

(10 m) 4 G. (b) Since 3 m >>15 cm, the approximation x a in Example 26.4 is justified. Then

2

0 /B Ia 3 7 2 2 3 5

2 | | (2 10 N/A )(2 A)(0.15 m) /(3 m) 1.05 10 G.x

75. INTERPRET This problem is about the magnetic field of a current-carrying conducting bar. Symmetry holds

approximately in certain limits.

DEVELOP Very near the conductor, but far from any edge, the field is like that due to a large current sheet.

On the other hand, very far from the conductor, the field is like that due to a long, straight wire.

EVALUATE (a) Approximating the bar by a large current sheet with / ,sJ I w Equation 26.19 gives 0

2.

I

wB

(b) Approximating the bar by a long, straight wire. Equation 26.17 gives 0 /2 .B I r

ASSESS The conductor exhibits different approximate symmetries, depending on where the field point is chosen.

76. The current distribution is similar to a solenoid, where the number of turns per unit length and the current in each

turn are related to the total current in the pipe by .tnLI I Therefore (see Section 26.8), the field is approximately

that of an infinite solenoid:0 0 /tB nI I L inside, and 0B outside, directed parallel to the axis to the left in

Fig. 26.52.

77. INTERPRET The system is a solid conducting wire having a non-uniform current density. We are interested in the

magnetic field strength both inside and outside the wire. The problem involves Ampère’s law.

DEVELOP The total current in the wire can be obtained by integrating the current density over the cross sectional

area. The magnetic field of a long conducting wire is approximately cylindrically symmetric, as discussed in

Section 26.8. The magnetic field can be found by using Ampère’s law:

0 encircledB dr I r r

r

EVALUATE (a) Using thin rings as the area elements with 2 ,dA rdr the total current in the wire (z axis out of

the page) is

2 32

0 0 00 0

0

1 1 2 2

2 3 3

RR R r r r

I J d A J r dr J R JR R

(b) A concentric amperian loop outside the wire encircles the total current, so Ampère’s law gives

26.14 Chapter 26

2

0 0 0

12

3rB I R J

or2

0 0 /6 .B J R r

(c) Inside the wire, Ampère’s law gives0 encircled2 .rB I The calculation in part (a) shows that within a loop of

radius ,r R

2 32

encircled 0 00

0

2 2 1

2 3 3

rr r r rI JdA J J r

R R

Therefore, 1 20 02 3

(1 ).r

RB J r

ASSESS At ,r R both equations give0 0 /6.B J R Since from part (a)

2

0 3 / ,J I R the magnetic field can also

be written as0 /2 .B I R The form is the same as that shown in Equation 26.17.

78. The disk may be considered to be composed of rings of radius r, thickness dr, and charge 2 .dq r dr Each

ring represents a circular current loop, / /(2 / ) ,dI dq T dq rdr which produces a magnetic field

strength 10 02

/2dB dI r dr at the center of the disk, directed out of the page, as sketched for positive

charge density. The total field strength is 1 10 0 0 02 2

.a a

B dB dr a

79. INTERPRET The problem is about magnetic dipole moment formed by a current-carrying wire of length L.

DEVELOP The number of turns of radius r that can be formed from a wire of length L is

2 2

L LN r

r N

The magnitude of the magnetic dipole moment of such a coil is2 .NIA NI r

EVALUATE (a) Substituting the expression for r into the equation for , we find

2 2

2 4

L ILNI

N N

(b) The magnetic dipole moment is clearly a maximum when N is a minimum, and the smallest value of N is, of

course, one.

ASSESS Since the radius is proportional to1/ ,N the area of the circular coil is proportional to21/ .N This makes

sense because with the length L kept fixed, the greater the number of turns, the smaller the area of the coil. Thus,

with2 ,NIA NI r we see that 1/ .N

80. (a) The velocity component in the direction of Br

is|| 15 cm/2 s 75 km/s,v which is also the numerical value

of ,v the speed perpendicular to .Br

Thus, the speed along the spiral trajectory is2 2

|| ||2 ,v v v v and the

actual length traveled in 2 s is 2(75 km/s)(2 s) 21.2 cm. (b) The cyclotron period is 2 / ,T m eB so the

number of orbits completed in19 31 3

2 s is 2 s/T (1 s)(1.6 10 C)(0.1 T)/ (9.11 10 kg) 5.59 10 .


81. INTERPRET Two forces are involved in this problem: gravitational and magnetic. At equilibrium, the two forces

cancel exactly.

DEVELOP If the height h is small compared to the length of the rods, we can use Equation 26.11 for the repulsive

magnetic force between the horizontal rods (upward on the top rod)

2

0

2B

I LF

h

The rod is in equilibrium when this equals its weight, .gF mg

EVALUATE The equilibrium condition B gF F gives

2 7 2 2

0

2

(4 10 N/A )(66 A) (0.95 m)3.84 mm

2 2 (0.022 kg)(9.8 m/s )

I Lh

mg

ASSESS The height h is indeed small compared to 95 cm. So our assumption is justified.

82. The magnitude of the force per unit length on a thin strip of ribbon, of width dx, carrying current / ,I dx w is given by

Equation 26.11:0/ ( ) 2 ( ),dF L I I dxLw L a x where x is the distance from the near edge. Integrating from 0x

to w, we find:

2 2

0 0

0 ln

2 2

wI IF dx a w

L w a x w a

The force is attractive since the currents are parallel.

83. INTERPRET We find the spacing between two current loops at which the second derivative of the magnetic field

becomes zero. We will use the results of Example 26.4, which give us the field for a single loop, along the axis of

the loop, and use superposition to extend this to two loops.

DEVELOP The field along the axis of a single loop is

2

0

2 2 3 / 22( )

IaB

x a

(Equation 26.9.) The field due to two such loops each at a distance2

bx from the center is then

2

0

2 2 3 /2

2(( ) )b

IaB

x a

From this equation for the field at the center of these two loops, we will find the second derivative of B and then

find the value of b for which the second derivative is zero.

EVALUATE

22 2 2 2 220 2 0 0

2 7 / 2 5 / 2 2 2 7 / 22 22 2

2 2

15 3 384 ( ( 2 ) )

(4 ( 2 ) )

b

b b

Ia x Ia Ia a b xd B

dx a b xa x a x

At 0,x

2 2 222 20

2 2 2 7/2

0

384 ( )0

(4 )x

Ia a bd Ba b

dx a b

26.16 Chapter 26

ASSESS The spacing between the coils must equal the radius of the coils. This is a standard method of producing

a uniform magnetic field in a region.

84. INTERPRET Given four long straight current-carrying wires at the corners of a square, we are asked to find the

magnetic field at the center of the square. We will use the magnetic field of a long straight current, and use vector

addition to find the total field.

DEVELOP We start with a sketch, as shown below. The four wires create the four contributions to the net field, as

shown. Since the arrangement is symmetric, and all the currents are the same, each contribution must have the

same magnitude and the direction of the net field must be in the –x direction.

The magnetic field of a long straight wire is 0

2,

I

rB

and .075 2 mr for each wire in this case. The y

components of the field cancel, so we’ll add only the x components to find the answer.

EVALUATE the x component of field due to one wire is

0 0 02cos(45 )

2 2 2 4 (0.75)

I I IB

r r

There are 4 wires, so the net field at the center is 0 1.33 T(0.75)

IB

in the negative y direction.

ASSESS We only have to find one component of the field due to one wire, since they’re all the same distance

away and the configuration is symmetric.

85. INTERPRET We find the net force on a wire due to three other wires. All wires are parallel and carry the same

current, although the direction of the currents varies. We will use the equation for force between two parallel wires

to find the force due to each wire, then add these force vectors.

DEVELOP The force per length between two parallel wires is 0 1 2

2.

I IF

L d

This force is attractive if the currents flow

in the same direction, so the forces on the wire at the lower left corner of the square are as shown in the figure later.

We will add these three force vectors. The currents are all 2.5 A,I and the distance along the edge of the square

is 0.15 m.d

EVALUATE

2 2 2

0 0 3 01 2 1 1늿 늿, and2 2 2 22 ( 2)

I I F IF Fj i j i

L d L L dd

rr r

so the net force per length is

2

6 5 50 1 1늿 늿1 1 (5.39 10 N) (1.13 10 N) or 1.25 10 N2 2 2 2 2

IF Fi j i j

L d L

r r

at an angle of 64.5° below the +x axis.


ASSESS This is a very small force, even with this relatively large current.

86. INTERPRET We adapt Example 26.5 to the case of a finite wire, and find the magnetic field at a point due to this wire.

DEVELOP We can adapt Example 26.5 to our needs by simply changing the limits of integration. Instead of

integrating from , we integrate from0x to

0 .x L

EVALUATE The magnetic field is in the same direction as before, and has magnitude

0

0

0

0

0 0 0 0 0

2 2 3 / 2 2 2 2 2 2 2

0 04 4 4( ) ( )

x L

x L

x

x

I I I x L xydx xB

yx y y x y x L y x y

ASSESS We can check this by taking the limit as0x L and .L The answer in this limit should be just

the result from Example 26.5: 0 .2

IB

y

87. INTERPRET We use the result from Problem 86 to find the magnetic field at the center of a square current loop.

Most of the work has been done already—we just need to replace0x and L with values appropriate to this problem,

for each side, then add the fields due to each side.

DEVELOP We’ll find the magnitude of the field due to one side of the square, then multiply by 4 since all sides

contribute the same field in the same direction. The field due to one side is obtained by substituting 0 2,ax

2,aL and

2

ay in the results from Problem 86,

0 0 0

2 2 2 2

0 04 ( )

I x L xB

y x L y x y

EVALUATE For one side,

0

24 a

IB

2

a

0 0

222 2

21

4 4( ) ( )aa a

I I

a

The net field at the center is then 0 2.

I

aB

ASSESS We could extend this method to find the magnetic field due to any configuration of short wire

segments—which is what integrating the Biot-Savart law does for us in the first place!

88. INTERPRET We find the magnetic field at the center of a “real” solenoid of finite length, treating the solenoid as

a stack of individual coils. We use the formula for the magnetic field due to a single loop, and integrate.

DEVELOP We’ll treat the current as a cylindrical sheet of current with current surface density ,sJ nI where n is

the number of turns per length. The currentlI through each infinitesimal loop of width dx is then

l sI J dx and the

magnetic field due to each loop, at the center of the solenoid, is2 2 2 3 / 2

0 ( ) /2( ) .sdB J dx a x a We will integrate

from2

Lx to2

Lx to find the total field at the center.

EVALUATE

26.18 Chapter 26

2

2

2 2 2

0 0 02 22 2 3 / 2 2 2 3 / 2 2 2

2 24

0

2

4

( )

2 22( ) ( )

2

L L

s s sL L

L

L

J dx a J a J adx LB

x a x a a a

nILB

a

ASSESS We can check this by taking the limit as .L

2

0 0

02

4

2lim

22L L

nIL nI LB nI

La

This is the same as Equation 26.20.

89. INTERPRET We find the force on a magnetic dipole located on the axis of a current loop by differentiating the

potential energy. We will use the magnetic field equation developed in Example 26.4.

DEVELOP The potential energy of a dipole in a magnetic field is given by ,U B rr

where in this case

2

0

2 2 3/2ˆ

2( )

IaB i

x a

r

and ˆ.i r

The force is .x

dUF

dx

EVALUATE

2

0

2 2 3/ 22( )

IaU

x a

so

2

0

2 2 5 / 2

3

2 ( )

Ia xF

x a

At ,x a

2 3

0 0 0

2 5/2 5 2

3 33

2 2(2 ) 4 2 8 2

Ia I Ia aF

a a a

ASSESS This force is opposite the direction of x, so it is an attractive force in this case.

90. INTERPRET We find the radii of the spiral paths of protons in a given magnetic field, for different proton energy.

We’ll need the speed of the proton, which we can obtain from the kinetic energy, and we’ll use the equation for the

radius of a charged particle’s path in a magnetic field.

DEVELOP The radius of path curvature for a charged particle in a magnetic field is given by Equation 26.3,

.mv

qBr We’ll need the speed v of the particle, which we can obtain from the energy

21

2.E mv The mass of a

proton is27

1.67 10 kg,m and the magnetic field is

510 T.B

The energies we need radii for are 0.1 MeV,

1 MeV, and 10 MeV.

EVALUATE

21 2 2 2, so

2

E m E mEE mv v r

m qB m qB

For 0.1 MeV, 4.57 km.r For 1 MeV, 14.4 km.r At 10 MeV, 45.7 km.r

ASSESS At 10 MeV, the proton’s speed is more than 1

10of the speed of light and we should consider relativistic

effects. Stay tuned—we’ll learn more about this later in the course!

91. INTERPRET What current would it take to generate a planet-sized magnetic field? Here we approximate the

Earth’s geodynamo with a single loop, and find the current necessary. We use the results of Example 26.4 to find

the magnetic field along the axis of a loop.

DEVELOP We will use


2

0

2 2 3 / 22( )

IaB

x a

where the radius of our loop is6

3 10 ma and6

62 10 T.B We solve for I, given that Ex R

66.38 10 m.

EVALUATE

2 2 3 / 29

2

0

2 ( )3.8 10 A

B x aI

a

ASSESS This seems like a very high current—but remember that’s the current flowing through an entire planet, so

the actual current density in this simplified model is quite low.

92. INTERPRET We need to find the magnetic field necessary to create a certain force on a wire loop. We will model

this as a force on a wire in a uniform magnetic field.

