Ch23

PART 4 ELECTROMAGNETISM

CHAPTER 23 ELECTRIC CHARGE, FORCE, AND FIELD

ActivPhysics can help with these problems: Activities 11.1–11.8

Section 23-2: Electric Charge

Problem1. Suppose the electron and proton charges differed by one part in one billion. Estimate the net charge you would carry.

SolutionNearly all of the mass of an atom is in its nucleus, and about one half of the nuclear mass of the light elements in livingmatter (H, O, N, and C) is protons. Thus, the number of protons in a 65 kg average-sized person is approximately12

27 2865 167 10 2 10( ) ( . ) kg kg = × ×− ¼ , which is also the number of electrons, since an average person is electrically

neutral. If there were a charge imbalance of q q eproton electron− = −10 9 , a person’s net charge would be about

± × × ×−2 10 1028 9 1 6 10 3 219. . ,× = ±− C C or several coulombs (huge by ordinary standards).

Problem2. A typical lightning flash delivers about 25 C of negative charge from cloud to ground. How many electrons are

involved?

SolutionThe number is Q e= == × = ×−25 16 10 1 56 1019 20 C C. . .

Problem3. Protons and neutrons are made from combinations of the two most common quarks, the u quark and the d quark. The

u quark’s charge is + 23 e while the d quark carries − 1

3 e. How could three of these quarks combine to make (a) aproton and (b) a neutron?

Solution(a) The proton’s charge is 1 2

323

13e e e e= + − , corresponding to a combination of uud quarks; (b) for neutrons, 0 =

23

13

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

Problem4. A 2-g ping-pong ball rubbed against a wool jacket acquires a net positive charge of 1 Cµ . Estimate the fraction of the

ball’s electrons that have been removed.

SolutionIf half the ball’s mass is protons, their number (equal to the original number of electrons) is 1 g=mp . The number ofelectrons removed is 1 Cµ =e, so the fraction removed is

( )( )

..

.11

10 167 1016 10 1

104 106 24

1911 C

C g

C gµ ==

eg mp

=× ×

× ×= ×

− −

−−

(a hundred billionth).

CHAPTER 23 541

Section 23-3: Coulomb’s Law

Problem5. If the charge imbalance of Problem 1 existed, what would be the approximate force between you and another person

10 m away? Treat the people as point charges, and compare the answer with your weight.

SolutionThe magnitude of the Coulomb force between two point charges of 3.2 C (see solution to Problem 1), at a distance of 10 m,is kq r2 2 9 2 2 2 89 10 3 2 10 9 22 10= == × ⋅ = ×( / )( . ) . N m C C m N. This is approximately 1.45 million times the weight of anaverage-sized 65 kg person.

Problem6. Find the ratio of the electrical force between a proton and an electron to the gravitational force between the two. Why

doesn’t it matter that you aren’t told the distance between them?

SolutionAt all distances (for which the particles can be regarded as classical point charges), the Coulomb force is stronger than thegravitational force by a factor of:

FF

ker

rGm mp e

elec

grav =

FHG

IKJFHG

IKJ

2

2

2

=× ⋅ ×

× ⋅ × ××

−

− − −

( / )( . )10 / )( . )( . )

. .119 10 16 10

167 10 911 102 3 10

9 2 2 19 2

2 2 27 3139 N m C C

(6.67 N m kg kg kg ¼

The spacial dependence of both forces is the same, and cancels out.

Problem7. The electron and proton in a hydrogen atom are 52.9 pm apart. What is the magnitude of the electric force between them?

Solutiona0 52 9= . pm is called the Bohr radius. For a proton and electron separated by a Bohr radius, F ke aCoulomb = 2

02= '

( / )( . . ) .9 10 16 10 5 29 10 8 23 109 2 2 19 11 2 8× ⋅ × × = ×− − − N m C C m N.=

Problem8. How far apart should an electron and proton be so the force of Earth’s gravity on the electron is equal to the electric

force arising from the proton? Your answer shows why gravity is unimportant on the molecular scale!

SolutionThe electric force between a proton and an electron has magnitude ke r2 2= , while the weight of an electron is m ge . These areequal when

r ke m ge= =× ⋅ ×

×=

−

−2

9 2 2 19 2

31 29 10 1 6 10

911 10 9 85 08=

( / )( . )( . )( . / )

. N m C C kg m s

m

(almost fifty billion atomic diameters).

542 CHAPTER 23

Problem9. Two charges, one twice as large as the other, are located 15 cm apart and experience a repulsive force of 95 N. What is

the magnitude of the larger charge?

SolutionThe product of the charges is q q r F k1 2

2 2 9 2 2 10 20 15 95 9 10 2 38 10= = × ⋅ = × −Coulomb m N N m C C= =( . ) ( ) ( / ) . . If one charge

is twice the other, q q1 22= , then 12 1

2 102 38 10q = × −. C and q1 218= ± . . Cµ

Problem10. Earth carries a net charge of − ×4 3 105. . C The force due to this charge is the same as if it were concentrated at Earth’s

center. How much charge would you have to place on a 1.0-g mass in order for the electrical and gravitational forces onit to balance?

SolutionThe mass could be suspended at the Earth’s surface if the electric force were repulsive (q same sign as qE) andkqq R mgE E= 2 = . Thus,

q =×

× ⋅ − ×= −

−( )( . / )( . )( / )( . )10 9 8 6 37 109 10 4 3 10

1033 2 6 2

9 2 2 5 kg m s m

N m C C C.µ

Problem11. A proton is on the x-axis at x = 16. nm. An electron is on the y-axis at y = 0 85. nm. Find the net force the two exert on

a helium nucleus (charge +2e ) at the origin.

SolutionA unit vector from the proton’s position to the origin is −î, so the Coulomb force of the proton on the helium nucleus isF î îP,He nm nN.= = −k e e( )( )( ) ( . ) .2 16 01802−− = (Use Equation 23-1, with q1 for the proton, q2 for the helium nucleus, andthe approximate values of k and e given.) A unit vector from the electron’s position to the origin is −−$j , so its force on thehelium nucleus is Fe,He nm nN.= − =k e e( )( )( $) ( . ) . $2 0 85 0 6382−− j j= The net Coulomb force on the helium nucleus is the sumof these. (The vector form of Coulomb’s law and superposition, as explained in the solution to Problems 15 and 19, providesa more general approach.)

Problem12. Let V î= +V Vx y

$j be an arbitrary vector, with V its magnitude. Show that V=V is a unit vector—i.e., that itsmagnitude is 1.

SolutionThe magnitude of V=V is always V V= =V V= = 1, since the magnitude of V is V = V . (The components of V are notneeded here.)

Problem13. A charge q is at the point x y= =1 0 m , . Write expressions for the unit vectors you would use in Coulomb’s law

if you were finding the force that q exerts on other charges located at (a) x y= =1 1 m m, ; (b) the origin;(c) x y= =2 3 m m, . Note that you don’t know the sign of q. Why doesn’t this matter?

CHAPTER 23 543

SolutionA unit vector from rq = ( )1 0 m, , the position of charge q, to any other point r = ( , )x y is $ ( )n r r r= − −r q q= =

( ) ( ) .x y x y− − +1 1 2 2 m, m= The sign of q doesn’t affect this unit vector, but the signs of both charges dodetermine whether the force exerted by q is repulsive or attractive, i.e., in the direction of + $n or − $n. (a) When the othercharge is at position r n= = + = =( ), $ ( , ) ( ) ( , ) $1 1 0 1 0 1 0 12 m, m m m = j . (b) When r n= = − ÷( , ), $ ( , )0 0 1 0 m

( ) ( , ) .− + = − = −1 0 1 02 m î (c) Finally, when r n= = + = =( ), $ ( ) ( ) ( ) ( , )2 3 1 3 1 3 1 3 102 2 m, m m, m m m = =

0 316 0 949. . $.î + j

Problem14. A proton is at the origin and an electron is at the point x y= =0 0 36.41 , . . nm nm Find the electric force on the proton.