DEVELOP The force on a wire in a magnetic field is .F IL B r r r

We will optimize our speaker design by making

Lr

and Br

perpendicular, so .F ILB From the coil diameter 0.035 md and number of turns 100,n we can find

the length of wire L. The current in the coil is given as 2.1 A,I and the force is 14.8 N,F so we will simply

solve for B.

EVALUATE

0.641 T( )

F FB

IL I nd

ASSESS This is a fairly high field strength, but reasonable for currently available permanent magnets.

93. INTERPRET We find the magnetic field of a sheet of current by integrating a row of line currents, instead of by

using Ampère’s law.

DEVELOP We start with a sketch, as shown in the figure later. Since it is an infinite sheet of current, there is

an equal amount of current to the right and to the left of the point P, and the vertical components of B cancel.

The field due to the line current dI is

0 0

2 2 1/22 2 ( )

sdI J dxdB

r x y

We need only the horizontal component of this field,

0

2 2 1/2 2 2 1/2sin

2 ( ) ( )

s

x

J dx ydB dB

x y x y

We integrate this over all x to find the field due to an infinite sheet.

EVALUATE

10 0 0

2 2 2 2

0

1tan

2 22 ( )

2

s s s

x x

s

x

J y J y J ydx xdB dx B

y yx y x y

JB

26.20 Chapter 26

ASSESS This is the same as the result obtained by use of Ampère’s law.

27.1

ELECTROMAGNETIC INDUCTION

EXERCISES

Sections 27.2 Faraday’s Law and 27.3 Induction and Energy

15. INTERPRET In this problem we are asked to verify that the SI unit of the rate of change of magnetic flux is volt.

DEVELOP We first note that the left-hand-side of Equation 27.2, / ,Bd dt represents the induced emf which

has units of volt. From the definition of magnetic flux given in Equation 27.1a, we see that it has SI units of 2

T m .

EVALUATE The reasoning above shows that the units of /Bd dt are

2 2T m /s (N/A m)(m /s) (N m/A s) J/C V

ASSESS Faraday’s law relates the induced emf to the change in flux. It is the rate of change of flux, and not the

flux or the magnetic field that gives rise to an induced emf.

16. For a stationary plane loop in a uniform magnetic field, the integral for the flux in Equation 27.1a is just B B A rr

2 4

cos (80 mT) (2.5 cm) cos30 1.36 10 Wb.BA (The SI unit of flux,2T m , is also called a weber, Wb.)

17. INTERPRET This problem is about the rate of change of magnetic flux through a loop due to a changing

magnetic field.

DEVELOP For a stationary plane loop in a uniform magnetic field, the magnetic flux is given by Equation 27.1b,

cos .B B A BA rr

Note that the SI unit of flux,2T m , is also called a weber, Wb. The rate of change of

magnetic flux is / / .B Bd dt t

EVALUATE (a) The magnetic field at the beginning1( 0)t is

2 2 4

1 1 1

1 1(40 cm) (5 mT) 6.28 10 Wb

4 4B A d B

(b) The magnetic field at2 25 mst is

2 2 3

2 2 2

1 1(40 cm) (55 mT) 6.91 10 Wb

4 4B A d B

(c) Since the field increases linearly, the rate of change of magnetic flux is

3 3

2 1

2 1

6.91 10 Wb 0.628 10 Wb0.251 V

25 ms

B Bd

dt t t t

From Faraday’s law, this is equal to the magnitude of the induced emf, which causes a current

| | 0.251 V= 2.51 mA

100 I

R

in the loop.

sol

coil sol sol coil sol| | ( )B dBd dN B A N A

dt dt dt

(d) The direction must oppose the increase of the external field downward, hence the induced field is upward and

I is CCW when viewed from above the loop.

ASSESS Since / ( / )B t B t A with the area of the loop kept fixed, the induced emf and hence the current

scale linearly with the value / .B t

27

27.2 Chapter 27

18. The flux through a stationary loop perpendicular to a magnetic field is B BA (see Exercise 16), so Faraday’s law

(Equation 27.2) and Ohm’s law (Equation 24.5) relate this to the magnitude of the induced current: | / |I R 2

| / |/ | / |/ . Therefore | / | / (320 mA)(12 )/(240 cm ) 160 T/s.Bd dt R A dB dt R dB dt IR A

19. INTERPRET The problem asks for the number of turns the coil must have in order to produce a given emf when it

is placed in a time-varying magnetic field.

sol

coil sol sol coil sol| | ( )B dBd dN B A N A

dt dt dt

DEVELOP When the coil is wrapped around the solenoid, all of the flux in the solenoid (Bsol Asol, for a long thin

solenoid) goes through each of thecoilN turns of the coil. Then the induced emf in the coil is

sol

coil sol sol coil sol| | ( )BdBd d

N B A N Adt dt dt

EVALUATE Substituting the values given in the problem statement, the number of turns in the coil is

coil 2

sol sol

| | 15 V199 turns

| / | (2.4 T/s) (0.1 m)N

dB dt A

ASSESS The number of turns is proportional to the induced emf, but inversely proportional to the rate of change

of magnetic field.

Section 27.4 Inductance

20. Equation 27.4 gives2 7 3 2 2

0 / (4 10 H/m)(10 ) (2 cm) /(50 cm) 3.16 mH.L N A l (The long thin solenoid

approximation is valid here.)

21. INTERPRET This problem asks for the self-inductance of an inductor.

DEVELOP The induced emf in an inductor is given by Equation 27.5: / .L LdI dt WithL and /dI dt given, we

can use the equation to compute the self-inductance L.

EVALUATE From Equation 27.5, we find the self-inductance to be

40 V0.4 H

/ 100 A/s

LLdI dt

ASSESS Our value of self-inductance is reasonable; inductances in common electronic circuits usually range from

micro-henrys to several henrys.

22. Assume that the current changes uniformly from 2 A to zero in 1 ms (or consider average values). Then

/ 2 A/ms,dI dt and Equation 27.5 gives (20 H)( 20 A/ms) 40 kV. (The emf opposes the decreasing

current.)

23. INTERPRET The problem asks for the number of turns in the inductor which in our case is a long solenoid.

DEVELOP The inductance of a long solenoid is given by Equation 27.4 (in Example 27.6):

2

2 0

0

N AL n Al

l

The equation allows us to determine N, the number of turns.

EVALUATE From the equation above, we find the value of N to be

3

7 2

0

(5.8 mH)(15 cm)1.35 10

(4 10 H/m) (1.1 cm)

LlN

A

ASSESS This corresponds to3/ 1.35 10 /(15cm) 90/cm.n N l That’s a lot of turns in one centimeter. So the

inductor is very tightly wound.

24. Since the inductive time constant in an RL circuit is / , (2.2 ms)(100 ) 220 mH.L LL R L R

25. INTERPRET This problem is about the resistance in a series RL circuit.

DEVELOP The buildup of current in an RL circuit with a battery is given by Equation 27.7:

/( ) (1 )

Rt LI t I e

where0/I R is the final current. This is the equation we employ to solve for R.

Electromagnetic Induction 27.3

EVALUATE Substituting the values given, one finds the resistance to be

1.8 mHln 1 ln(1 0.20) 130

3.1 s

L IR

t I

ASSESS We find R to be inversely proportional to t. This means that the greater the value of R, the shorter the

time it takes for the current to increase to 20% of its final value.

Section 27.5 Magnetic Energy

26. Equation 27.9 gives2 21 1

2 2(5 H)(35 A) 3.06 kJ.U LI

27. INTERPRET This problem is about the magnetic energy stored in an inductor. The object of interest is the current.

DEVELOP The amount of energy stored in an inductor is given by Equation 27.9:2/2.U LI This equation

allows us to determine the current I.

EVALUATE From the equation above, we find the current to be

2 2(50 J)0.1 A

10 mH

UI

L

ASSESS Since the energy stored in the inductor is small, we expect the current (which is proportional to U ) to

be a small one as well.

28. From Equation 27.9,2 2 2 21 1

2 2( ) (220 mH)[(800 mA) (350 mA) ] 56.7 mJ.f iU L I I

29. INTERPRET This problem is about the magnetic energy stored in a solenoid.

DEVELOP We first note that the inductance of a solenoid is given by Equation 27.4:2 2

0 0 / .L n Al N A l

Equation 27.9,2/2,U LI can then be used to find the amount of energy stored in the solenoid.

EVALUATE Combining Equations 27.4 and 27.9, we find the stored energy to be

2 2 7 2 2 2 22 01 1 (4 10 N/A )(500) (1.5 cm/2) (65 mA)

0.510 J2 2 2(23 cm)

N AIU LI

l

ASSESS This inductance of the solenoid is about 240 H. The stored energy is typical for a small inductor with a

small current.

30. From Equation 27.10,2 7 2 3

(50 T) /(8 10 N/A ) 995 MJ/m .Bu (This is about 2.8% of the energy density

content of gasoline; see Appendix C.)

31. INTERPRET We are given the magnetic field strength and asked to compute the magnetic-energy density.

DEVELOP Equation 27.10,2

0/2 ,Bu B provides the connection between the magnetic energy density and the

magnetic field strength.

EVALUATE From Equation 27.10, the energy density is

2 23

7 2

0

(50 T)995 MJ/m

2 2(4 10 N/A )B

Bu

ASSESS This is about 2.8% of the energy density content of gasoline,3

gasoline 35,000 MJ/m .u

32. From Equation 27.10,7 3

02 (8 10 H/m)(7.8 J/cm ) 4.43 T.BB u

Section 27.6 Induced Electric Fields

33. INTERPRET We’re given the induced electric field and asked to find the rate of change of the magnetic field.

DEVELOP The connection between the induced electric field and the rate of change of the magnetic field is given

by Equation 27.11:

BdE dr

dt

r r

The geometry of the induced electric field from the solenoid is described in Example 27.11, where

22( )

2 | |d R B dB

r E Rdt dt

27.4 Chapter 27

EVALUATE The rate of change of the magnetic field is

3

2 2

2 | | 2(12 cm)(45 V/m)1.08 10 T/s 1.08 T/ms

(10 cm)

dB r E

dt R

ASSESS The magnetic field is changing at a very fast rate. Note that the sign of /dB dt and the direction of the

induced electric field are related by Lenz’s law.

34. INTERPRET There is a changing magnetic field inside a solenoid with changing current, so there must be

an electric field as well. We use Faraday’s law to find the electric field as a function of r inside a solenoid.

DEVELOP We’ll use a circular Amperian loop, of radius r, centered inside the solenoid. The flux through this

loop is2

.BA r B We are told that the field in the solenoid is .B bt Faraday’s law, integrated around this

loop, gives us .d

dtE dr r r

r By symmetry, the electric field is constant, which makes the integration easy.

EVALUATE

2 22 [ ( )]

2

d d rbE dr rE r bt r b E

dt dt

r r

ASSESS Just as in previous problems using Gauss’s and Ampère’s laws, it is important to choose our symmetry to

make things easy on ourselves.

PROBLEMS

35. [NOTE: The third sentence of the problem statement should read: “Subsequently, the current increases according

to2,I bt where b is a constant with the units

2/ .A s ”]

INTERPRET This problem is about the rate of change of magnetic field strength, given the induced current as a

function of time.

DEVELOP The flux through a stationary loop perpendicular to a magnetic field is ,B BA so Faraday’s law

(Equation 27.2) and Ohm’s law (Equation 24.5) relate this to the magnitude of the induced current:

| / || | | / |Bd dt A dB dtI

R R R

EVALUATE From the above expression, we find the rate of change of the magnetic field to be

2dB IR bRt

dt A R

Integration yields3

( ) ( /3 ) ,B t bR A t where (0) 0B was specified.

ASSESS Since the current increases as2,t we expect the magnetic field strength to increase as

3.t

36. The reasoning in the solution to Exercise 18 shows that2 2| | | / |/ (2 )/ (0.15 m )(2 2 T/s ) /I A dB dt R A at R t

1(6 ) 10 (A/s).t

(a) For 3 s, | | 0.3 A,t I and (b) for2

/ 8 T/(2 T/s ) 2 s, | | 0.2 A.t b a I (Note: In

this

problem, there is enough information to also specify the direction of I. Choose the x-y axes as shown and the z axis

out of the page. For positive normal to the loop along the z axis, a positive sense of circulation for the induced

current is CCW. Then / (1/ ) ( )/ ( / )( / ) ( / )(2 ).zI R R d B A dt A R dB dt A R at r

Negative currents are CW

around the z axis.)


37. INTERPRET This problem is about the work done by an external agent to move a loop at a constant speed across

a region with uniform magnetic field.

DEVELOP The loop can be treated analogously to the situation analyzed in Section 27.3, under the heading

“Motional EMF and Lenz’s Law”; instead of exiting, the loop is entering the field region at constant velocity.

All quantities have the same magnitudes, except the current in the loop is CCW instead of CW, as in Fig. 27.13.

EVALUATE Since the applied force acts over a displacement equal to the side-length of the loop, the work done

can be calculated directly:

2

app app ( )W F l IlB l Il B rr

Since the induced current is

| / || | 1( )Bd dt d Blv

I BlxR R R dt R

the work applied by the external agent is

2 32 2

app

Blv B l vW Il B l B

R R

ASSESS Alternatively, the work can be calculated from the conservation of energy:

2 2 2 32

diss app diss

( ) ( )

Blv Blv l B l vP I R W P t

R R v R

38. The peak voltage induced in a loop or coil rotating about an axis perpendicular to a uniform magnetic field is

calculated in Example 27.5, peak .NA B Therefore2 11

peak 4/ 360 V/[5 (1 cm) (20 s )] 14.6 mT.B NA

39. INTERPRET This problem is about the magnetic flux through a square due to a non-uniform field.

DEVELOP We take elements of area to be0

ˆ(2 ) ,dA x dx kr

which are rectangular strips parallel to the y axis.