SolutionThe magnitude of the force is

Fpker

= =× ⋅ ×

+ ×= ×

−

−−

2

2

9 2 2 19 2

2 2 18 2109 10 16 10

0 0 36 107 74 10( / )( . )

( .41 . ). , N m C C

m N

and its direction is from the proton (at rp = 0) to the electron (at r îe = +( .41 . $) ),0 0 36j nm for an attractive force, at an

angle θ = = °−tan ( . .41) .1 0 36 0 413= to the x-axis. The vector form of Coulomb’s law, F r r r rp p e p eke= − 2 3( )− −= (see

solution of next problem) gives the same result:

F î

îp = − × ⋅ × − − +

= + ×

− −

−

( / )( . ) ( .41 . $) ( .41 . ) ( )

( . . $)

9 10 16 10 0 0 36 0 0 36 10

5 82 511 10

9 2 2 19 2 2 2 3 2 9 2

10

N m C C m

N.

j

j

= =

Problem15. A 9 5. - Cµ charge is at x y= =16 5 0 cm cm, . , and a −3 2. - Cµ charge is at x y= =4 11.4 , . cm cm Find the force on

the negative charge.

SolutionDenote the positions of the charges by r î1 16 5= ( $)+ j cm for q1 9 5= . , Cµ and r î2 4 11= +( .4 $)j cm for q2 3 2= − . C.µThe vector from q1 to q2 is r r r= −2 1 , and a unit vector in this direction is $ ( ) .r r r r r= − −2 1 2 1= The vector form of

Coulomb’s law for the electric force of q1 on q2 is F r r r r12 1 2 2 1 2 13

= − −kq q ( ) .= (This gives the Coulomb force betweentwo 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:

F î î

î

12

9 2

2 2 2 3 2 39 10 9 5 3 2 4 11 16 5

4 16 11 5

14 2 7 37

=× ⋅F

HGIKJ − ×

+ − −

− + −

= −

N mC

C C cm cm

N,

( . )( . ) ( .4 $ $)[( .4 ) ( ) ]

( . . $)

µ µj j

j

=

with magnitude 16.0 N and direction θ = − °27 3. to the x-axis (negative angle measured CW).

Problem16. A charge 3q is at the origin, and a charge −2q is on the positive x-axis at x a= . Where would you place a third charge

so it would experience no net electric force?

544 CHAPTER 23

SolutionThe reasoning of Example 23-3 implies that 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 isF kQ qx q x ax = − −− −[ ( ) ],3 22 2 so Fx = 0 implies that 3 22 2( ) ,x a x− = or x xa a2 26 3 0− + = . Thus,x a a a= ± − =3 9 32 2 ( ) .3 6± a Only the solution ( ) .453 6 5+ =a a is to the right of x a= .

Problem17. A 60- Cµ charge is at the origin, and a second charge is on the positive x-axis at x = 75 cm. If a third charge placed at

x = 50 cm experiences no net force, what is the second charge?

SolutionIn order for the net force to be zero at a position between the first two charges, they must both have the same sign, i.e.,q1 60= Cµ at x1 0= and q2 0> at x2 75= cm. (Then the separate forces of the first two charges on the third are inopposite directions.) Therefore, for the third charge q3 at x F kq q x x q x xx3 3 3 1 3 1

22 2 3

250= = − − − =− − cm, [ ( ) ( ) ]kq q3

22

260 50 25 0[ ( ) ( ) ] C cm cmµ − −− = implies q2260 25 50 15= = C Cµ µ( ) .=

Problem18. You have two charges +4q and one charge −q . (a) How would you place them along a line so there’s no net force on

any of the three? (b) Is this equilibrium stable or unstable?

SolutionBy 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,

k q a k q q a( ) ( ) ( ) ,4 2 42 2 2= == −

which holds for any a. The equilibrium is unstable, since if −q is displaced slightly toward one charge, the net force on itwill be in the direction of that charge.

Problem 18 Solution.

Problem19. In Fig. 23-39 take q q1 68 34= = − C, C2µ µ , and q3 15= Cµ . Find the electric force on q3.

SolutionDenote the positions of the charges by r r î1 2 2= =$, ,j and r î3 2 2= + $j (distances in meters). The vector form ofCoulomb’s law (in the solution to Problem 15) and the superposition principle give the net electric force on q3 as:

F F F r rr r

r rr r

î

î

3 13 231 3 3 1

3 13

2 3 3 2

3 23

9 6 6 69 10 15 10 68 10 2 5 5 34 10 2 8

164 0 326

= +−

−

−

−+= + = × × × + − ×

= −

− − −kq q kq q( ) ( ) ( )( )[( )( $) ( ) $ ]

( . . $)

N

N,

j j

j

= =

CHAPTER 23 545

or F F Fx y3 32

32 167= + = . N at an angle of θ = = − °−tan ( ) .1

3 3 112F Fy x= to the x-axis.

FIGURE 23-39 Problem 19 Solution.

Problem20. In Fig. 23-39 take q1 25= Cµ and q2 20= Cµ . If the force on q1 points in the −x direction, (a) what is q3 and

(b) what is the magnitude of the force on q1?

SolutionThe positions of the charges are the same as in the previous problem, so the net force on q1 is

F r rr r

r rr r

î î1 1

2 1 2

1 23

3 1 3

1 33 1

23 2 2

33 2 2

25

2= +

LNMM

OQPP=

−+

−LNMM

OQPPkq q q kq q q( ) ( ) ( $) ( $)−

−

−

−

+ −

5

j j= = m m

.

(a) If F y1 0= , then q q2 3 0− = , or q3 20= C.µ (b) Then F î î19 2 3 29 10 25 20 4 1 61= × × − = −−( )( )( )5 . N C N.µ =

Problem21. Four identical charges q form a square of side a. Find the magnitude of the electric force on any of the charges.

SolutionBy symmetry, the magnitude of the force on any charge is the same. Let’s find this for the charge at the lower left corner,which we take as the origin, as shown. Then r1 0= , r2 = a$,j r î r î3 4= =a a( $), ,+ j and

F r rr r

r rr r

r rr r

î î î12 1 2

1 23

1 3

1 33

1 4

1 43

23 3 3

2

221 1

2 2= + +

LNMM

OQPP=

−− −

LNM

OQP= − +F

HGIKJkq kq a

aa

aaa

kqa

−

−

−

−

−

−

++

$ ( $)2

( $) ,j jj

(Use the vector form of Coulomb’s law in the solution to Problem 15, and the superposition principle.) Since î j+ $ ,= 2

F12 2 2 2 1

22 22 1 1 2 2 2 191= + = + =( ) ( ) ( )( ) . .kq a kq a kq a= = = =


546 CHAPTER 23

Problem22. Three identical charges +q and a fourth charge −q form a square of side a. (a) Find the magnitude of the electric force

on a charge Q placed at the center of the square. (b) Describe the direction of this force.

SolutionThe magnitudes of the forces on Q from each of the four charges are equal to kqQ a kqQ a= = =( ) .2 2 22 2= But the forcesfrom the two positive charges on the same diagonal are in opposite directions, and cancel, while the forces from the positiveand negative charges on the other diagonal are in the same direction (depending on the sign of Q) and add. Thus, the netforce on Q has magnitude 2 2 2( )kqQ a= and is directed toward (or away from) the negative charge for Q > 0 (or Q < 0).

Problem23. Three charges lie in the x-y plane: q1 55= Cµ at x y q= =0 2 0, . m; 2 at x y= =3 0 0. ; m, and q3 at x = 4 0. m,

y = 3 0. m. If the force on q3 is 8 0 15. $ ,î + j N find q2 and q3.

SolutionUsing the vector form of Coulomb’s law explained in the solution to Problem 15, and superposition, we can write the force onq3 as F r r r r r r3 3 1 3 1 3 1

32 3 2 3 2

3= +

− −kq q r r q[ ( ) ( ) ]− − − − . Substituting the given values, F î r3 18 15 2= +( $) , $j j N m,=

r î2 = 3 m, and r î3 4 3= ( $) ,+ j m we find ( $) [ ( $)( ) ( $)( ) ] .8 15 4 4 1 3 1 323 1

2 2 3 22

2 2 3 2î î î+ +j j j N m⋅ = + + + +− −kq q q= =

Equating x and y components, we get 8 4 17 1023 1

3 22

3 2 N m⋅ = +− −= = =kq q q and 15 N 17 3 1023 1

3 22

3 2⋅ = +− −m = = =kq q q .Dividing these equations and solving for q2 in terms of q1 55= Cµ , we find q q2

3 2110 17 52 9 143= =( ) ( )= == C.µ

Substituting this into either component equation, we get q k q q k32

13 2

23 2 1 28 4 17 10 15= ⋅ + = ⋅− − −( / )[ ] ( / ) N m N m= =

[ ] .q q13 2

23 2 117 3 10 116− − −+ == = Cµ

Problem24. Two identical small metal spheres initially carry charges q1 and q2, respectively. When they’re 1.0 m apart they

experience a 2.5-N attractive force. Then they’re brought together so charge moves from one to the other until theyhave the same net charge. They’re again placed 1.0 m apart, and now they repel with a 2.5-N force. What were theoriginal values of q1 and q2?

SolutionThe charges initially attract, so q1 and q2 have opposite signs, and 2 5 11 2

2. . N m= −kq q = When the spheres are broughttogether, they share the total charge equally, each acquiring 1

2 1 2( ).q q+ The magnitude of their repulsion is2 5 11

4 1 22 2. ( ) . N m= +k q q = Equating these two forces, we find a quadratic equation 1

4 1 22

1 2( ) ,q q q q+ = − orq q q q1

21 2 2

26 0+ + = , with solutions q q1 23 8= − ±( ) . Both solutions are possible, but since 3 8 3 8 1+ = − −( ) , theymerely represent a relabeling of the charges. Since − = ⋅ × ⋅ =q q1 2

2 9 2 2 22 5 9 10 16 7. ( / ) ( . ) , N m N m C C= µ the solutions areq1 3 8 16 7 40 2= ± + = ±( . ) . C Cµ µ and q2 40 2 3 8 6 90= + =m m. ( ) . , C Cµ µ= or the same values with q1 and q2

interchanged.

Section 23-4: The Electric Field

Problem25. An electron placed in an electric field experiences a 61 10 10. × − N electric force. What is the field strength?

CHAPTER 23 547

SolutionFrom Equation 23-3a, E F e= = × × = ×− −= =6 1 10 1 6 10 3 81 1010 19 9. . . / . N C N C (The field strength is the magnitude of thefield.)

Problem26. What is the magnitude of the force on a 2 0. - Cµ charge in a 100 N C/ electric field?

SolutionFrom Equation 23-3b, F qE= = = × −( )( / )2 100 2 10 4 C N C N.µ

Problem27. A 68-nC charge experiences a 150-mN force in a certain electric field. Find (a) the field strength and (b) the force that

a 35- Cµ charge would experience in the same field.

SolutionEquations 23-3a and b give (a) E = =150 68 2 21 mN nC MN C= . / , and (b) F = =( )( . / ) .35 2 21 77 2 C MN C N.µ

Problem28. A −10. - Cµ charge experiences a 10î-N electric force in a certain electric field. What force would a proton experience

in the same field?

SolutionThe electric field is E î î= − = −10 1 10 N C MN C=( ) / .µ The force on a proton is eE î= × − =−( . )( / )1 6 10 1019 C MN C−16. î pN.

Problem29. The electron in a hydrogen atom is 0.0529 nm from the proton. What is the proton’s electric field strength at this

distance?

SolutionThe proton in a hydrogen atom behaves like a point charge, for an electron one Bohr radius away (see solution toProblem 7), so Equation 23-4 gives E ke a= = × ⋅ × × = ×− −= =0

2 9 2 2 19 11 2 119 10 1 6 10 5 29 10 515 10( / )( . ) ( . ) . / . N m C C m N C

Problem30. A 65- Cµ point charge is at the origin. Find the electric field at the points (a) x y= =50 0 cm , ; (b) x = 50 cm,

cm c cm cmy x y= = − =50 25 75; ( ) , .

SolutionThe electric field from a point charge at the origin is E r r r( ) $ ,= =kq r kq r= =2 3 since $ .r r= =r (a) For r î= 0 5. mand q = = × ⋅ =65 9 10 65 0 5 2 349 2 2 2 C N m C C m MN Cµ µ, ( / )( ) ( . ) . / .E î î= (b) At r î= 0 5. ( $), m + jE î î= × × ⋅ =( / )( . )( $) ( . ) ( / )( $).9 65 10 0 5 0 5 2 8273 2 3 N m C m m kN C+ +j j= (The field strength is 117. / MN C at 45°

to the x axis.) (c) When r î E î= − = × ⋅ − − + =( . . $) ( . / )( . . $) [( . ) ( . ) ]0 25 0 75 5 85 10 0 25 0 75 0 25 0 755 2 2 2 3 2 3+ +j j m, N m C m m= =

( $) / ( / , ).− = = °296 888 936 108î E+ j kN C kN C θ x

548 CHAPTER 23

Section 23-5: Electric Fields of Charge Distributions

Problem31. In Fig. 23-40, point P is midway between the two charges. Find the electric field in the plane of the page (a) 5.0 cm

directly above P, (b) 5.0 cm directly to the right of P, and (c) at P.

SolutionTake the origin of x-y coordinates at the midpoint, as indicated, and use Equation 23-5. Let r± = ±( . )$2 5 cm j denote thepositions of the charges, and r that of the field point. A unit vector from one charge to the field point is ( ) ,r r r r− −± ±=

so the spacial factors in Coulomb’s law are $ ( )r r r r r ri i i ir r= = =2 3 3= = − −± ± . (a) For r r r r= +( . )$,5 0 1 cm j = − =

( . )$ ( . )$ ( . )$,5 0 2 5 2 5 cm cm cmj j j− = and r r r r2 2 7 5= = − − = ( . )$ cm j . Then

E r r= +

FHG

IKJ= ×

⋅FHG

IKJ −

LNM

OQP=k q

rqr

1 1

13

2 2

23

92

2 29 10 22 5 7 5

25 6N mC

C( . cm) cm

MN C2( )$ $

( . )( . / )$.µ

j jj

(b) For r î= ( . ) ,5 0 cm

E î î= ×

⋅FHG

IKJFH IK + −

−+

+

LNM

OQP= −9 10 2 5 0 2 5

5 0 2 55 0 2 5

5 0 2 55159

2

2 2 2 2 3 2 2 2 3 2N m

C C

cm MN Cµ ( . . $)

( ( . ) )( . . $)

( . . )( . / )$−−

..

j jj

= =.

(c) For r = 0,

E = ×⋅F

HGIKJFH IK

−−

LNM

OQP= −9 10 2

2 5 2 557 69

2

2 2 2 2N m

C C

cm MN Cµ $

( . )

$

( . )( . / )$j j

j.


Problem32. A 10. - Cµ charge and a 2 0. - Cµ charge are 10 cm apart, as shown in Fig. 23-41. Find a point where the electric field is

zero.


SolutionThe field can be zero only along the line joining the charges (the x-axis). To the left or right of both charges, the fields dueto each are in the same direction, and cannot add to zero. Between the two, a distance x > 0 from the 1 Cµ charge, the

CHAPTER 23 549

electric field is E î î= + − −k q x q x[ ( ) ( ) ],12

2210= = cm which vanishes when 1 2 102 2 C C cmµ µ= =x x= −( ) , or

x = + =10 2 1 cm=( ) 414. cm.

Problem33. A proton is at the origin and an ion is at x = 5 0. nm. If the electric field is zero at x = −5 nm, what is the charge on

the ion?

SolutionThe proton, charge e, is at rp = 0, and the ion, charge q, is at rI = 5î nm. The field at point r î= −5 nm is given byEquation 23-5, with spacial factors written as in the solutions to Problems 15 or 31:

E r r rr r

î î î( ) ( ) ( )( )

( ) .=−

−=

−+

− −∑ kq ke kqii

i

i3 3

55

5 5 nm nm

nm nm(10 nm)3

Therefore, E = 0 implies 2 10 53 3q e= =( ) ( ) ,= − or q e= −4 . (Note how we used the general expression for the electric field,at position r, due to a distribution of static point charges at positions ri.)

Problem34. For the situation of Example 23-3, (a) write an expression for the electric field as a function of x for points to the right

of the charge −q shown in Fig. 23-13. (b) Taking q = 1 0. Cµ and a = 1 0. m, plot the field as a function of position forx = 5 m to x = 25 m.

Solution(a) For points on the x-axis with x a> , the electric field from the charge +2q is in the x direction ($ )r î= , and that ofthe charge −q is in the negative x direction ($ )r î= − . Thus, E î( ) [ ( ) ( ) ]x a kq x a x a> = + − −− −2 2 2 (which is F=Q inExample 23-3). (b) For q = 1 Cµ and a = 1 m, E î( ) ( / )[ ( ) ( ) ],x x x> = + − −− −1 9 2 1 12 2 kN C with x in meters. A graph ofE for 5 25≤ ≤x is shown.