The flux through the loop is then equal tosquare .B B dA

rr

EVALUATE The integral gives

0

3 22

20 0 0 0 0

02square 000

2 (2 ) 162

3 3

x

B

B B x B xB dA x x dx

xx

rr

ASSESS The quantity2

0 016 /3B x has units of 2T m , which is what we expect of a magnetic flux.

40. (a) To shine at full brightness, the potential drop across the bulb must be 6 V. This is equal to the induced emf, if

we neglect the resistance of the rest of the loop circuit. From Faraday’s law, | | | / | | ( )/ |Bd dt d BA dt

| | .A B t Thus,2| |/| | (3 m) (2 T)/6 V 3 s.t A B (b) The direction of current opposes the decrease of B

r

into the page, and thus must act to increase Br

into the page. From the right-hand rule, this corresponds to a

clockwise current in Fig. 27.38.

41. INTERPRET This problem is about the current induced in a rectangular loop due to a nearby time-varying current

source.

DEVELOP The normal to the loop in Fig. 27.7 was taken to be in the direction of the magnetic field of the wire, or

into the page, so the positive sense of circulation around the loop is clockwise (from the right-hand rule). Faraday’s

and Ohm’s laws give an induced current in the loop of

27.6 Chapter 27

| / || | Bd dtI

R R

The magnetic flux has been calculated in Example 27.2: 0

2ln( ).

Il a wB a

EVALUATE Combining the above expressions gives

7 2

0 ( / )| / || | (4 10 N/A )(6 cm)(25 A/s) 4.5 cmln ln

2 2 (50 m ) 1.0 cm

9.02 A

Bl dI dtd dt a w

IR R R a

By Lenz’s law, the induced current is counterclockwise in the loop.

ASSESS As the inward flux increases (because the current in the long wire is increasing), by Lenz’s law, the

induced current must flow in the counterclockwise direction to produce an outward flux to oppose the change.

42. (a) The magnetic field inside the solenoid is 0 ,B nI so the flux through the loop is21

loop 0 loop4.B BA n D I

From Faraday’s and Ohm’s laws, the magnitude of the induced current is

2 2 7 2

loop 0 loop 2

| | 1 1 1 N 2000 (1.0 kA/s)10 (0.1 m) 1.97 mA

4 2 m (5 )A

Bd N dII D

R R dt L R dt

(b) If the loop encloses the solenoid, thensolenoid,B BA and the induced current would increase to

2

solenoid loop/ (1.5)A A times the value in part (a), or 4.44 mA.

43. INTERPRET This problem is about the current induced in a loop due to a time-varying magnetic field.

DEVELOP Take the normal to the loop along the z axis, so that the positive sense of circulation for the induced

emf and current is CCW (looking down on the x-y plane) as shown. The flux through the loop is

0 0늿( sin ) ( ) sinB B A B t k Ak B A t

rr

so the induced current is

0 cos/ /B AB td dt AdB dtI

R R R R

EVALUATE (a) Using the expression above, the current at 0t is

1 2

0 (10 s )(2 T)(150 cm )(0) 60 mA

5

B AI

R

The negative sign implies that the induced current is clockwise.

(b) Similarly, at 0.1 s,t the induced current is

1 210 (10 s )(2 T)(150 cm )

(0.1 s) cos cos[(10 s )(0.1 s)]5

60 mA cos(1 radian) 32.4 mA

B AI t

R

ASSESS During the time interval in which the outward flux through the loop increases ( | cos | 0,t as viewed

from above the loop), by Lenz’s law, the induced current must flow in the clockwise direction to produce

an inward flux to oppose the change.

44. A normal to the loop parallel to the z axis corresponds to CCW positive circulation (via the right-hand rule), when

viewed from the positive z axis. Faraday’s and Ohm’s laws give the current in the loop: / ( / )/BI R d dt R 2( / )/ / (240 cm )(0.35 T/s)/(0.2 ) 42.0 mA,A dB dt R Ab R the minus sign indicating a CW circulation

viewed from the positive z axis.

45. INTERPRET We have a conducting coil rotating in a fixed magnetic field. The induction is due to the change in

the orientation of the loop.

DEVELOP The magnetic flux through one turn of coil is (see Example 27.5)

1 cos cos(2 )B A BA BA ft rr

Therefore, using Faraday’s law, the induced emf is

[ 2 sin(2 )]BdNBA f ft

dt


The peak output voltage of an electric generator is21

peak 42 , where f NBA A d is the loop area in this case.

EVALUATE Substituting the values given, we find

2

peak 2 2 (1000/60 s)(250)(0.1 T) (0.1 m/2) 20.6 Vf NBA

ASSESS The value is typical of a car alternator, and this is attained when |sin(2 )| 1.ft

46. In the typical configuration for a generator shown in the Application on page 471, Example 27.5 showed that

peak 2 ,fNBA so2

peak /2 (6.7 kV)/[2 (60 Hz)(0.14 T)(0.75 1.3 m )] 130 turns.N fBA

47. INTERPRET We have a circuit formed by the rails, the resistance, and the conducting bar. As the bar slides along the

rails the circuit area increases and a current is induced. We are interested in the rate of work done by an external agent

to move the conducting bar.

DEVELOP To find the direction of the current in the loop, we note that since the area enclosed by the circuit, and the

magnetic flux through it, are increasing, Lenz’s law requires that the induced current opposes this with an upward

induced magnetic field.

To answer part (b), we make use of the result obtained in Example 27.4, which analyzed the same situation.

Therefore, the current in the bar was found to be | |/ / .I R Blv R

Since this is perpendicular to the magnetic field, the magnetic force on the bar ismagF IlB (to the left in Fig. 27.39).

The agent pulling the bar at constant velocity must exert an equal force in the direction of v.

EVALUATE (a) From the right-hand rule, the induced current must circulate CCW. (Take the positive sense of

circulation around the circuit CW, so that the normal to the area is in the direction of B, into the page.)

An alternative way to determine the direction of the current is to note that the force on a (hypothetical) positive

charge carrier in the bar, ,F qv B r rr

is upward in Fig. 27.39, so current will circulate CCW around the loop

containing the bar, the resistor, and the rails (i.e., downward in the resistor). (The force per unit positive charge is

the motional emf in the bar.)

(b) The rate of work done by the external agent is2( ).

Blv

RP F v IlBv

r r

ASSESS The conservation of energy requires that the work done by the agent be equal to the rate energy is

dissipated in the resistor (we neglected the resistance of the bar and the rails),2 2 2( / ) ( ) / .I R Blv R R Blv R

48. (a) The motional emf (mentioned in part (a) of the previous solution) is upward in the moving bar, and so acts like

the positive terminal of a battery. Thus, the positive terminal of the voltmeter should be connected to the top rail in

Fig. 27.39. (b) When an ideal voltmeter replaces the resistor, no current flows (since its resistance is infinite), and

no work must be done moving the bar. (However, during a brief instant when charge is separating in the bar, work

is done.)

49. INTERPRET The circuit consists of the rails, the resistance, the conducting bar, and the battery. As the bar slides

along the rails the circuit area increases and a current is induced. We are interested in the subsequent motion of the

conducting bar.

DEVELOP To analyze the subsequent motion of the bar, we first note that the battery causes a CW current

(downward in the bar) to flow in the circuit composed of the bar, resistor, and rails. (For positive circulation CW,

the right-hand rule gives a positive normal to the area bounded by the circuit into the page, so that the flux

B B A Blx rr

is positive. The length of the circuit is x, as in Example 27.4.) Thus, there is a magnetic force

magF Il B IlB rr r

to the right, which accelerates the bar in that direction. (Any other forces on the bar are

assumed to cancel, or be negligible.)

EVALUATE (a) As in Example 27.4, there is an induced emf / ,i Bd dt Blv opposing the battery (with the

negative sign indicating a CCW sense in the circuit). The instantaneous current is

( ) i BlvI t

R R

Thus, as the speed v increases, the induced current I (and the accelerating force) decreases.

(b) Eventually ( ), ( ) 0,t I and the magnetic forcemag ( ) 0.F I lB So the velocity v stays constant.

(c) When ( ) 0, 0,I Blv which implies that / .v Bl

27.8 Chapter 27

Although v doesn’t depend on the resistance, the value of R does affect how rapidly v approaches .v For large R,

I charges slowly and v takes a long time to reach .v

ASSESS In this problem, the equation of motion of the bar (mass m) is

2 2( )( ) v v l Bdv Blv lBm IlB

dt R R

which can be rewritten as

2 2dv l Bdt

v v mR

For0 0,v this integrates to

2 2ln(1 / ) / ,v v l B t mR or

2 2 /( ) (1 ).l B t mRv t v e The time constant,

2 2/ ,mR l B

depends on the resistance.

50. The situation is like that described in Example 27.4 and the solution to Problem 47. (a) / /I R Blv R

(0.5 T)(0.1 m)(2 m/s)/4 25 mA. (Neglect the resistance of the bar and rails.) (b)mag (25 mA)F IlB

3(0.1 m) (0.5 T) 1.25 10 N. (c)2 2

J (25 mA) (4 ) 2.5 mW.P I R (d) The agent pulling the bar must

exert a force equal in magnitude tomagF and parallel to .v Therefore, it does work at a rate (1.25Fv

310 N)(2 m/s) 2.5 mW. The conservation of energy requires the answers to parts (c) and (d) to be equal.

51. INTERPRET The problem involves a changing magnetic field that induces an electric field. Both fields exert a

force on a proton.

DEVELOP The induced electric field inside a long thin cylindrical solenoid, whose magnetic field is increasing with

time as ˆ,B btkr (take k into the page as in Fig. 27.30), can be found by modifying the argument in Example 27.11 for

radius .r R The magnitude of Er

is

1

2 2

dB rbE r

dt

The induced electric field circulates CCW around the direction of ,Br

or ˆ( /2)E bx jr for the given point on the

x axis ( 5 cm, 0)x y z inside the solenoid (whose axis we assume is the z axis). At 0.4 s,t the uniform

magnetic field is ˆ(0.4 s) .B b kr The electromagnetic force on a proton with the given velocity is

( ).F e E v B r r rr

EVALUATE Substituting the values given, we find the net force on the proton to be

19 6

18 16

1 ˆ늿( ) (1.6 10 C) (2.1 T/ms)(5 cm)( ) (4.8 10 m/s) (2.1 T/ms)(0.4 s)2

늿(8.40 10 N) (6.45 10 N)

F e E v B j j k

j i

r r rr


ASSESS In the region ,r R the induced electric field rises linearly with r. Since the proton is moving, the net

force on the proton is a vector sum of electric and magnetic force (i.e., the electromagnetic force).

52. The electric field has magnitude / .E F e If we suppose that this is the electric field induced by the changing

magnetic field in the solenoid, and use the axial symmetric approximation in Example 27.11 (modified as in the

accompanying exercise,2

B r B inside the solenoid, and2 1

22 ( / )), then | / | / ,rE r dB dt E r dB dt F e

19or | / | 2 / 2(1.3 fN)/(28 cm 1.6 10 C) 58.0 T/ms.dB dt F re

53. INTERPRET This is a problem about induction in a circular loop, with the flux change caused by a changing

magnetic field.

DEVELOP The flux through a stationary loop perpendicular to a magnetic field is ,B BA so Faraday’s law

(Equation 27.2) and Ohm’s law (Equation 24.5) relate this to the magnitude of the induced current:

2| / || | | / | | / |Bd dt A dB dt a dB dtI

R R R R

In terms of the charge moving through the circuit during the interval the magnetic field is changing, / ,I dq dt or

.dq Idt

EVALUATE Carrying out the integration, we find the total charge moving around the loop to be

2 22 2

2 11 1

( )a a

Q dq dB B BR R

ASSESS In the situation where the magnetic field is constant,1 2 ,B B there would be no induced current, and the

charge moved around the loop would be zero.

54. Initially, the flux through the flip coil is ,B NBA but is reversed to NBA when the coil is rotated180 , soB

2 .NBA The total charge which flows isav ,Q I t where Iav is the average induced current and t the time for

the

rotation. From Faraday’s and Ohm’s laws,av ( / )/ 2 / ,BI t R NBA R t hence 2 / .Q NBA R If the

properties of the coil are known and the total charge is measured, one can find the magnetic field

strength, /2 .B R Q NA

55. INTERPRET This problem is about the time constant of a series RL circuit.

DEVELOP In a series RL circuit, the current as a function of time is given by Equation 27.7:

/( ) (1 )Rt LI t I e

where0/I R is the final current.

EVALUATE From Equation 27.7, the time constant is

7.6 s 7.6 s11.0 s

ln(1 / ) ln(1 1/ 2) ln 2L

L t

R I I

ASSESS The time constant is inversely proportional to the resistance R. The physical meaning of the time

constant /L L R is that significant changes in current cannot occur on time scales much shorter than .L

56. Starting with/ ,Rt L

oI I e we have ( / ) ln(1 / ).t L R I I Here,0/ 45 V/3.3 13.6 A,I R so t

(2.1 H/3.3 ) ln(1 9.5/13.6) 0.759 s.

57. INTERPRET In this problem we are asked about the inductance and the long-time behavior of a series RL circuit.

27.10 Chapter 27

DEVELOP In a series RL circuit, the rising current as a function of time is given by Equation 27.7:

/( ) (1 )

Rt LI t I e

where0/I R is the final current.