Problem35. (a) Find an expression for the electric field on the y-axis due to the two charges q in Fig. 23-11. (b) At what point is the

field on the y-axis a maximum?

550 CHAPTER 23

Solution(a) The electric field is the force per unit charge, so Example 23-2 shows that E( ) $( ) .y kqy a y= + −2 2 2 3 2j = (b) Themagnitude of the field, a positive function, is zero for y = 0 and y = ∞, hence it has a maximum in between. Setting thederivative equal to zero, we find 0 22 2 3 2 3

22 2 5 2= + − +− −( ) ( ) ( ),a y y a y y= = or a y y2 2 23 0+ − = . Thus, the field strength

maxima are at y a= ± = 2 (the directions at these points, of course, are opposite, by symmetry).

Problem36. Write an expression for the dipole moment vector of the dipole shown in Fig. 23-18.

SolutionThe vector separation of the positive charge from the negative charge is r r î î î+ −− = − − =a a a( ) ,2 and the dipole momentvector is q times this, p î= 2qa .

Problem37. A dipole lies on the y axis, and consists of an electron at y = 0 60. nm and a proton at y = −0 60. nm. Find the electric

field (a) midway between the two charges, (b) at the point x y= =2 0 0. , nm, and (c) at the point x = −20 nm, y = 0.

SolutionWe can use the result of Example 23-6, with y replaced by x, and x by −y (or equivalently, $j by î , and î by −$)j . ThenE( ) $( ) ,x kqa a x= + −2 2 2 3 2 j = where q e= = × −16 10 19. C and a = 0 6. nm. (Look at Fig. 23-18 rotated 90° CW.) Theconstant 2 2 9 10 1 6 10 2 889 2 2 19 2kq = × ⋅ × =−( / )( . ) ( . / ( ) . N m C C GN C) nm (a) At x kq a= = =0 0 2 2, ( ) $ E j=( . / )$ ( . ) ( . / )$.2 88 0 6 8 002 GN C GN Cj j= = (b) For x = 2 nm, E = + =−( . / )$( . )( . ) ( / )$.2 88 0 6 0 6 2 1902 2 3 2 GN C MN Cj j=

(c) At x = = + =−20 2 88 0 6 0 6 20 2162 2 3 2 nm GN C kN C, ( . / )$( . )( . ) ( / )$ .E j j=

Problem38. What is the electric field strength 10 cm from a point dipole with dipole moment 3.8 µC m⋅ (a) on the dipole’s

perpendicular bisector and (b) on its axis?

Solution(a) Equation 23-7a gives E kp y= = × ⋅ == =3 9 2 2 39 10 3 8 0 1 34 2( / )( . ) ( . ) . / . Nm C C m m MN Cµ (b) Equation 23-7b, for thesame distance, gives a field strength twice as large, E kp x= =2 683= .4 / . MN C

Problem39. The dipole moment of the water molecule is 6 2 10 30. × ⋅− C m. What would be the separation distance if the molecule

consisted of charges ±e? (The effective charge is actually less because electrons are shared by the oxygen and hydrogenatoms.)

SolutionThe distance separating the charges of a dipole is d p q= = × ⋅ × =− −= =6 2 10 16 10 38 830 19. . . C m C pm.

Problem40. You’re 1.5 m from a charge distribution whose size is much less than 1 m. You measure an electric field strength of

282 N C./ You move to a distance of 2.0 m and the field strength becomes 119 N C/ . What is the net charge of thedistribution? Hint: Don’t try to calculate the charge. Determine instead how the field decreases with distance, and fromthat infer the charge.

CHAPTER 23 551

SolutionTaking the hint, we suppose that the field strength varies with a power of the distance, E rn' . Then 282 119 15 2= == ( . ) ,n

or n = = −ln( ) ln( . ) . .282 119 0 75 3 00= = A dipole field falls off like r −3, hence the net charge is zero.

Problem41. Three charges form an equilateral triangle of side a. At one vertex is a charge +2q; at the other two vertices are charges

−q. The triangle is oriented with the charge 2q on the positive x axis and both charges −q on the y-axis. (a) Find anexpression for the electric field on the x-axis, in the approximation x aÀ . (b) Compare with Equation 23-7b to showthat your result in (a) is a dipole field, and give an expression for the magnitude of the triangle’s dipole moment.

Solution(a) With the charges positioned as shown, the electric field on the positive x-axis, due to the two negative charges at( , )0 2 ±a= , matches the field found in Example 23-2 (replace q with −q, a with a=2, y with x, and $j with î):F î= = =Q k q x a= − × + −2 42 2 3 2( ) ( ) . For the positive charge at ( , ),3 2 0a= the electric field on the x-axis to the right( )x a> 3 2= is just k q x a( )( )2 3 2 2− −= î (the unit vector in Equation 23-5 is î and the distance is x a− 3 2= ). The totalfield is the sum of these, E î( ) [( ) ( ) ].x kq x a x x a= − − +− −2 3 2 42 2 2 3 2= = = For x aÀ , one can use the binomialapproximation (see Appendix A): ( ) ( )x a x a x− = + +− −3 2 1 32 2= = . . . and x x a x( ) ( . . . ),2 2 3 2 24 1+ = +− −= = where. . . indicates terms of order a x2 2= or higher. Therefore, E î î( ) [ . . . ( . . . )] .x kq x a x kqax= + + − +− −2 1 3 1 2 32 3= '(b) This field is the same as Equation 23-7b, with p qa= 3 . (In general, the dipole moment for a distribution of pointcharges is ∑ ri iq . Also note that the field in part (a) can also be found from Equation 23-5, with $ ( )r r r r ri i i ir= =2 3

= − − asin the solution to Problem 15.)


Problem42. Three identical charges q form an equilateral triangle of side a, with two charges on the x-axis and one on the positive

y-axis. (a) Find an expression for the electric field at points on the y-axis above the uppermost charge. (b) Show thatyour result reduces to the field of a point charge 3q for y aÀ .

552 CHAPTER 23

Solution(a) The electric field on the y-axis ( )y a> 3 2= due to the two charges on the x-axis, follows from Example 23-2: F=Q =

2 2 14

2 3 2kqy y a$( )j + − = . That from the charge on the y axis is just kq y a$( )j − −3 2 2= (see Equation 23-4). The total field isE( ) $[ ( ) ( ) ],y kq y a y y a= + + −− −4 4 4 2 32 2 3 2 2j = for y a> 3 2= . (b) For y a y a y y aÀ , [ ( ) ( ) ] 4 4 2 32 2 3 2 2+ + − →− −=

4 2 2 3 23 2 2y y y y( ) ( ) ( ) ,− − −+ = so E( ) $ ( ) ( )$ ,y kq y k q y→ =−4 3 2 32 2j j = like the field from a point charge of magnitude 3q.


Problem43. A 30-cm-long rod carries a charge of 80 µC spread uniformly over its length. Find the electric field strength on the rod

axis, 45 cm from the end of the rod.

SolutionApplying the result of Example 23-7, at a distance a = 0.45 m from the near end of the rod, we get E kQ a a= + == ( )l( / )( ) ( .45 )( .45 . ) . / .9 10 80 0 0 0 30 2139 2 2× ⋅ + = N m C C m m m MN Cµ =

Problem44. A thin rod of length l carries charge Q distributed evenly over its length. A point charge with the same charge Q lies a

distance b from the end of the rod, as shown in Fig. 23-42. Find a point where the electric field is zero.

FIGURE 23-42 Problem 44.

SolutionAt a point between the right end of the rod and the point charge, the fields from each oppose one another, and can cancel.At such a point, a distance x from the right end of the rod, the electric field is zero when kQ x x kQ b x= =( ) ( )+ = −l 2, orx x b bx x x b b2 2 2 22 2+ = − + = +l l, ( ). or =

Problem45. A thin rod of length l has its left end at the origin and its right end at the x = l . It carries a line charge density given

by λ λ π= 02 2( ) sin( )x x= =l l , where λ 0 is a constant. Find the electric field strength at the origin.