EVALUATE (a) The current has risen to half its final value in 30 s. Thus, the above equation gives

/ /( ) 11 1

2

Rt L Rt LI te e

I

which can be solved to yield(2.5 k )(30 s)

ln 2 ln 2108 mH.RtL

(b) After a long time ( ),t the exponential term in Equation 27.7 is negligible, and the current is

0 50 V20 mA

2.5 kI

R

ASSESS After many time constants, there is no induced emf in the inductor and we can simply think of the

inductor is a conducting wire connecting the different parts of the circuit. Note that 0/I R is what you would get

by neglecting the inductance and using Ohm’s law.

58. INTERPRET We use the equation for current decay in an inductor, and energy stored in an inductor, to find the

time it takes to lose 90% of the energy stored in an inductor when the circuit becomes resistive.

DEVELOP The energy initially stored in the inductor is21

0 02.U LI The current through an RL circuit is given

by/

0 .Rt L

I I e For this problem, the initial current is 0 2.4 kA,I the inductance is 0.53 H,L and the resistance

is 31 m .R We want to calculate the time required to dissipate 90% of the initial energy.

EVALUATE

2 2 2 / 2 /

0 0

1 1

2 2

Rt L Rt LU LI LI e U e

so the time we’re looking for is when

2 / 2100% 90% 0.10. ln(0.10) ln(0.10) 20 s

2

Rt L Rt Le t

L R

ASSESS The initial energy stored is 1.5 MJ, so the average power loss is nearly 69 kW! Note also that we didn’t

actually use the initial current in our calculations.

59. INTERPRET This problem is about the rate of change of current in a series RL circuit at different instants.


/( ) (1 )

Rt LI t I e

where0/I R is the final current. The rate of change of current is 0 / .Rt LdI

dt Le

EVALUATE (a) For 0,t we have

0 60 V40 A/s

1.5 H

dI

dt L

(b) Similarly, for 0.1 s,t the rate is

/ / (22 )(0.1 s)/(1.5 H)0

0

(40 A/s) 9.23 A/sRt L Rt LdI dI

e e edt L dt

ASSESS The rate of change of current decreases exponentially with time. After many time constants, the current

is approximately equal to0/ ,I R and the rate of change becomes negligibly small.

60. As explained in Example 27.9, when the switch is opened (after having been closed a long time), the voltage across

R2 (which equals the inductor emf ) is2 2 2 0 2 1/ .V I R R R If we choose to limit this to no more than100 V, then

2 (100 V)(180 )/45 V 400 .R

61. INTERPRET This problem is about the buildup and decay of current in a series RL circuit.


/( ) (1 )

Rt LI t I e


where 0/I R is the final current. On the other hand, as the switch is thrown back to position B, the battery is

removed and the current decays according to Equation 27.8:

/

0( )Rt L

I t I e

EVALUATE (a) When the current is building up from zero, Equation 27.7 gives

/ (2.7 )(5 s)/(20 H)0 12 V(5 s) (1 ) 1 2.18 A

2.7

Rt LI e eR

(b) At 10 s,t the current has built up to

(2.7 )(10 s)/(20 H)12 V(10 s) 1 3.29 A

2.7 I e

This current decays when the switch is thrown back to B. Equation 27.8 (where t is the time since 10 s) gives

(2.7 )(15 s 10 s)/(20 H)(3.29 A) 1.68 AI e

ASSESS The time constant of the circuit is / (20 H)/(2.7 ) 7.4 s,L L R so 10 st corresponds to only

1.35 .L At this instant, the current is only 74% of its long-time value 0/ (12 V)/(2.7 ) 4.44 A.I R

One needs to wait many time constants for the current to approach .I

62. (a) As explained in Example 27.9, the inductor current is zero just after the switch is closed. At this instant, the

currents can be determined from the circuit with the inductance open-circuited, so that its branch can be removed.

Thus, 3 0,I and 1 2 1 2/( ) 12 V/(4 8) 1 A.I I R R

(b) After the currents have been flowing a long time, they reach steady values ( / 0),dI dt and the voltage across

the inductance is zero. The currents can be found by short-circuiting the inductance (see Example 27.9 again, and

refer to Chapter 25):

1

1 2 3 2 3

3 12 1 15

2 3

43 1 2 2 3 15

12 A2.14 A,

/( ) 4 8 2/10

0.429 A, and

/( ) 1.71 A.

IR R R R R

RI I I

R R

I I R R R I

(c) When the switch is reopened, no current flows through the battery’s branch,1 0,I which can be removed from

the circuit to calculate2 3I I at this instant. The induced emf acts to keep the current flowing at its value in part (b)

(as explained in Example 27.9), so2 3 1.71 A.I I

63. INTERPRET This problem is about the magnetic energy stored in a series RL circuit as a function of time.

DEVELOP The magnetic energy stored in an inductor is given by Equation 27.9,2( ) ( ) /2,U t LI t where

/( ) (1 ).Rt LI t I e Combining the two equations, the stored magnetic energy as a function of time is

27.12 Chapter 27

2 / 2 / 21( ) (1 ) (1 )

2

Rt L Rt LU t LI e U e

where2/2U LI is the steady-state value of the magnetic energy. When the stored energy is half its steady-state

value, ( )/ 1/2,uU t U we have

/ /2( )1 1(1 ) 1

2 2

u uRt L Rt LuU te e

U

EVALUATE Solving the above equation forut yields

2ln 1.28

2 1u

L Lt

R R

On the other hand, the current is half its steady-state value when1 ( / ) ln 2 1 ms .t L R Dividing these results, we

find

1

1.28 1.28(1 ms) 1.77 ms

ln 2 ln 2ut t

ASSESS Since1,ut t the magnetic energy reaches half its steady-state value after the current has already

surpassed half its steady-state value. In fact, the current atut t is

/( ) (1 ) 0.707

2

uRt L

u

II t I e I

64. If 2 21 1 1 1

0 02 4 4 2( ),U LI U LI then 1

02.I I Substituting this value for I in

/

0

Rt LI I e

gives /

1/2Rt L

e or

( / ) ln 2 (1.0 H/3.6 s) ln 2 193 m .R L t

65. INTERPRET This problem is about the magnetic energy stored and dissipated in a superconducting solenoid.

DEVELOP The magnetic energy stored in the superconducting solenoid can be found by using Equation 27.9,

2( ) ( ) /2.U t LI t To answer parts (b) and (c), we note that the current in an inductor cannot change instantaneously,

and the power dissipated in the copper is2

.P I R The power drops to one half its maximum original value, when

the current drops to1/ 2 times its initial value.

EVALUATE (a) In its superconducting state, the solenoid’s stored energy is

2 21

2

1(3.5 H)(1.8 kA) 5.67 MJ

2U LI

(b) If the power dissipated in the copper just after a sudden loss of superconductivity must be less than 100 kW, then

2 2

100 kW30.9 m

(1.8 kA)

PR

I

The maximum resistance is therefore equal to 30.9 m .

(c) The decay current as a function of time is given by Equation 27.8,/

0( ) .Rt L

I t I e Thus, the decay time for

0( )/ 1/ 2I t I is

0 3.5 Hln ln 2 39.3 s

30.9 m

ILt

R I

ASSESS Since the resistance of the superconducting solenoid is very small, its time constant is large

( / (3.5 H)/(30.9 m ) 113 s).L L R This means that it takes much longer for the current to decay, compared

to

that in an ordinary conductor.

66. The energy density in a field of this strength is2 8 2 7 21 3

0/2 (10 T) /(8 10 H/m) 3.98 10 J/mBu B (see

Equation 27.10). This is about (a)11

1.1 10 times the energy density content of gasoline3(44 MJ/kg 800 kg /m

10 33.52 10 J/m ), and (b) 2600 times that of pure

235 13 3 3 18 3U (8 10 J/kg 19 10 kg/m 1.52 10 J/m ).

67. INTERPRET In this problem we are asked to compare the magnetic energy density between a current-carrying

loop and a solenoid.


DEVELOP The magnetic field at the center of the loop is loop 0 /2B I R (see Equation 26.9). On the other hand,

the magnetic field inside a solenoid is solenoid 0 .B nI The energy density can be found by using Equation 27.10: 2

0/2 .Bu B

EVALUATE Using the equations above, we find the energy density at the center of the loop to be

22 2loop(loop) 0 0

2

0 0

1

2 2 2 8B

B I Iu

R R

In a long thin solenoid of the same radius,

2 2 2

(solenoid) 2solenoid 0

0

0 0

1( )

2 2 2B

B n Iu nI

so the ratio is

2 2(loop)

0

(solenoid) 2 2 2 2

0

/8 1

/2 4

B

B

I Ru

u n I n R

ASSESS As expected, the ratio is dimensionless (recall that n represents the number of turns per unit length of a

solenoid). The result shows that the energy density inside a solenoid is greater than that at the center of a loop by

a factor of2 2

4 .n R

68. The field inside such a wire is2

0 /2B Ir R (see Equation 26.18) so the energy within a unit length is

(see previous solution)

2 2 2 22 4

0 0 0

2 4 400 0

22 4 168 4

RR

BI r I IU dVU B r

dV rdrL L L R R

where we used 2 .dV rL dr

69. INTERPRET In this problem we want to verify that the time integral of the power dissipated through the resistor

in a series RL circuit is equal to the energy initially stored in the inductor.

DEVELOP Equation 27.8,/

0( ) ,Rt L

I t I e gives the current decaying through a resistor connected to an inductor

carrying an initial current0 .I The instantaneous power dissipated in the resistor is

2.RP I R

EVALUATE (a) Using the two equations above, the power dissipated in the resistor as a function of time is

2 2 2 /

0

Rt L

RP I R I R e

(b) In a time interval dt, the energy dissipated is ,RdU P dt so the total energy dissipated is

22 /2 2 / 2 20

0 0 00

0

1

( 2 / ) 2 2

Rt LRt L I RLe

U I Re dt I R LIR L R

ASSESS This is precisely the energy initially stored in the inductor. The result must hold true by energy

conservation.

70. The combination of Equations 23.7 and 27.10 implies that if2 21

0 02/2 ,E Bu E u B then 0 0/ 1/ .E B

Numerically,7 2

0 4 10 N/A and9 2 2

0(1/4 ) 9 10 N m /C , so0 01/

9 2 2 7 2 8(9 10 N m /C )/(10 N/A ) 3 10 m/s,

which is, in fact, the speed of light (see Section 29.5).

71. INTERPRET This is an induction problem with changing magnetic flux due to a change in area.

DEVELOP The magnetic flux through the loop is proportional to the vertical distance, y, it falls into the field

region, .B BA Bwy The rate of change of the flux is

( )Bd dBwy Bwv

dt dt

where / .v dy dt

EVALUATE (a) As long as the flux through the loop is changing, there is an upward magnetic force on the

induced current in the bottom wire, which may cancel the downward force of gravity on the loop.

(b) Faraday’s and Ohm’s laws give a magnetic force proportional to the vertical speed,

27.14 Chapter 27

2 2| | | / |i Bd dt wB vw BF IwB wB

R R R

When this equals mg, the terminal speed is

2 2t

mgRv

w B

(c) The flux of Br

is positive (into the plane of Fig. 27.42) for clockwise circulation around the loop, so the induced

current must be negative, or counterclockwise. The motional emf, ,dqv Brr

on positive charge carriers in the

bottom wire, is to the right, and the magnetic force on them, ,IdL Br r

is upward.)

ASSESS The terminal speed is proportional to the resistance R. This is not surprising because if R is very small,

then the induced current would be very large, giving rise to a large upward magnetic force to counter the

gravitational force mg, and hence the terminal speed would also be small.

72. INTERPRET A conductive disk is in a changing magnetic field, and we are asked to find the current density in the

disk and the rate of power dissipation in the disk. We will use Faraday’s law and the resistance of the individual

loops that make up the disk.

DEVELOP We will treat the disk as a set of infinitesimal loops with radius r, thickness h, resistivity , and width dr.

The resistance of each such loop, using ,L

AR is 2 .r

hdrR The magnetic flux through each loop is

2 2,B r bt r so the induced emf around the loop is2.d

dtb r The current density is given by

.RI

A hdrJ

To find the total power, we will integrate the power in each infinitesimal loop:0 .a

dP dI P dI

EVALUATE

(a)

2

2 2rhdr

br brJ

Rhdr hdr

(b)

2

0

2 22 3

2 0 0

2 4

2 2 2

8

a

a a

rhdr

P br dI

br brhdr brh b hdI P br dr r dr

R

b haP

ASSESS There are several interesting aspects of this problem. First, the current density is linear with r, and is

independent of the thickness h. This makes sense: a thicker disk would have more current, but the current density

would be the same. Second, the power actually depends on the fourth power of disk radius a, so increasing the size

of this disk increases the power dissipation dramatically. This phenomenon is used in metal detectors, and explains

why large metal objects are easier for metal detectors to find than small ones.

73. INTERPRET We qualitatively sketch current and power in a loop as a magnet passes through the loop.

DEVELOP As the magnet is approaching the loop, the flux through the loop is increasing to the right, which

creates a current opposing this change in flux. As the magnet leaves the loop, the flux is in the same direction but

decreasing, so the current is in the opposite direction.

Power depends on current squared, so it is positive in both cases.

EVALUATE See figure below.


ASSESS The graph is symmetric because the magnet is moving at a steady rate.

74. INTERPRET In Problem 47, we are shown a movable bar which completes a circuit in a magnetic field. That

problem uses Faraday’s law to find the direction of the current and the power. Here we extend the problem to find

the speed of the bar as a function of time with a constant force pulling the bar.