CHAPTER 23 553

SolutionThe electric field at the origin, due to a small element of charge, dq dx= λ , located at position x, is d k dx xE î= − λ = 2.Using λ λ π= = =x x2

02= ( )sin( )l l and integrating from x = 0 to x = l , we find

E î î= −FHG

IKJ

FH IK = FH IKz0 02

02

0

ll

l l l

l

l k x dx k xλ π λ

ππsin cos

= − − = −î î( )( ) .k kλ π λ π0 01 1 2= =l l


Problem46. Two identical rods of length l lie on the x-axis and carry uniform charges ±Q, as shown in Fig. 23-43. (a) Find an

expression for the electric field strength as a function of position x for points to the right of the right-hand rod. (b)Show that your result has the 1 3=x dependence of a dipole field for x À l . (c) What is the dipole moment of thisconfiguration? Hint: See Equation 23-7b.


Solution(a) The field due to each rod, for a point on their common axis, can be obtained from Example 23-7: E kQ x x+ = −= ( ),l tothe right, and E kq x x− = += ( ),l to the left. The resultant field (positive right) is

E E E kQx x x

kQx x

= − =−

−+

FH IK=−

+ −1 1 2

2 2l l

l

l( ).

(b) For x À l, E kQ x¼ 2 3 l= . (c) Comparison with Equation 23-7b shows that the rods appear like a dipole with momentp Q= l.

Problem47. A uniformly charged ring is 1.0 cm in radius. The electric field on the axis 2.0 cm from the center of the ring has

magnitude 2.2 MN/C and points toward the ring center. Find the charge on the ring.

SolutionFrom Example 23-8, the electric field on the axis of a uniformly charged ring is kQx x a( ) ,2 2 3 2+ − = where x is positive awayfrom the center of the ring. For the given ring in this problem, − = + −2 2 2 4 12 2 3 2. / ( )( ) , MN C cm cm cmkQ = or Q =

( . / )( . ) ( / ) . .− × ⋅ = −2 2 5 59 9 10 01372 9 2 MN C cm N m C C= µ

554 CHAPTER 23

Problem48. Figure 23-44 shows a thin, uniformly charged disk of radius R. Imagine the disk divided into rings of varying radii r, as

suggested in the figure. (a) Show that the area of such a ring is very nearly 2π r dr. (b) If the surface charge density onthe disk is σ C m/ 2, use the result of (a) to write an expression for the charge dq on an infinitesimal ring. (c) Use theresult of (b) along with the result of Example 23-8 to write the infinitesimal electric field dE of this ring at a point onthe disk axis, taken to be the positive x axis. (d) Integrate over all such rings (that is, from r = 0 to r R= ), to showthat the net electric field on the disk axis is

E k xx R

= −+

FHG

IKJ2 1

2 2π σ .


Solution(a) The area of an anulus of radii R R1 2< is just π( ).R R2

212− For a thin ring, R r1 = and R r dr2 = + , so the area is

π π[( ) ] ( ).r dr r r dr dr+ − = +2 2 22 When dr is very small, the square term is negligible, and dA r dr= 2π . (This is equalto the circumference of the ring times its thickness.) (b) For surface charge density σ σ πσ, . dq dA r dr= = 2 (c) FromExample 23-8, dE k dq x x r k xr x r drx = + = +− −( ) ( ) ( ) ,2 2 3 2 2 2 3 22= =π σ which holds for x positive away from the ring’scenter. (d) Integrating from r = 0 to R, one finds E dEx

Rx= z0 , or

E k x r drx r

k xx r

k xx

xx Rx

RR

=+

=−

+= −

+

LNM

OQPz2 2 1 22 2 3 20 2 2

02 2 1 2π σ π σ π σ

( ) ( )

.= =

(Note: For x x x> =0, and the field is E k x x Rx = − + −2 1 2 2 1 2π σ[ ( ) ].= However, for x x x< = −0, andE k x x Rx = − + + −2 1 2 2 1 2π σ[ ( ) ]= . This is consistent with symmetry on the axis, since E x E xx x( ) ( ).= − − )

Problem49. Use the result of the preceding problem to show that the field of an infinite, uniformly charged flat sheet is 2π σk ,

where σ is the surface charge density. Note that this result is independent of distance from the sheet.

SolutionAn infinite flat sheet is the same as an infinite flat disk (as long as the dimensions are infinite in all directions, the shape isirrelevant). Thus, we can find the magnitude of the electric field from a uniformly changed infinite flat sheet by lettingR → ∞ in the result of the previous problem. Then, the limit of the second term is zero, and the magnitude is constant,E k= 2π σ . (The direction is perpendicularly away from (towards) the sheet for positive (negative) σ .)

CHAPTER 23 555

Problem50. A semicircular loop of radius a carries positive charge Q distributed uniformly over its length. Find the electric field at

the center of the loop (point P in Fig. 23-45). Hint: Divide the loop into charge elements dq as shown in Fig. 23-45,and write dq in terms of the angle dθ . Then integrate over θ to get the net field at P.


SolutionThis problem is the same as Problem 73, with θ 0 0= . Thus, E î( ) .P kQ a= 2 2=π

Problem51. The electric field 22 cm from a long wire carrying a uniform line charge density is 1.9 kN/C. What will be the field

strength 38 cm from the wire?

SolutionFor a very long wire ( )l À 38 cm , Example 23-9 shows that the magnitude of the radial electric field falls off like 1=r.Therefore, E E( ) ( )38 22 22 38 cm cm cm cm;= == or E( ) ( ) . / . / .38 22 38 1 9 110 cm kN C kN C= ==

Problem52. What is the line charge density on a long wire if the electric field 45 cm from the wire has magnitude 260 kN/C and

points toward the wire?

SolutionIf the electric field points radially toward the long wire ( )l À 45 cm , the charge on the wire must be negative. Themagnitude of the field is given by the result of Example 23-9, E k rr = 2 λ= , so λ = − ÷( / )( .45 )260 0 kN C m( / ) . / .2 9 10 6 509 2× × ⋅ = − N m C C mµ

Problem53. A straight wire 10 m long carries 25 Cµ distributed uniformly over its length. (a) What is the line charge density on

the wire? Find the electric field strength (b) 15 cm from the wire axis, not near either end and (c) 350 m from the wire.Make suitable approximations in both cases.

Solution(a) For a uniformly charged wire, λ µ= =Q=l 2 5. / . C m (b) Since r = =15 10 cm m¿ l and the field point is farfrom either end, we may regard the wire as approximately infinite. Then Example 23-9 gives E k rr = =2 λ=

556 CHAPTER 23

( / )( . / ) ( . ) / .2 9 10 2 5 0 15 3009 2 2× × ⋅ = N m C C m m kN Cµ = (c) At r = =350 10 m mÀ l, the wire behaves approximatelylike a point charge, so the field strength is kQ r= =2 9 6 2 29 10 25 10 350 184= × × × ⋅ =−( / ) ( ) . / . N m C m N C

Problem54. Figure 23-46 shows a thin rod of length l carrying charge Q distributed uniformly over its length. (a) What is the line

charge density on the rod? (b) What must be the electric field direction on the rod’s perpendicular bisector (taken to bethe y axis)? (c) Modify the calculation of Example 23-9 to find an expression for the electric field at a point P adistance y along the perpendicular bisector. (d) Show that your result for (c) reduces to the field of a point charge Q fory À l.


Solution(a) λ = Q=l. (b) The x components of the fields from symmetrically placed elements of charge, dq dx= λ at ±x, cancel, sothe net field is along the y axis (see Fig. 23-23). (c) Proceed exactly as in Example 23-9, except that the limits of integrationare from − +l l= =2 2to . Thus,

E k y dxx y

k y xy x y

ky y

y =+

=+

=+−

+

−

zλ λλ

( ).2 2 3 22

2

2 2 22

2

2 2 4==

=

=

=

=l

l

l

l

l

l

(d) For y À l, we can neglect l in the square root, so we obtain E k y kQ yy = =λl= =2 2 as for a point charge. (Of course, forl → ∞ , the result of Example 23-9 is recaptured.)

Section 23-6: Matter in Electric Fields

Problem55. In this famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil

drops in an electric field. With a field strength of 20 MN/C, what mass drop can be suspended when the drop carries anet charge of 10 elementary charges?

SolutionIn equilibrium under the gravitational and electrostatic forces, mg qE= , or m = × × ×−( . )10 1 6 10 19 C( / ) ( . / ) . .2 10 9 8 3 27 107 2 12× = × − N C m s kg= (Because this is so small, the size of such a drop may be better appreciatedin terms of its radius, R m= ( )3 4 1 3= =πρoil . Millikan used oil of density 0 9199 3. / g cm , so R = 9.46 mµ for this drop.)