DEVELOP There are two forces on the bar: the constant external force eF and the Lorentz force due to the current

and the magnetic field. We write the total force as ,eF F IlB where l is the spacing between the rails and B is

the magnetic field. The current I is created by Faraday’s law:

d

dt

1so

dI

R R dt

Also,

dvF ma m

dt

We will use these equations to find a differential equation involving velocity, then solve for velocity.

EVALUATE The current is

1 1( ) ( )

d d BlvI BA Blx

R dt R dt R

The net force equation is then

2 2

e e

dv Blv B lm F lB F v

dt R R

We rearrange this to

2 2

eFdv B lv

dt m Rm

and then guess at a solution of the form ( ) :Ctv t A De

2 2

2 2 2 2

2 2

2 2

( )

and

Ct Cte

e e

Fdv B lDCe A De

dt m Rm

B l D B lDC C

Rm Rm

F F RB lA A

m Rm B l

The initial condition we are given is that (0) 0,v so D A and

2 2

2 2( ) 1

B lRm

teF Rv t e

B l

ASSESS This problem is quite interesting in that there is a terminal velocity: the highest speed reached by the bar

is2 2

( ) / .ev F R B l

27.16 Chapter 27

75. INTERPRET We find the total flux through a toroidal coil with circular cross-section by finding the flux through

the coil and using the equation for self-inductance. To find the flux through the coil, we use the magnetic field in a

toroidal coil from Chapter 26, and integrate.

DEVELOP Since the magnetic field varies across the interior of the coil, we’ll have to integrate. A diagram will

be helpful to show how we’re doing the integration, as shown below. We will use x for our integration variable,

integrating from x R to .x R The field inside the coil is given by Equation 26.21 as 0 ,2

NIB

r

where in this

case .r A x We find the total flux through one loop by integrating the flux d through each strip of width dx

shown in the figure, 0( ) (2 ),2

NId B r dA ydr

r

where2 2 2 2

( ) .y R x R r A Finally, we multiply

this flux by N loops in the toroid and find the self-inductance by using Equation 27.3, .I

L

EVALUATE

2 2 22

00

2

2 20

2 2 2

0

( )2

2

[ ( )]

( )

A R A R

A R A R

N I R r AN IN d ydr dr

r r

N IA A R

L N A A RI

ASSESS The more common approach is to treat the field as approximately constant and equal to the value at

.r A For further details, see Problem 27.76.

76. INTERPRET In Problem 27.75, we obtained an exact value for the inductance. Now we’ll apply the

approximation 1,R

A and show that this gives us the same answer as for a long, thin, straight solenoid with a length

equal to the circumference of this toroid, and an area equal to the cross-sectional area of the toroid.

DEVELOP We can factor2A out of the radical in the previous answer to obtain

2 2 1/ 2(1 (1 ( ) ) ),R

o AL N A then

use the binomial approximation.

EVALUATE

2 222 0

2

1

2 2o

N RRL N A

AA

A straight solenoid of radius R and length 2l A would have inductance

2 2( ) , whereo

NL n R l n

l

and

so

2 2 22

2 , so (2 )2 2

o

o

N RNl A L R A

A A

ASSESS Our approximate answer is exactly what we would obtain by use of Equation 27.4 directly.


77. INTERPRET We use Kirchhoff’s laws to find the current in a circuit element. The circuit includes an inductor,

so the equations we obtain using Kirchhoff’s laws will be differential equations.

DEVELOP We start by drawing our loops and nodes as shown in the figure below. Node A and node B give the same

information, so we will use only loop 1, loop 2, and node A. From node A we get 1 2 3 0.I I I From loop 1, we

obtain 1 1 2 2 0,I R I R and from loop 2 we have 3

2 2 0.dI

dtI R L We want to find the current 2I as a function

of time.

EVALUATE From node A: 1 2 3.I I I

Substitute this into loop 1:

3 1

2 3 1 2 2 2

1 2

( ) 0I R

I I R I R IR R

Substitute into loop 2:

3 1 3 3 3 1 2 2 1 2

2 3

1 2 1 2 1 2 1 2

0( ) ( )

I R dI dI I R R R R RR L I

R R dt dt R R L L R R L R R

We guess at a solution of the form3 ( ) ,CtI t A Be with initial condition

3 (0) 0I since the inductor acts as an

open circuit at first.

3 2 1 2

1 2 1 2

1 2

1 2

1 2 2

1 2 1 2 1

3

1

1

( )( ) ( )

( )

( ) ( )

(0) 0 0

Ct CtdI R R RBCe A Be

dt L R R L R R

R RC

L R R

R R RA A

L R R L R R R

I BR

BR

So

1 2( )1 2

3

1

( ) 1R R

L R Rt

I t eR

ASSESS The current gradually increases, with a time constant 1 2 1 2/ ( ).R R L R R

78. INTERPRET We calculate the self-inductance per length of a coaxial cable, using the flux through the area

between the two conductors. We’ll use Ampère’s law to find the magnetic field, and integrate over a rectangular

area between the inner and outer conductor.

DEVELOP We find the flux through the area between the conductors by integrating the flux d over a series of

strips of length l and width dr. The field at distance r from the center is 0

2.

I

rB

Once we find the flux, we use

IL to find the inductance.

EVALUATE

27.18 Chapter 27

2

2

ln2

o

bo

a

o

Id BdA ldr

r

Il dr

r

lL b

l a

ASSESS Note that the inner conductor must have a finite size or the inductance becomes infinite.

79. INTERPRET We find the energy stored in a toroidal coil with square cross-section, given the geometry of the

toroid and the current it carries. We will use Amprère’s law to find the magnetic field inside the coil, integrate to

find the total flux, and then find the inductance and thus the energy stored.

DEVELOP The energy stored in an inductor is21

2.U LI We will check the manufacturer’s claims that the energy

stored with a current 63 AI is 100 J.U First, we integrate the flux through strips as shown in the figure below.

The field at a distance r from the center is 0 ,2

NIB

r

and .d Bldr The dimensions of the coil are 1.5 m,R

0.228 m,l 2500,N and 63 A.I

EVALUATE

2 2

0

2

0

2 2

21

2

ln2 2

ln 12

ln 1 80 J4

R lo

R

o

N I N Ildr R lN d l

r R

N l lL

R

N lI lU LI

R

ASSESS The inductor will not perform as specified.

80. INTERPRET We determine the voltage induced across the wingtips of a plane flying through a magnetic field,

using Faraday’s law. The wings of the plane sweep out area at a certain rate, and the resulting change in flux

generates a voltage.

DEVELOP We use the rate at which area is swept out by the wings of the plane, times the magnetic field, to find

the change in flux .d

dt

From Faraday’s law, ,

d

dt

we then find the voltage induced across the wingtips.

The wingspan is 60 m,l the speed of the plane is 600 mi/h 268 m/s,v and the magnitude of the magnetic

field is 5

0.2 G 2 10 T.B We will assume that the plane, and the plane’s velocity, are both perpendicular

to .Br

EVALUATE The rate at which the wings sweep out area is ,dA

lvdt

so the change in flux is

d dAB Blv

dt dt

and the induced voltage is 0.322 V.Blv This is not sufficient to run much of anything,

even if it could be used.

ASSESS This voltage cannot be used by anything in the plane, anyway. If you hooked up wires to the wingtips,

those wires would have the same induced voltage, and no current would flow.


81. INTERPRET Here we investigate the feasibility of using the Earth’s magnetic field to induce an electric field

along the blades of a windmill. The blades sweep out area at a certain rate, and the resulting change in flux creates

a voltage by Faraday’s law.

DEVELOP Each blade sweeps out a circle once per second, so the rate of change in area is2

1 s

dA R

dt

and the voltage

induced between center and edge is2

1 s.R B The size of the windmill is 50 m,R and the horizontal component

of the Earth’s magnetic field is51 10 T.B

EVALUATE

2

79 mV1 s

R B

ASSESS Go ahead and use a generator: this voltage is not worth the effort.

82. INTERPRET We find the magnetic field change rate necessary to induce a given voltage in a coil. We use

Faraday’s law.

DEVELOP We are given the diameter and the number of turns of the coil: 2 mm 0.001 m,d r and

5000.N The induced voltage is 7.07 mV. The area of the coil does not change, so the change in flux is due

entirely to the changing magnetic field,

2( )

d dBN r

dt dt

We need to solve for .dB

dt

EVALUATE

20.45 T/s 450 T/ms

dB

dt N r

ASSESS The size of the field actually does not matter: it’s how fast the field changes.

28.1

ALTERNATING-CURRENT CIRCUITS

EXERCISES

Section 28.1 Alternating Current

14. Use of Equations 28.1 and 28.2 allows us to write rms2 2(230 V) 325 V,pV V and 2 f 1

2 (50 Hz) 314 s . Then the voltage expressed in the form of Equation 28.3 is1

( ) (325 V)sin[(314 s ) ].V t t

15. INTERPRET We’re given the rms voltage and asked to find the peak voltage.

DEVELOP The rms voltagermsV and the peak voltage

pV are related by Equation 28.1,rms / 2.pV V

EVALUATE Equation 28.1 gives rms2 2(6.3 V) 8.91 V.pV V

ASSESS The rms (root-mean-square) voltage is obtained by squaring the voltage and taking its time average, and

then taking the square root. Therefore, it is smaller than the peak voltage by a factor of 2.

16. (a) 2(208 V) 294 VpV (Equation 28.1), and (b)3 12 (400 Hz) 2.51 10 s (Equation 28.2).

17. INTERPRET We’re given the AC current in terms of sinusoidal function, and asked to deduce the rms current and

the frequency of the current.

DEVELOP As shown in Equation 28.3, the AC current can be written as

p sin( )II I t

wherepI is the peak current amplitude, is the angular frequency, and

I is the phase constant. Comparison of the

current with Equation 28.3 shows that its amplitude and angular frequency arep 495 mAI and

19.43 (ms) .

EVALUATE (a) Applying Equation 28.1 gives rms p/ 2 495 mA/ 2 350 mA.I I

(b) Similarly, using Equation 28.2 we have1/2 9.43 ms /2 1.50 kHz.f

ASSESS The phaseI is zero in this problem. Note that since the rms (root-mean-square) current is obtained by

squaring the current and taking its time average, and then taking the square root, it is smaller than the peak current

by a factor of 2.

18. The phase constant is a solution of Equation 28.3 for 0,t i.e., (0) sin( ).p VV V Since sin( ) sin( ),V V one

must also consider the slope of the sinusoidal signal function at 0.t In addition, the conventional range forV

usually runs from 180 to 180 , or .V Thus,1sin [ (0)/ ]V pV V when

0( / ) 0,dV dt butV

1sin [ (0)/ ]pV V when0( / ) 0dV dt according as (0) 0.V

(Here, we are taking

1|sin [ (0)/ ]| /2pV V or 90 , as

common on most electronic calculators, since the sine function is one-to-one only in such a restricted range.) For

signal (a) in Figure 28.25, we guess that (0) / 2pV V (since that curve next crosses zero about halfway

between /2 and ) and the slope at zero is positive, so1

sin (1/ 2) /4 or 45 .a (This signal is

sin( ) sin( /4),p a pV t V t which leads a signal with zero phase constant by 45 .) For the other signals,

(b) (0) 0,V 0( / ) 0,dV dt so 0;b (c)0(0) , ( / ) 0,pV V dV dt so

1 1sin (1) sin (1) /2c or 90 ;

(d) (0) 0,V 0( / ) 0,dV dt so d or 180 ; and (e)0(0) , ( / ) 0,pV V dV dt so

1sin ( 1)e

1sin ( 1) /2 or 90 .

Section 28.2 Circuit Elements in AC Circuits

19. INTERPRET In this problem we want to find the rms current in a capacitor connected to an AC power source.

28

28.2 Chapter 28

DEVELOP The amplitude of the current in a capacitor is given by Equation 28.5:

p p

p p1/C

V VI CV

X C

Using Equation 28.1, the corresponding rms current is rms rms .I CV

EVALUATE Substituting the values given in the problem statement, we find the rms current to be

rms rms (2 60 Hz)(1 F)(120 V) 45.2 mAI CV

ASSESS The capacitive reactance is31/ 2.65 10 .CX C In this circuit, the current in the capacitor leads

the voltage across the capacitor by 90 .

20. The equations in Table 28.1 (expressed in rms values) give,rms ,rms6.3 V/470 13.4 mA; 2 (60 Hz)R CI I

, rms(10 F) (6.3 V) 23.8 mA; (6.3 V)/[2 (60 Hz)(750 mH)] 22.3 mA.LI

21. INTERPRET This problem is about capacitive reactance at various angular frequencies.

DEVELOP From Equation 28.5, we see that the capacitive reactance is

1 1

2CX

C f C

EVALUATE (a) For 60 Hz,f the capacitive reactance is

6

1 1804

2 2 (60 Hz)(3.3 10 F)CX

f C

(b) For 1.0 kHz,f the capacitive reactance is

6

1 148.2

2 2 (1.0 kHz)(3.3 10 F)CX

f C

(c) Similarly, for 20 kHz,f the capacitive reactance is

6

1 12.41

2 2 (20 kHz)(3.3 10 F)CX

fC

ASSESS One can see that a capacitor has the greatest effect (largest reactance) at low frequency.

22. We could take the minimum safe voltage equal to the peak voltage. Thenrms2 / 2(1.4 A)/p p CV I X I C

[2 (60 Hz)(15 F)] 248 V at 60 Hz, and 14.9 V at 1 kHz.

23. INTERPRET This problem is about the capacitance of a capacitor that’s connected across an AC power source.

DEVELOP The fact that the capacitor and the resistor both pass the same current implies that

p p

p

1C

C

V VI R X

R X C

Therefore, the capacitance is 1/ .C R

EVALUATE Substituting the values given, we obtain

1 1 11.47 F

2 2 (60 Hz)(1.8 k )C

R fR

ASSESS Since ,CR X the greater the value of resistance R, the greater the capacitive reactance, and hence the

smaller the capacitance.

24. From Equation 28.7, /2 /2 .p pf V I L Since the ratio of the peak values of voltage and current is the same as

that of the rms values, 10 V/2 (2 mA)(50 mH) 15.9 kHz.f

Section 28.3 LC Circuits

25. INTERPRET This problem is about the resonant frequency of an LC circuit.

DEVELOP Using Equations 28.2 and 28.10, the resonant frequency can be written as

Alternating-Current Circuits 28.3

1

2 2f

LC

EVALUATE Substituting the values given, we find the resonant frequency to be

1 18.23 kHz

2 2 (1.7 mH)(0.22 F)f

LC

ASSESS The mechanical analog of the LC circuit is the mass-spring system whose angular frequency is

/ .k m Thus, the correspondence between the two systems is: L m and 1/ .C k

26. From Equation 28.10,2 2 1

1/ (2 /2.4 s) (18 mF) 8.11 H.L C

27. INTERPRET In this problem we want to know the capacitance range of a variable capacitor, given the frequency

range covered.


1

2 2f

LC

which can be solved to give2 2 2

1 1

4.

L f LC

EVALUATE The frequencies1 550 kHzf and

2 1600 kHzf correspond to

21 2 2 2

1

22 2 2 2

2

1 141.9 pF

4 4 (550 kHz) (2 mH)

1 14.95 pF

4 4 (1600 kHz) (2 mH)

Cf L

Cf L

Thus, the range of the capacitance is from 41.9 pF down to 4.95 pF.

ASSESS We find the capacitance to be inversely proportional to2.f Therefore, lower capacitance covers the

higher end of the frequency band.

28. (a) The inductance can be calculated from Equation

28.10:2 2 21/ ( /2 ) / (5 ms/2 ) /20 F 31.7 mH.L C T C (b) Figure 28.11 and the expressions for the

electric and magnetic energies for the LC circuit in the text imply

that21

2 pCV 21

2,pLI so / (25 mA) 31.7 mH/20 F 995 mV.p pV I L C

29. INTERPRET We find the inductance needed to make an LC circuit to have a desired resonance frequency with a

given value of capacitance.

DEVELOP The resonance frequency of an LC circuit is0 1/ ,LC so

0 1/2 .f LC The frequency desired is

689.5 10 Hz,of and the capacitance is12

47 10 F.C We solve for L.

EVALUATE 2 2

0 01/2 1/4 67.3 nH.f LC L f C

ASSESS This is a very small inductor. At high frequencies, small inductances such as this can have a large effect:

so much so that it becomes important to design circuits so as to minimize the inductance of individual lead wires!

Section 28.4 Driven RLC Circuits and Resonance

30. (a) From the expression for resonance in an RLC circuit,2 2 1

01/ (2 4 kHz) (20 mH) 79.2 nF.C L

(b) At resonance, 0,L CX X so 75 .Z R (c) At 3 kHz, (1/ ) (2 3 kHz 20 mH)L CX X L C

1(2 3 kHz 79.2 nF) 293 , and

2 2 2 2( ) (75 ) ( 293 ) 303 .L CZ R X X

31. INTERPRET Here we find the impedance of an LRC circuit at a given frequency.

DEVELOP 2 2( ) ,L CZ R X X where 1

C CX and .LX L The frequency given is

310 10 Hzf

3 120 10 s . The values of the circuit elements are 1.5 k ,R 65.0 10 F,C and

350 10 H.L

EVALUATE 2 2( ) 3.48 k .L CZ R X X

28.4 Chapter 28

ASSESS Note that at this frequency the capacitor has almost no effect compared to the other two circuit elements.

32. INTERPRET We use the equation for impedance to find the frequency for minimum impedance of a series RLC

circuit, and the value of that impedance.

DEVELOP 2 2( ) ,L CZ R X X where 1

C CX and .LX L The frequency at which Z is minimum will be

when .C LX X At that resonance frequency, the impedance will be just .Z R

The component values in this circuit are 18 k ,R 14 F,C and 0.20 H.L

EVALUATE

(a) 1 1 1

95 Hz2

C LX X L fC LC LC

(b) 18 k .Z R

ASSESS At resonance, the effects of the inductor and the capacitor cancel out, leaving only resistance.

33. INTERPRET We find the peak current through an RLC circuit at resonance, for three values of R in the circuit. We

use the fact that at resonance, .Z R

DEVELOP The peak voltage ismax 100 V.V The value of R is 10 k .R Since Z R at resonance, the peak

current will be max

max .V

RI

EVALUATE For max1max2

, 20 mA.V

RR I Similarly, for R

max 10 mAI and formax2 5 mA.R I

ASSESS At frequencies off the resonance peak, these calculations become somewhat more complicated. But at

resonance, Z R and everything becomes easy.

Sections 28.5 Power in AC Circuits and 28.6 Transformers and Power Supplies

34. The average power consumed by an AC circuit is given by Equation 28.14, av rms rms cos (120 V)P V I

(4.6 A)cos( 25 ) 500 W.

35. INTERPRET We use the average power of a lamp, as well as the RMS voltage and the power factor, to calculate

the RMS current.

DEVELOP cos ,RMS RMSP I V where cos 0.85 is the given power factor. 40 WP and 120 V,RMSV so

we simply solve for .RMSI

EVALUATE cos

390 mA.RMS

P

RMS VI

ASSESS The power factor actually matters with fluorescent lamps. With incandescent lamps, the impedance is

almost entirely resistive, so the power factor is almost exactly one.

36. INTERPRET We compare the power consumption of two circuits: one is purely resistive and in the other the

voltage leads the current. The difference in the power usage by these two circuits will be due to the difference in

power factors between the two circuits.

DEVELOP The average power consumption of a circuit is cos .RMS RMSP I V In the first circuit, the power factor

is cos 1, since the circuit is purely resistive. In the second, 20 . In each case, 20 ARMSI and 240 V.RMSV

EVALUATE For the first circuit, cos 4.8 kW.RMS RMS RMS RMSP I V I V For the second,

cos 4.5 kW.RMS RMSP I V

ASSESS This is a fairly direct application of a power calculation.

37. INTERPRET We use the transformer equation, and conservation of energy, to design a transformer that converts

230 VAC to 120 VAC.

DEVELOP The transformer equation is 2

12 1.

N

NV V The voltages are

1 230 VV and2 120 V,V and

1 1000.N We

solve for2 .N For the second part, we use the idea of conservation of energy and let

1 2 ,P P whereRMS RMSP I V and

1 7.3 A.I

EVALUATE

(a) 2

11 1 522 V

VN N turns.

(b) 1

21 1 2 2 2 1 14 V.

V

VIV I V I I

ASSESS Conservation of energy still holds!


PROBLEMS

38. (a) Equation 28.7 gives, rms rms / 120 V/[2 (60 Hz)(2.2 H)] 145 mA.LI V L (b) A similar calculation with

European values gives 333 mA.

39. INTERPRET This problem is about capacitive and inductive reactances, and how varying frequency affects them.

DEVELOP From Equations 28.5 and 28.7, the capacitive and inductive reactances can be obtained as

1 1, 2

2C LX X L f L

C fC

EVALUATE (a) From the above equation, the frequency of the applied voltage is

1 179.6 Hz

2 2 (1 k )(2 F)C

fX C

(b) The equalityL CX X implies

1.0 k2 H

2 2 (79.6 Hz)

C CX XL

f

(c) Doubling doublesLX and halves

CX soLX would be four times

CX at 159 Hz.f

ASSESS Capacitive reactanceCX is inversely proportional to , while inductive reactance

LX is proportional

to . A larger capacitor has lower reactance and a larger inductor has higher reactance.

40. Inductance is defined as the ratio of flux to current (Equations 32.1 and 27.3), and capacitance as that of charge to

potential difference (Equation 31.5). Thus, the units of inductive reactance ( )LX L are1 2s ( )/ /T m A V A

(the middle step following from Faraday’s law), and for capacitive reactance ( 1/ )CX C the units are 1/ s /V C V A (the middle step following from the definition of current).

41. INTERPRET In this problem a capacitor is connected across an AC generator. We are given the AC voltage and

asked about the current across the capacitor.

DEVELOP The current across the capacitor is given by Equation 28.4:

p p( ) cos sin( /2)I t CV t CV t

EVALUATE (a) The above equation shows that the peak current is

1

p p C (22 V)(377 s )(1.2 F) 9.95 mAI V

(b) The voltage at 6.5 mst is (remember that t is in radians)

1(6.5 ms) sin (22 V)sin[2 (60 s )(6.5 ms)] 14.0 VpV V t

(c) Similarly, the current is

1

p(6.5 ms) cos (9.95 mA)cos[2 (60 s )(6.5 ms)] 7.67 mAI I t

The magnitude of the current is 7.67 mA.

ASSESS In a capacitor, the current leads the voltage by 90.

42. We are given that1 110(1/ ),L C or

2

110/ .LC The reactances are equal if21/ ;LC therefore

1/ 10,

or 10 kH 3.16 kHz.f

43. INTERPRET In this problem an inductor and a lamp are connected across an AC source. We are given the AC

voltage and asked about the rms current across the lamp.

DEVELOP In a series circuit, the same current flows through the inductor and lamp. Since the ratio of the rms

quantities for a given circuit element equals that of the peak values, Equation 28.7 gives

rms rms

rms =L

V VI

X L

EVALUATE Substituting the values given, we find the rms current to be

rms

rms

90 V318 mA

(2 60 Hz)(0.75 H)

VI

L

28.6 Chapter 28

ASSESS The current in the inductor and the lamp lags the voltage by 90 .

44. Capacitors in parallel add, so the reactance of the combination is1

1 2[ ( )]CX C C and the rms current

is ,rmsCI 1 2 rms( ) .C C V (a) At a frequency of 1 kHz, 1 2 (3.4 mA)/[2 (1 kHz)(10 V)] 54.1 nF.C C

Thus,2C (54.1 2.2)nF 51.9 nF. (b) Dividing the rms currents at two frequencies, we get

2 1 rms,2 rms,1/ / ,f f I I

or 2 (1.2 mA/3.4 mA)f (1 kHz) 353 Hz.

45. INTERPRET This problem asks for the inductance that satisfies the resonance condition for a given range of

capacitances and frequencies.


1

2 2f

LC

which can be solved to give2 2 2

1 1

4.

C f CL

EVALUATE Using either condition,1 88 MHzf with

1 16.4 pF,C or2 108 MHzf with

2 10.9 pF,C we find

the inductance to be

22 2 2

1 1

22 2 2

2 2

1 10.199 H

4 4 (88 MHz) (16.4 pF)

1 10.199 H

4 4 (108 MHz) (10.9 pF)

Lf C

Lf C

ASSESS For a given inductance L, the capacitance is inversely proportional to2.f Thus, lower capacitance covers

the higher end of the frequency band.

46. (a) In an LC circuit, the peak current and voltage are related by / /p p p p pI q CV CV LC V C L (see

Example 28.3). Thus, (190 V) 0.025 F/340 H 1.63 A.pI (b) The current peaks one quarter of a period

after the voltage, or 1 1 1

4 4 2(2 / ) ( /2) 0.025 F 340 H 4.58 s.t T LC

47. INTERPRET This problem involves an oscillation in which energy is transferred back and forth between electric

and magnetic fields.

DEVELOP The electric energy stored in the capacitor is given by2

( ) ( )/2 ,EU t q t C where p( ) cosq t q t

(see Equation 28.9). Similarly, the magnetic energy stored in the inductor is2

( ) ( )/2,BU t LI t where

p p( ) / sin cos( /2)I t dq dt q t I t

The quantities are evaluated at 2

8 4( /8) 45t T (i.e., 1

8cycle). Note that phase constant zero

corresponds to a fully charged capacitor at 0.t

EVALUATE (a) From Equation 28.9, we obtain 1

2cos 45 .

p

q

q

(b) From the equation for electric energy, the ratio is

2 22 2

2 2

, p p

( ) ( )/2 ( ) 1cos cos 45

2/2

E

E p

U t q t C q tt

U q C q

(c) The ratio of the current is 1

2( )/ cos( /2) cos135 .pI t I t The direction of the current is away from

the positive capacitor plate at 0.t

(d) From the equation for magnetic energy,2 2 2 1

, p 2( )/ ( )/ cos 135 .B B pU t U I t I

ASSESS At one-eighth of a cycle, half of the total energy is magnetic and half is electric. This is illustrated in

Figure 28.11.

48. (a) The energy initially stored in the first capacitor is21

2(2 mF)(200 V) 40 J. First close switch B for one quarter

of a period of the LC circuit containing the 2000 F capacitor, or 1 1 1

4 4 2(2 / )B B B Bt T LC

1

2(100 H)(2 mF) 702 ms. This transfers 40 J to the inductor. Then open switch B and close switch A for one

quarter of a period of the LC circuit containing the 500 F capacitor, or 1 1

2 2(100 H)(0.5 mF) 351 ms.A Bt t


This transfers 40 J to the second capacitor from the inductor. Finally, open switch A. (b) When the second

capacitor has 40 J of stored energy, its voltage is 2(40 J)/(0.5 mF) 400 V.

49. INTERPRET This problem is about an LC circuit with damping due to the resistance. We are interested in the time

it takes for the peak voltage to be halved.