Problem56. How strong an electric field is needed to accelerate electrons in a TV tube from rest to one-tenth the speed of light in a

distance of 5.0 cm?

CHAPTER 23 557

SolutionFor uniform acceleration, a eE m= = , electrons, starting from rest, reach speed v2 2= ax, traversing a region of length x.Therefore, ( ) ( ) ,c eE m x= =10 22 = or

E mcex

= =× ×

×= ×

−

−

2 31 8 24

200911 10 3 10

0 05512 10( . ( / )

)( . ). / . kg) m s

200(1.6 10 C m N C19

Problem57. A proton moving to the right at 3 8 105. /× m s enters a region where a 56 kN/C electric field points to the left. (a) How

far will the proton get before its speed reaches zero? (b) Describe its subsequent motion.

Solution(a) Choose the x axis to the right, in the direction of the proton, so that the electric field is negative to the left. If theCoulomb force on the proton is the only important one, the acceleration is a e E mx = −( ) .= Equation 2-11, with vox =

3 8 105. /× m s and vx = 0, gives a maximum penetration into the field region of x x a m eEox x ox− = − = =02 22 2v = v =

( . )( . / )( . )( / )

. .167 10 3 8 102 16 10 56 10

13527 5 2

19 3× ×

× ×=

−

−

kg m s C N C

cm

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

Problem58. An oscilloscope display requires that a beam of electrons moving at 8.2 Mm/s be deflected through an angle of 22° by a

uniform electric field that occupies a region 5.0 cm long. What should be the field strength?

SolutionReferring to Example 23-10, one sees that v =v = vy x y xqE x m= =tanθ ∆ 2 . Therefore, in this case, Ey = × ×−( . )911 10 31 kg( . / ) tan ( . )( . ) . / .8 2 10 22 16 10 0 05 3 096 2 19× ° × =− m s C m kN C=

Problem59. An ink-jet printer works by “steering” charged ink drops to the right place on the page by passing moving drops

through a uniform electric field that deflects them by the appropriate amount. Figure 23-47 shows an ink dropapproaching the field region, which has length l and width d between the charged plates that establish the field. Findan expression for the minimum speed a drop with mass m and charge q must have if it is to get through the regionwithout hitting either plate.


SolutionIf they enter the field region midway, moving horizontally, the maximum vertical deflection, during the transit time t = l=v ,can be d=2, for the ink drops to pass through. Thus, y = 1

22 1

22 1

2at qE m d= <( )( ) ,= =vl or v => l qE md .

558 CHAPTER 23

Problem60. An electrostatic analyzer like that of Example 23-11 has b = 7 5. cm. What should be the value of E0 if the device is to

select protons moving at 84 km/s?

SolutionFrom the analysis in Example 23-11, E m eb0

2 27 3 2 19167 10 84 10 1 6 10 0 075 982= = × × × =− −v = =( . )( / ) ( . )( . ) / . kg m s C m N C

Problem61. An electron is moving in a circular path around a long, uniformly charged wire carrying 2.5 nC/m. What is the

electron’s speed?

SolutionThe electric field of the wire is radial and falls off like 1=r (see Example 23-9). For an attractive force (negative electronencircling a positively charged wire), this is the same dependance as the centripetal acceleration. For circular motionaround the wire, the Coulomb force provides the electron’s centripetal acceleration, or − = − = −eE m ke mr r= = v =2 2λ . Thus,v = = == = × ⋅ × × × =− − −2 2 9 10 16 10 2 5 10 911 10 2 819 2 2 19 9 31 1 2ke mλ [ ( / )( . )( . / ) ( . )] . N m C C C m kg Mm/s.

Problem62. Figure 23-48 shows a device its inventor claims will separate isotopes of a particular element. (Isotopes of the same

element have nuclei with the same charge but different masses.) Atoms of the element are first stripped completely oftheir electrons, then accelerated from rest through an electric field chosen to give the desired isotope exactly the rightspeed to pass through the electrostatic analyzer (see Example 23-11). Prove that the device won’t work—that is, that itwon’t separate different isotopes.


SolutionWhen the device is in operation, an isotope, of nuclear charge q and mass m, is accelerated from rest to a speed v, in adistance d, by the field E1, where v =2

1 12 2= =a d qE m d( ) . This will be the proper speed to pass through the analyzer ifa qE m r qE m d r2 2

212= = == v = = =( ) , or E E d r2 12= = . 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 areproportional to q m= . )

Problem63. What is the line charge density on a long wire if a 6 8. - gµ particle carrying 2.1 nC describes a circular orbit about the

wire with speed 280 m/s?

CHAPTER 23 559

SolutionThe solution to Problem 61 reveals that λ = − = − × × ⋅ × =− −m kqv = =2 9 2 9 2 2 92 6 8 10 280 2 9 10 2 1 10( . )( / ) ( / )( . ) kg m s N m C C−141. / . C mµ (In this case, the force on a positively charged orbiting particle is attractive for a wire with negative linearcharge density.)

Problem64. The electron in a hydrogen atom has kinetic energy 2 18 10 18. .× − J Assuming the electron is in a circular orbit around

the central proton, estimate the size of the atom. (Although this problem gives a reasonable answer, the simple modelof an electron orbiting a proton is not consonant with the quantum mechanical description of the atom.)

SolutionIn a circular orbit, the electron’s acceleration ( )−v =2 r is provided by the Coulomb force of the proton ( ).F m ke mr= == − 2 2

Thus, r ke m ke Ek= = = × ⋅ × × = ×− − −2 2 12

2 12

9 2 2 19 2 18 119 10 16 10 2 18 10 5 28 10= v = =( / )( . ) ( . ) . N m C C J m (essentially aBohr radius, see Problem 7).

Problem65. A dipole with dipole moment 1.5 nC m⋅ is oriented at 30° to a 4.0-MN/C electric field. (a) What is the magnitude of

the torque on the dipole? (b) How much work is required to rotate the dipole until it’s antiparallel to the field?

Solution(a) The torque on an electric dipole in an external electric field is given by Equation 23-11; τ θ= × = =p E pE sin( . )( . / ) sin . .15 4 0 30 3 0 nC m MN C mN m⋅ ° = ⋅ (b) The work done against just the electric force is equal to the change in thedipole’s potential energy (Equation 23-12); W U pEf i= = − ⋅ − − ⋅ = ° − ° = ⋅ ×∆ ( ) ( ) (cos cos ) ( . )p E p E 30 180 1 5 nC m( . / )( . ) . .4 0 1866 112 MN C mJ=

Problem66. A molecule has its dipole moment aligned with a 1.2-kN/C electric field. If it takes 31 10 27. × − J to reverse the

molecule’s orientation, what is its dipole moment?

SolutionFrom Equation 23-12, the energy required to reverse the orientation of such a dipole is ∆U pE= 2 , therefore p =12

12

27 3 3031 10 12 10 129 10∆U E= == × × = × ⋅− −( . ) ( . / ) . . J N C C m

Problem67. Two identical dipoles, each of charge q and separation a, are a distance x apart as shown in Fig. 23-49. By considering

forces between pairs of charges in the different dipoles, calculate the net force between the dipoles. (a) Show that, in thelimit a x¿ , the force has magnitude 6 2 4kp x= , where p qa= is the dipole moment. (b) Is the force attractive orrepulsive?


560 CHAPTER 23

SolutionAll 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 thedirection of the right-hand dipole in Fig. 23-49. The right-hand dipole has charges +q at x a+ =2, −q at x a− =2 , each ofwhich experiences a force from both charges of the left-hand dipole, which are +q at a=2 and −q at −a=2. (There are forcesbetween four pairs of changes.) The Coulomb force on a charge in the right-hand dipole, due to one in the left-hand one, iskq q x x x xr r rl l l( )− −î= 3 (see solution to Problem 15), so the total force on the right-hand dipole is

F kqx x a x a x

kq a x ax x ax = −

+−

−+

LNM

OQP= −

−

−2

2 2 2 2

2 2 2 2

2 2 2 21 1 1 1 2 3î î

( ) ( )( )

( ).

(a) In the limit a x F kq a x x kq a x kp xx¿ , ( ) ,→ − = − = −2 3 6 62 2 2 6 2 2 4 2 4î î î= = = where p qa= is the dipole moment of bothdipoles. (b) The force on the right-hand dipole is in the negative x direction, indicating an attractive force.