DEVELOP Equation 28.11 gives the charge as a function of time. Since ( ) ( )/ ,V t q t C the voltage as a function of

time can be written as

/ 2

p( ) cosRt LV t V e t

The peak voltage decays with time constant 2 / .L R Half the initial peak value is reached after a time

(2 / ) ln 2t L R (when/ 2 1

2).Rt Le

EVALUATE Since the period of oscillation is 2 2 ,T LC the number of cycles represented by t is

(2 / ) ln 2 ln 2 ln 2 20 mH50.4

(1.6 ) 0.15 F2

t L R L

T R CLC

ASSESS We have an oscillation which is underdamped. The larger the resistance, the more rapidly the oscillation

will decay.

50. If only half the energy is lost after 15 cycles, the damping is small and the energy varies like the square of

Equation 28.11, namely/ 2

tot cos .Rt L

pU U e t (The energy time constant is / ,L R one half the charge time

constant.) After 15 cycles, 15 15(2 / ),t T the fraction of energy remaining is15 / 2 15 /1

2cos 30 .RT L RT Le e

Take logarithms and use 1/ LC to get ln 2 15 30 ,L RT R LC from which we find

2 2ln 2 ln 2

(100 mH) 0.216 F30 30 5

C LR

51. INTERPRET This problem is about a series RLC circuit at resonance. We want to find the smallest resistance that

still keeps the capacitor voltage under its rated value.

DEVELOP In a series RLC circuit at resonance, the peak capacitor voltage is

p p

,p p

0

/C C

V R V LV I X

C R C

where 0 1/ LC is the resonant angular frequency.

EVALUATE The condition that, 400 VC pV implies

p

,p

32 V 1.5 H6.20

400 V 250 FC

V LR

V C

ASSESS Our results shows that,pCV is inversely proportional to R. This means that a larger resistor would be

required if the capacitor has a lower voltage rating.

52. (a) The peak capacitor voltage at resonance3 1

0( 1/ 1/ 50 mH 1.5 F 3.65 10 s )LC is

, res 0( / )(1/ )C p C pV I X V R C ( / ) / (100 V/10 ) 50 mH/1.5 F 1.83 kV.pV R L C (b) At other frequencies,

0

,2 2

(1.83 kV)( / )1

1 ( ) /

p

C p

L C

VV

Z C X X R

where we used Equation 28.12 and the answer to part (a). In terms of and0,

0 0

0 0

1L CX X LQ

R R C

where 18.3Q for this circuit (see the solution to Problem 45(b) and Problem 71). If we let0/ ,y then

,C pV

28.8 Chapter 28

2 2 2 2(1.83 kV)/ ( 1) .y Q y This function has a similar shape to the resonance curve in Figure 28.17 for the high-Q

circuit, except for starting at a positive value at the origin. Because the curve is so sharply peaked in this case, it may be

difficult to get good numerical results from a graph. One can see that, 1.2 kVC pV except for a narrow range of

frequency around0, which can be determined from the solutions of the equation

, 1.2 kV.C pV This is equivalent

to2 2 2 2 2( 1) ( 1) (1.83/1.2) 1 1.31,Q y y or

22 21 [ 1 1 4 (1.31)]/2 .y Q Q Thus, 1 0.0643y

0.967 or 1 0.0613 1.03. The corresponding unsafe frequency range is:3 1

0 00.967 3.53 10 s 1.03

3 13.76 10 s ,

or 562 Hz 599 Hz.f

53. INTERPRET This problem involves analyzing a phasor diagram for a driven RLC circuit.

DEVELOP Our diagram has three phasors,p p, ,R LV V and

p,CV representing the voltages across the resistor, the

inductor, and the capacitor, respectively. Since the resistor voltage is in phase with the current,pI is in the same

direction as p .RV The resonant frequency is 0 1/ .LC

EVALUATE (a) From the observation that / ,Lp p Cp pV I L V I C we conclude that2 2

01/ ,LC the

frequency is above resonance. (b) The applied voltage phasor is the vector sum of the resistor, capacitor, and

inductor voltage phasors, as shown below. The current is in phase with the voltage across the resistor, in this case

lagging the applied voltage (since1tan [( )/ ] 0)Lp Cp RpV V V by approximately 50 (as estimated from the

figure).

ASSESS Our circuit is inductive since .Lp CpV V Note that a positive means that voltage leads current, and

a negative means voltage lags current; at resonance,L CX X and 0.

54. (a) In the solution to Problem 45(b), we showed that, in an RLC circuit,

1/ 22

02

2 2res 0

11

1 ( ) /

p

L C

I L

I R CX X R

where 0 1/ .LC Since this expression is not changed when0/ is replaced by its reciprocal, the assertion

in this problem is true. (That is, 102

and02 give the same .pI In fact, for this particular circuit,

res/pI I

2 2 1/21 1

2 2[1 ( / )(2 ) ] ,L R C implies

2 4

3/ ,L R C which we can use in part (b).) (b) For

0 (at resonance),

res2 / 3, / ,C L pX X R I V R and 0 (the current is in phase with the applied voltage). For 102

(below

resonance), 1res2

4 4 / 3, ,C L pX X R I I and 60 (the current leads the applied voltage). For02

(above resonance), p p p4 4 / 3, / /L CX X R I V Z V R and 60 (the current lags the applied voltage). These

relationships are shown in the following phasor diagrams.


55. INTERPRET This problem is about power factor in a series RLC circuit.

DEVELOP The power factor of the circuit is

cosRp p

p p

V I R R

V I Z Z

The average power in the circuit is given by Equation 28.14:

2

rms rms rms rms rmscos ( )( / )P I V I I Z R Z I R

EVALUATE (a) Substituting the values given, we find the power factor to be

100 cos 0.333

300

R

Z

(b) The above equation gives2 2

rms (200 mA) (100 ) 4 W.P I R Note that the average AC power is given

by the same expression as the DC power if the rms current is used.

ASSESS The power factor must be between zero and 1. A purely resistive circuit has a power factor of 1, while a

circuit with only capacitance or inductance has a power factor of zero.

56. (a) The geometry of Figure 28.16, with the peak voltages expressed in terms of the peak current, shows that

,cos / ( )/( ) / .R p p p pV V I R I Z R Z (Alternatively, use Equations 28.12 and 28.13 and the trigonometric identity 2sec 21 tan .) Therefore, cos (100 )(0.8) 80 .R Z (b) The reactance of the circuit can be

expressed in terms of the inductance, the resonant frequency, and the given values by the use of Equation 28.12:

2 2 2 2 2

2 22 2

0 0

100 80 60 1/ (1 1/ )

(1 / ), or / 1 (60 / )

L CX X Z R L C L LC

L L

Since (2 60 Hz)(0.1 H) 37.7 60 ,L we can discard the unphysical solution (with2 2

0 / 0) to

find 0 1 (60/37.7), or

0 60 Hz 2.59 96.6 Hz.f (If the value of L had been 60 , there would have

been two physically acceptable solutions.)

57. INTERPRET This problem is about the relationship between power lost in transmission and power factor.

28.10 Chapter 28

DEVELOP We assume the average power supplied by the city to berms rms cos ,P I V where cos is the power

factor. The average power lost in the transmission line is2

rms .P I R

EVALUATE (a) With cos 1, the average power supplied to the city is

rms rms cos (200 A)(365 kV)(1) 73.0 MWP I V

and that the average power lost in the transmission line is

2 2

rms (200 A) (100 ) 4 MWP I R

Thus, the percent lost is / (4/73) 100% 5.5%.P P

(b) If the power factor in part (a) were 0.6 instead of 1.0, the percent lost would be / (5.5%)/0.6 9.1%.P P

(c) Since P is a constant, the larger P (which is proportional to cos ), the smaller the fraction of power lost,

/ .P P A large power factor is therefore better for the power plant owners.

ASSESS The power factor cos is between zero and 1. Our problem demonstrates that low power factors result in

wasted energy.

58. (a) The DC power output,out (10 A)(14 V) 140 W,P is 80% of the average AC power input,

rms(120 V)inP I

(where cos 1 is assumed for the charger). Thus,rms 140 W/(0.8 120 V) 1.46 A.I (b) The cost for 10 h of

operation is (140 W/0.8)(10 h)(9.5? kWh) 16.6?

59. INTERPRET This problem deals with DC power supplies. If the time constant RC is long enough, the capacitor

voltage will only decrease slightly before the AC voltage from the transformer rises again to fully charge the

capacitor.

DEVELOP The scenario is depicted in Figure 28.24. From the given DC output, we find the load resistance

to be (22 V)/(150 mA) 147 .R In one period of the input AC ( 1/ ),T f the capacitor voltage must decay

by less than 3%, or/

0.97.T RC

e

EVALUATE The above condition implies that

13.73 mF

ln(0.97) (60 Hz)(147 ) ln(0.97)

TC

R

ASSESS If the capacitance is large enough, the load current and voltage can be made arbitrarily smooth with

negligible decay.

60. For the damped oscillations of an RLC circuit, the voltage decays according to Equation 28.11, ( ) ( )/V t q t C

/2 cos ,Rt L

pV e twith frequency given by Equation 28.10. If in ten cycles ( 10 10(2 / ) 20 )t T LC the

peak voltage has decayed from 35 V to 28 V, then ln(35/28) /2 10 / ,Rt L R C L or

1(10 ) 27 mH/3.3 FR ln(35/28) 642 m .

61. INTERPRET We have an AC generator connected to a series RLC circuit, and we want to know its maximum peak

voltage when the circuit is at resonance.

DEVELOP The peak capacitor voltage isp p .C CV I X At resonance, the impedance is Z R and

p p/ .I V R The

capacitive reactance is01/ / .CX C L C

EVALUATE The condition thatp p( / ) / 600 VCV V R L C implies

p

0.33 F(600 V)(1.3 ) 2.73 V

27 mHV

ASSESS The inductor voltage at resonance is

p p p

p p 0L L

V V VL LV I X L

R R R CLC

which is the same asp.CV The two voltages cancel exactly at resonance. Note that

pCV andpLV are higher than p .V

62. INTERPRET Starting with the charge on the capacitor in an LC circuit, we find the current and the voltage, then

the energy stored in the capacitor and in the inductor. We sum the two to find the total energy, and show that it’s

constant.

DEVELOP We start with cos ,pq q t and differentiate with respect to time to find the current. We also use


q CV to find the voltage. The energy stored in the electric field of the capacitor is21

2,CU CV and the energy

stored in the magnetic field of the inductor is21

2.LU LI

EVALUATE 2

2

2cos cos .p pq qq

CC C Cq CV V t U t

2 2 21

2cos sin sin .

dq

p p L pdtq q t I q t U L q t 1 ,

LC so

22 2 21 1

2 2( ) sin sin .pq

L pLC CU L q t t

The total energy is2 2

2 2

2 2(sin cos ) .p pq q

C L C CU U U t t

ASSESS We have shown that the energy in this circuit is a constant.

63. INTERPRET In Example 28.4, we found a frequency at which the current in an RLC circuit is half the maximum.

Here we find a second frequency at which the current will be half the maximum. We use the equation for Z.

DEVELOP From the example, we have 11.5 F,C 8.0 ,R and 2.2 mH.L We also know that

2 2( ) ,L CZ R X X where 1

C CX and .LX L The current is given by Ohm’s law, ,V

ZI and we are

looking for a value of such that 2 .Z R

EVALUATE

2

2

22 2 22 2

2 2

24 2 4 22 2 2 2

2 2

12

1 14 3 2

1 13 2 2 3 0

? 3882, 10181}

Z R L RC

LR R L R L

C C C

L LR L L R

C CC C

The sign of does not matter. We need to convert to frequency using2

,f so {618 Hz,1620 Hz}.f

ASSESS The 618-Hz answer was given in the example, so the solution we need is 1620 Hz.f

64. For constantp p CV I X (i.e., independent of frequency), the same peak current will be supplied if the capacitive

reactances for the two connections are equal, i.e.,1 1 2 22 2 .f C f C Thus, for parallel and series combinations of

two equal capacitors,2

2 1 parallel series 1 1( / ) ( ) / 4 .a b a bf f C C f C C C C f

65. INTERPRET We have two capacitors connected first in series and then in parallel with an AC generator, and we

want to know their capacitances.

DEVELOP Equation 28.5 gives the rms current when capacitors are connected to an AC generator,

rms rms rms/ .CI V X CV In the parallel connection,1 2 ,C C C and we have

5

1 230 mA (2 10 V/s)( )C C

while in the series connection,1 2 1 2/( ),C C C C C and Equation 28.5 gives

5 1 2

1 2

5.5 mA (2 10 V/s)C C

C C

The two equations can be used to solve for1C and

2 .C

EVALUATE Simplifying the above two equations leads to1 2 47.7 nFC C and

2

1 2 (20.4 nF) .C C Eliminating

1C from the second equation and substituting into the first equation, we obtain the following quadratic equation:

2 2

2 2(47.7 nF) (20.4 nF) 0C C

Since the initial two equations are symmetric in1C and

2 ,C eliminating2C gives the same equation as the above, but

with2C replaced by

1.C So the solutions for1C and

2C are

2 21[(47.7 nF) (47.7 nF) 4(20.4 nF) ] 11.5 nF

2 and 36.2 nF

ASSESS The parallel connection yields a greater capacitance, and hence a larger current compared to the series

combination.

66. From Equation 28.11, the voltage is ( ) cos ,pV t V t so the frequency of this LC circuit is1 1cos ( ( )/ )pt V t V

1 1 1(35 ms) cos (8.5/14) 26.2 s .

(a) The peak current (there is only one current in an LC circuit, and the

28.12 Chapter 28

capacitance is given) is1/ (26.2 s )(2000 F)(14 V) 735 mA.p p C pI V X CV (b) If the voltage is peak

at 0,t the current is peak after one quarter of a period, at /2 59.9 ms.t

67. INTERPRET This problem involves designing a circuit to be used in our “black box.”

DEVELOP We want the output voltage to lead the input voltage (by 45 ). By inspecting Figure 28.16, we readily

notice that the voltage across a resistor leads the voltage across a capacitor in series with it. We also see that the

voltage across an inductor leads the voltage across a resistor in series with it. Both circuits can be adapted to the

criteria of the “black box” in this problem.

EVALUATE Case (1): RC circuit with AC input. In this circuit, it can be seen thatR CV V V and .C RI I I

In the corresponding phasor diagram shown below, CV lags I by 90°, RV and I are in phase, and V is the vector sum

of these (see Table 28.1). We drew I horizontally for convenience. The impedance is

2 2

2 2 2

2 2

1Rp Cpp

C

p p

V VVZ R X R

I I C

and the phase angle is

1tan

Cp C

Rp

V X

V R RC

The current I always leads V, because 0. (Recall that is defined by sinI I t pwhen

p sin( ).)V V t

The result implies that in a series RC circuit,RV leads the applied voltage, V, by an angle

1tan (1/ ),RCwhich may

be adjusted to 45° if 1.RC The peak voltage across the entire resistance is p p cos 45 / 2,R pV V V so if we

divide the resistance into two parts,1 2 ,R R R with

2/ 1/ 2,R R then the peak voltage across2R will

be 1p p2

(1/ 2) ,RV V as desired (rms voltages have the same ratio as peak voltages).

Case (2): RL circuit with AC input. When a capacitor is replaced with an inductor, the phasors forRV and I are still

parallel, butLV leads I by 90°. The voltage V is the vector sum of

RV and ,LV so the impedance is

2 2

p 2 2 2 2 2

p p

Rp Lp

L

V VVZ R X R L

V V

and the phase angle is

tanLp L

Rp

V X L

V R R


In this case, I always lags V, because 0. (Negative is in the same sense as ,t measured from V.)

In a series RL circuit,LV leads V by

190 tan ( / ),L R which equals 45° if .L R Again,p p / 2,LV V so if we

divide L into1 2 ,L L with 2 / 2,L L the peak voltage across

2L is /2.pV Both circuits are sketched below.

ASSESS We have shown how the circuit can be designed in two different ways to adapt to the criteria of the

“black box.” Our circuit conditions can be verified explicitly. Note thatoutV is the open-circuit output voltage. If a

load is connected across the output terminals, the magnitude and the phase of the voltage will be changed

accordingly.

68. The peak current in general is / ,p pI V Z while at resonance,res

/ .p pI V R Therefore,res1

2p pI I implies

2 22 ,R Z R X or | | | (1/ )| 3 .X L C R Since frequency is inherently positive, the minimum and

maximum 's are the positive roots of the quadratics derived from the above inequality,2 3 1/ 0,L R C or

2

1 3 3

2 2

R R

LC L L

For the given numerical values,6 2 1 1

[ 10 (163) 163] s 850 s (or 135 Hz), and

3 11.18 10 s

(or

187 Hz), respectively.

69. INTERPRET This problem involves a series RLC circuit at resonance. We are asked to find the resistance, the

inductance, and the capacitance.

DEVELOP At resonance, the impedance is ,Z R and the current isp p p/ /I V Z V R and 0.L CX X Away

from resonance,p p/ ,Z V I and

2 2| | .L CX X Z R

EVALUATE The resonance condition gives

p

p

20 V400

50 mA

VR

I

On the other hand, at half the resonant frequency, 1

21 kHz (2 kHz), the impedance is

p

p

20V1.33 k (10/3)

15 mA

VZ R

I

which gives

2 2 2 91| | (10/3) 1

3L CX X Z R R R

WithLX L and 1/ ,CX C we obtain the following conditions:

28.14 Chapter 28

0 010 02

1 1 1 910,

2 3L L R

C C

These equations can be solved for C and L, with the following result:

0

2 2

0

2 91 2 91(400 )67.5 mH

9 9(2 2 kHz)

1 193.8 nF

(2 2 kHz) (67.5 mH)

RL

CL

ASSESS Below resonance, capacitive reactance dominates, with .C LX X

70. In the parallel RLC circuit, the currents in each element add to give the total current, so we apply phasor currents to

the mode law,R L CI I I I as shown.

RI is in phase with V (which is the same across each element, i.e.,

),R L C CV V V V I leads V by 90°, andLI lags V by 90°. The peak values are

, ,/ , / ,R p p C p p CI V R I V X and

, / .L p p LI V X The peak total current is / .p PI V Z The pythagorean theorem applied to the phasor diagram gives

2 2

, , ,( ) ,p R p L p C pI I I I

or

2

2

1 1 1 1.

L CZ X XR

71. INTERPRET In this problem we are asked to derive the Q-factor of the RLC circuit.

DEVELOP To derive the expression for Q, we first need to know the power in the circuit. From Equations 28.12

and 28.14 (with rms values), and

p p

p p

cosRV I R R

V I Z Z

the average power in a series RLC circuit can be written as

2

rmsrms rms rms rms 2

cos ( / ) ( / )V R

P I V V Z V R ZZ

The above expression show the power falls to half its resonance value2

rms( / )V R when 2 ,Z R or when

| | .L CX X R In terms of the resonant frequency, 0 1/ ,LC this condition becomes

2

2 20

0

1

RL L R

C L

The solutions of these quadratics, with 0, are

22

02

14

2

R R

L L

The Q-factor is then equal to0/ , where .


EVALUATE If 0R

L (equivalent to / ),R L C we can neglect the first term under the square root sign

compared to the second, obtaining0 /2 .R L The difference between these two values of is / ,R L

from which we obtain0 0/ / .Q L R

ASSESS The Q-factor measures the “quality” of oscillation. The smaller the resistance, the higher the Q-factor.

In the absence of resistance ( 0),R the LC circuit can oscillate indefinitely.

72. (a) With no load across the capacitor,outV is just

CV in a series RC circuit driven byin in, sin .pV V t The peak

voltage across the capacitor is ,p CI X wherein, /p pI V Z and

2 2 21/Z R C (Equation 28.12 with no

inductance).

Thus,2 2 2 2

, out, in, in,(1/ )( / 1/ ) / 1 ( ) .C p p p pV V C V R C V RC (b) out , in ,/ 1/ 2p pV V implies

22 ( ) 1,RC or 1/ .RC (Alternatively, a phasor diagram has

,R pV parallel to the current,,C pV lagging by 90°,

and 2 2

in, , , .p R p C pV V V Since, , ,( / ) ,R p p C p C C pV I R V X R RCV the same results follow.)

73. INTERPRET We find the RMS voltage of a sawtooth wave by finding the average value of the square of the

voltage.

DEVELOP The voltage goes linearly frompV to

pV in period ,T so the equation for ( )V t (for one period) is 2( ) ( 1).t

p TV t V We integrate

2V over one period, then divide by the period to find the average value of2,V then

take the square root to find .RMSV

EVALUATE

22 222 2

20 0 0

2

22

2

1 1 2 4 41 1

4 12

3 3

3

T T Tp

p

p

p

p

RMS

Vt t tV V dt V dt dt

T T T T TT

VV T T T V

T

VV V

ASSESS This seems reasonable, since for a sine wave2.

pV

RMSV

74. INTERPRET We use the equation for charge on a capacitor in an RLC circuit, and the differential equation for

an RLC circuit from Kirchhoff’s laws, and using these two we find an expression for .

DEVELOP / 2( ) cos ,Rt L

pq t q e t and 2

2 0.d q dq q

dt CdtL R We take the derivatives of q, substitute into the

differential equation, and solve the resulting equation for .

EVALUATE

/ 2

2

2 2222

( ) cos

( ) ( cos 2 sin )2

( ) (( 4 )cos 4 sin )4

Rt L

p

Rtp L

Rtp L

q t q e t

qq t e R t L t

L

qq t e R L t RL t

L

28.16 Chapter 28

Substituting these into the differential equation gives us, after some algebraic steps,

22 22( 4 4 )cos 0.

Rtp L

q

CLe L CR CL t In order for this equation to be true for all values of t, the term in

parentheses must be zero.

22 2 2

2

14 4 0

4

RL CR CL

LC L

ASSESS This reduces to 1LC

if 0.R

75. INTERPRET We find the voltage across each element in an RLC circuit, using the total impedance, the impedance

of each element, and the phase angles.

DEVELOP The voltage generated is sin .pV V t The RMS voltage generated is2

120 V .pV

RMSV The

frequency of the AC signal is1377 s , impedance is

2 2( ) ,L CZ R X X and the current in the circuit

is .V

ZI Voltage in a resistor is in phase with the current, voltage in a capacitor is 90° behind the current, and

voltage in an inductor is 90° ahead of the current.

We can usepV

p ZI to find the current in the circuit, then use

p pV I X to find the peak voltage in each element with

reactance .X Once we have that peak value, we can use the phase of the generator’s voltage to find the phase of the

voltage in each element and thus the actual voltage in each element.

The component values in the circuit are 50 ,R 0.5 H,L and6

20 10 F.C

EVALUATE We need the reactance of each element, and the impedance of the entire circuit. 188.5 ,LX L

1 132.6 ,C CX and 75.0 .Z The peak current is then 2 2.26 A.pV V

p Z ZI The phase difference

between the peak voltagepV and peak current

pI is 1tan 48.2 .L CX X

R

Now we find a time that fits the description for the voltage from the generator: ( ) sin 45 VpV t V t

0.00762 s.t (Note that we had to take the second solution to t in this case, to meet the stipulation that the voltage

is decreasing.) The phase angle of the voltage at this time is 2.873 rad 164 .t

For the resistor, voltage is in phase with current, which is 48.2° degrees behind the generator voltage. So

sin sin(164.6 48.2 ) 101.2 V.R p R pV I R I R

For the capacitor, voltage is 90° behind the current, so 1sin sin(164.6 48.2 90 ) 133.3 V.C p C C p CV I X I

For the inductor, voltage is 90° ahead of the current, so sin sin(164.6 48.2 90 )L p L C pV I X I L

189.4 V.

ASSESS Note that the sum of the voltages on each element is 45 V, which is just the voltage provided by the

generator at that instant.

76. INTERPRET We find the frequency at which the voltage across a capacitor is maximized, and also the maximum

voltage. We use both the impedance of the RLC circuit, and the impedance of the capacitor alone.

DEVELOP We are given, in Example 28.4, the component values 2.2 mH, 11.5 F,L C and 8.0 .R The

peak voltage is 20 V.pV The peak voltage across the capacitor will be ,pc p CV I X wherepV

p ZI and 1 .C C

X

We want to find the maximum value ofpcV and the frequency at which it occurs.

EVALUATE

2 21

1

( )

p

pc

C

VV

CR L

so set the derivative equal to zero and solve for .


2 2

22 2 22 1

2 2

2

2

( 2 ( 1))0

(1 ( ( 2)))

0 2 ( 1)

1 rad5737 913 Hz

s2

pc p

C

dV V CR L CL

d C CR L CL R L

CR L CL

Rf

LC L

We substitute this value of into the equation forpcV to find the peak voltage across the capacitor, 36.1 V.pcV

ASSESS Although the peak voltage on the capacitor is higher than the peak supply voltage, that’s ok: the voltage

across the inductor will be negative when the capacitor hits this voltage so Kirchhoff’s loop law is not violated.

77. INTERPRET We plot two sine waves with given frequencies, amplitudes, and relative phases.

DEVELOP ( ) 2 sin( ).RMSV t V t The frequency of each wave needs to be 1.0 kHz, so1

2000 s for

each. The RMS amplitude of the first one is 10 V, and the second, 7.1 V. The phase of the second one is 90

relative to the first.

EVALUATE See figure below.

ASSESS The actual phases are not stated: you may use any you like as long as the difference between the two

is 90°.

78. INTERPRET Here we find the capacitance value such that an LC circuit goes from a fully-charged capacitor to

zero-charge on the capacitor in 15 seconds. This voltage change is 1

4of the change that occurs during an entire

period, so what we’re really looking for is an LC circuit with a period of 60 seconds. We are given the value of L,

and need to find C.

DEVELOP 21 2 .LC

T LC We substitute 60 sT and 25 H,L and solve for C.

EVALUATE 2

243.65 F.T

LC

ASSESS This is a rather large capacitance, although obtainable with currently-available technology.

79. INTERPRET We find the maximum current in an RLC circuit at resonance.

DEVELOP At resonance, the impedance Z is just the resistance R, and the current is the same in all series-circuit

elements, so the maximum current in the inductor is just max

max .V

RI The max voltage is

max 8.0 V.V The resistance

is 5.5 ,R and we really don’t care what the inductor and capacitor values are.

EVALUATE 8.0 Vmax 5.5

1.45 A.I

ASSESS This current is within the safe limit.

80. INTERPRET We design an LC oscillator that has the same frequency as a spring-mass system, using our

knowledge of the frequency of an LC oscillator.

DEVELOP For a spring-mass oscillator ,k

m and for an LC oscillator 1 .

LC 5 kg,m 3

1.44 10 N/m,k

and 2.5 H.L We set the two angular frequencies equal to each other and solve for C.

EVALUATE 31 1.39 10 F 1390 F.k m

m kLLCC

ASSESS In either case, the angular frequency is 17 Hz.

28.18 Chapter 28

ELECTRIC CHARGE, FORCE, AND FIELD

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