Problem68. A dipole with charges ±q and separation 2a is located a distance x from a point charge +Q, with its dipole moment

vector perpendicular to the x axis, as shown in Fig. 23-50. Find expressions for the magnitude of (a) the net torque and(b) the net force on the dipole, both in the limit x aÀ . (c) What is the direction of the net force?


Solution(a) In the limit x aÀ , the torque on the dipole is p E× , where E is the field from the point charge Q, at the position of thedipole. With x axis in the direction from Q to p and y axis parallel to the dipole in Figure 23-50, p = 2qa$j and E î= kQ x= 2 .Then ττ = × = −2 22 2qa kQ x kQqa x$ ( )$j î k= = (i.e., into the page, or CW, to align p with E). (b) and (c) The Coulomb forceobeys Newton’s third law. The field of the dipole at the position of Q is (Example 23-6 adapted to new axes)Edip = −( )$,2 3kqa x= j so the force on Q due to the dipole is Q kQqa xEdip = −( )$.2 3= j The force on the dipole due to Q is theopposite of this, +( )$2 3kQqa x= j (magnitude 2 3kQqa x= , parallel to the dipole moment).

Paired Problems

Problem69. An electron is at the origin and an ion with charge +5e is at x = 10 nm. Find a point where the electric field is zero.

SolutionThe electron’s field is directed toward the electron (a negative charge) and the ion’s field is directed away from the ion (apositive charge). Therefore, the fields can cancel only at points on the negative x axis ( ),x < 0 since the directions are

opposite there and the smaller charge is closer. The field from one point charge is E îq q qx kq x x x x( ) ( )= − −=3, where

CHAPTER 23 561

q e xq= − =, 0 for the electron, and q e xq= =5 10, nm for the ion. The total field is zero when 0 3= − +−k e x x[( )

5 10 10 3e x x( ) ].− − − nm nm (See note to solution of Problem 33.) Since x x x< = −0, and x x− = −10 10 nm nm ,so this implies x x− −− − =2 25 10 0( ) , nm or 4 2 10 10 02 2x x+ − =( ) ( ) . nm nm The negative solution to this quadratic is

x =− − +

= − + = −[ ( ) ( ) ] . ( ) .10 10 4 10

42 5 1 5 8 09

2 2 nm nm nm nm nm.

Problem70. A proton is at the origin and an ion is at x = 5 0. nm. If the electric field is zero at x = −6 83. nm, what is the charge on

the ion?

SolutionIf the field is zero on the negative x axis (the other side of the proton from the ion), the ion must have a negative charge ofgreater magnitude than the proton’s. (See the previous solution.) Thus, e qI( . ) ( . ) ,6 83 6 83 52 2 nm nm nm− −= − + qI =

− = − = −( . . ) .1183 6 83 3 00 32= e e e (since ionic charges are multiples of e).

Problem71. A thin rod of length l has its left end at x = −l and its right end at the origin. It carries a line charge density given by

λ λ= 0

2

2xl

,

where λ 0 is a constant. Find the electric field at the origin.

SolutionThe electric field at the origin, due to an element of charge dq dx= λ , located at x, where − ≤ ≤l x 0, isd k dx x k dxE î î= =λ λ = =2

02( ) .l The total field is the integral of this over the rod,

E î î î( ) ) ( ) [ ( )] ( ) .0 00

02

02

0= = − − =−z (k dx k kl

l l l lλ λ λ= = =

Problem

72. Repeat the preceding problem for the case when λ λ= 0

4

4xl

.

SolutionWith λ λ= =x x2

02 4= l in the previous solution,

E( ) [ ( ) ] .0 13

03

04

2 04

0 3 0=FHG

IKJ =

FHG

IKJ − − =

−z k x dx k kλ λ λî î îl l

lll

Problem73. A thin, flexible rod carrying charge Q spread uniformly over its length is bent into a quarter circle of radius a, as

shown in Fig. 23-51a. Find the electric field strength at the point P, which is the center of the circle. Hint: ConsultProblem 50.

562 CHAPTER 23

SolutionIt should be clear from the symmetry that the electric field is along the radius bisecting the arc, so take this as the x-axis,with P at the origin, and θ as defined in Fig. 23-45. The electric field at P, from each charge element, dq d= =λ l

λ θ θa d at ,, has the same magnitude, dE k dq a k d a= = = =2 λ θ , but direction $ $r = −î sin cos ,θ θj as sketched. λ = Q=lis constant, and the arc extends from θ 0 45= ° to π θ− = °0 135 , so the total field at P is the integral of dE$r from θ 0 to π θ− 0:

E( ) ( ) ( $ ) ( ) $

( ) .

0

2 20

0

0

0= − = − −

= =

− −zk a d k a

k a k a

λ θ θ θ λ θ θ

λ θ λ

θ

π θ

θ

π θ= =

= =

î î

î î

sin cos cos sin

cos 0

j j

Here, we used sin cos cos0 0θ π θ θ π θ= − = − −sin( ), ( ),0 0 and θ 0 45= °. In terms of the total charge, λ = =Q=l

Q a=( ),12 π so E( ) .0 2 2 2= kQ aî=π [Note: in general, l = −( ) .]π θ2 0 a

FIGURE 23-51 Problems 73 and 74 Solution.

Problem74. A thin, flexible rod carrying charge Q spread uniformly over its length is bent into a circular arc of radius a, as shown

in Fig. 23-50b. Find the electric field strength at the point P, which is the center of the circular arc.

SolutionThis problem is the same as the previous one, except θ 0 60= °. Thus, E( ) ( )(0 1

3= ° =k a Q a= = π )2 cos 60î 3 2kQ aî=π .

Problem75. Ink-jet printers work by deflecting moving ink droplets with an electric field so they hit the right place on the paper.

Droplets in a particular printer have mass 11 10 10. × − kg, charge 2.1 pC, speed 12 m/s, and pass through a uniform97-kN/C electric field in order to be deflected through a 10° angle. What is the length of the field region?

SolutionSuppose the ink droplets enter the field region perpendicular to the field, as in the geometry of Example 23-10. Then theanalysis of that example shows that v =v = vy x y xqE x m= =tan θ ∆ 2 , so ∆x m qEx y= = ×v =2 011 12 tan g m/s)2θ µ( . )(tan 10 pC kN/C) cm.° ==( . )( .2 1 97 137

Problem76. If the drop speed in the printer of Problem 75 is doubled, what should be done to the electric field to have the drops hit

the same point on the paper?

CHAPTER 23 563

SolutionSince tan θ » Ey x=v2 (for droplets of given q m= and fixed printer field region), droplets will hit the same point if ′ =E Ey y4when ′ =v vx x2 .

Supplementary Problems

Problem77. A spring of spring constant 100 N/m is stretched 10 cm beyond its 90-cm equilibrium length. If you want to keep it

stretched by attaching equal electric charges to the opposite ends, what magnitude of charge should you use?

SolutionThe repulsive force between like charges, kq r r2 2 90 10 1= ( = + = cm cm m), must balance the spring force, k x xs ( =10 cm is the stretch and ks is the spring constant). Thus,

q r k x ks= ± = ± × ⋅ = ±2 2 9 2 1 2100 01 9 10 33 3= = =[( ) ( . ) ( / )] . N/m)(1 m m N m C C.2 µ

Problem78. Two small spheres with the same mass m and charge q are suspended from massless strings of length l, as shown

in Fig. 23-52. Each string makes an angle θ with the vertical. Show that the charge on each sphere is q =±2l sin tan θ θmg k= .

SolutionThe magnitudes of the forces on either sphere (drawn acting on the righthand one in Fig. 23-52) are F mg Fgrav elec= =,kq2 22=( ) ,l sin θ and T (the unknown string tension). In equilibrium, ∑ = ∑ =F Fx y 0 0, or sin andelec= −F T θ

0 = −T mg cos θ . Eliminating T, we obtain tan sin or sin tan elecθ θ θ θ= = = ±F mg kq mg q mg k= = =2 22 2( ) , .l l


Problem79. A charge −q and a charge 4

9 q are located a distance a apart, as shown in Fig. 23-53. Where would you place a thirdcharge so that all three are in static equilibrium? What should be the sign and magnitude of the third charge?

564 CHAPTER 23

SolutionBecause of the vector nature of the forces, the third charge, Q, must be placed along the line joining the other two, as inExample 23-3. Q cannot go to the left of −q , since the magnitude of the force on it due to −q would always be greater thanthat due to 4

9 q . It cannot go between −q and 49 q , since the forces on it would always be in the same direction. Thus, Q

must go to the right of 49 q , as shown (x-axis to the right with origin at 4

9 q ). The net force on each charge must be zero, so,for Q kQ q x kQ q x a: ( ) ( ) , 0 4

92 2= + − +c h= = or 4

92 2( ) .x a x+ = For 4

949

2 49

20q k q q a k q Q x: ( ) , = − −c h c h= = or − =qx Qa2 2 .(The equation for the force on the third charge follows from the equations for the other two plus Newton’s third law.) Thesolution of these two equations (for x > 0) is x a Q q= = −2 4 and . (The equilibrium is unstable. A slight displacement ofthe positive charge to the right, for example, would cause it to be attracted more strongly to the right.)


Problem80. Two 34- Cµ charges are attached to the opposite ends of a spring of spring constant 150 N/m and equilibrium length

50 cm. By how much does the spring stretch?

SolutionSuppose that the Coulomb repulsion is the only force stretching the spring. When balanced with the spring force,kq x k xs

20

2=( ) ,l + = or x x( . ) ( / )( ) ( . .0 5 9 10 34 150 6 94 102 9 2 2 2 m N m C C N/m) m2 3+ = × ⋅ = × −µ = This cubic equationcan be solved by iteration or by Newton’s method to yield x = 15 95. . . . cm.

Problem81. A 3.8-g particle with a 4.0-µC charge experiences a downward force of 0.24 N in a uniform electric field. Find the

electric field, assuming that the gravitational force is not negligible.

SolutionIf gravity and Coulomb forces both act, then F Enet N= − + = −mg q$ ( . )$,j j0 24 where $j is upward. Thus, E =

( . )$ [( . )( . ) . $ ( . . $mg q− = × − = −−0 24 3 8 10 9 8 0 24 4 0 50 73 N kg m/s N] C) kN/C.2j j j= = µ

Problem82. A rod of length 2l lies on the x-axis, centered on the origin. It carries a line charge density given by λ λ= 0 ( ),x l=

where λ 0 is a constant. (a) What is the net charge on the rod? (b) Find an expression for electric field strength at allpoints x > l. (c) Show that your result has the 1 3=x dependence of dipole field when x À l. Hint: For l ¿ x,ln( ) ( )x x− +l l= becomes approximately − −2 2 33 3l l= =x x . (d) By comparing with Equation 23-7b, determine thedipole moment of the rod.

CHAPTER 23 565

Solution(a) Q dx x dx x= z = z = =− − −l

ll

l

l

ll lλ λ λ( ) ( ) .0 0

12

2 0= = (b) An element of charge, dq dx= ′λ , produces an electric field

d k dq x xE = − ′ î=( )2 at position x > l on the x-axis, where we are using ′x for the variable of integration. Thus,

E( )( ) ( ) ( )

x k x dxx x

k dx xx x x x

=FHG

IKJ

′ ′

− ′=FHG

IKJ ′

− ′−

− ′

LNM

OQP− −z zλ λ0

20

21

l ll

l

l

l

î î

=FHG

IKJ −

−+

+−+

FH IKLNM

OQP=

−+

−+

FH IKLNM

OQP

k xx

xx

xx

k xx

xx

λλ0

0 2 22 1

l l l

l

l l l

l

lî îln ln .

(c) For l=x ¿ 1,

2 2 1 2 12 2

2

2

1 2

2x

x x x x x−= −

FHG

IKJ = + +

FHG

IKJ

−

l

l l . . . ,

while

1 1 1 1 12

13

12

13

2

2

3

3

2

2

3

3l

l

l l

l l

l

l l l l l l ln ln ln 1 + . . . . . .xx x x x x x x x x−

+FH IK= −FH IK− FH IK

LNM

OQP= − − − + − − + +

FHG

IKJ

LNMM

OQPP

= − − +2 2

3

2

3x xl . . .

Thus,

E( )x kx x x x

kx

= + + − − +LNM

OQPλ

λ0

2

3

2

30

2

32 2 2 2

34

3î îl l l. . . . . . '

(d) On the axis of a point dipole, Equation 23-7b gives E( ) ,x kp x= 2 3î= so comparison reveals that p = 2 302λ l = . (Note:

Explicit use of the formula p x dq x dx= z = z− −ll

ll lλ 0

2 = gives the same result.)

Problem83. The electric field on the axis of a uniformly charged ring has magnitude 380 kN/C at a point 5.0 cm from the ring

center. The magnitude 15 cm from the center is 160 kN/C; in both cases the field points away from the ring. Find theradius and charge of the ring.

SolutionThe electric field on the axis of a uniformly charged ring is calculated in Example 23-8, so the data given in the questionimply 380 5 5 2 2 3 2 kN/C cm cm= + −kQ a( )[( ) ] ,= and 160 15 15 2 2 3 2 kN/C cm cm= + −kQ a( )[( ) ] .= Dividing these twoequations and taking the 2

3 root we get

380 15160 5

3 70 155

2 3 2 2

2 2×

×FH IK = =

+

+

=

. ( )( )

, cm cm

aa

which when solved for the radius, gives

a = − =[( ) ( . )( ) ] . .15 3 70 5 2 70 7 002 2 cm cm cm.=

566 CHAPTER 23

Substituting for a in either of the field equations allows us to find

Q =+

× ⋅=

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

380 5 79 10 5

5382 2 3 2

9 kN/C) cm cm

N m C cm nC.2 3

=

Problem84. Use the binomial theorem to show that, for x RÀ , the result of Problem 48 reduces to the field of a point charge

whose total charge is the charge density σ times the disk area.

SolutionThe result of Problem 48 for the field on the axis of a uniformly charged disk, of radius R, at a distance x > 0 along theaxis (away from the disk’s center) was E k x x R k R xx = − + = − + −2 1 2 1 12 2 2 2 1 2π σ π σ( ) [ ( ) ].= = = The binomialexpansion in Appendix A, for R x2 2 1= ¿ , gives

E k Rx

k Rx

kQxx = − − +

FHG

IKJ

LNMM

OQPP =2 1 1 1

22

2

2

2

2

2 2π σπ σ. . . , '

which is the field from a point charge Q R= π σ2 at a distance x.

Problem85. A molecule with dipole moment p is located a distance r from a proton, oriented with its dipole moment vector p as

shown in Fig. 23-54. (a) Use Equation 23-7b to find the force the molecule exerts on the proton. (b) Now find the netforce on the molecule in the proton’s nonuniform electric field by considering that the molecule consists of two oppositecharges ±q, separated by a distance d such that qd p= . Take the limit as d becomes very small compared with r, andshow that the resulting force has the same magnitude as that of part (a), as required by Newton’s third law.

SolutionTake the origin at the molecule, with the x-axis parallel to its dipole moment, as shown on Fig. 23-54. (a) Equation 23-7bgives the dipole’s electric field at the position of the proton, so the force on the latter is F Ep e e kpr= = =−

dip ( )2 3î2 3kepr − î. (b) The electric field of the proton, for points on the x-axis to its left ( )x r< is E p x ke r x( ) ( )= − − −2 î (seeEquation 23-4). If the molecular dipole is considered to consist of charges ±q located at x d= ± =2, where p qd= , thenthe force on it is F E Edip = − − = − − − + = − −− − −q d q d keq r d r d kepr r dp p( ) ( ) [ ) ( ) ] ( ) .= = = = =2 2 2 2 2 42 2 2 2 2î ( î In the limitd r¿ , this becomes F Fdip = − = −−2 3kepr pî , in agreement with the result of part (a) and Newton’s third law.


Problem86. An electric quadrupole consists of two oppositely directed dipoles in close proximity. (a) Calculate the field of the

quadrupole shown in Fig. 23-55 for points to the right of x a= , and (b) show that for x aÀ the quadrupole field fallsoff as 1 4=x .

CHAPTER 23 567


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

E( )( ) ( )

( )( )

x k qx a

qx

qx a

kqa x ax x a

=−

− ++

LNM

OQP=

−

−î î2 2 2

22 2

2 2 22 2 3

(b) For x a x kqa xÀ , ( ) .E → 6 2 4î= (The quadrupole moment of this “linear quadrupole” is Q qaxx = 4 2 .)

Problem87. Derive Equation 23-9 in Example 23-9 by making θ the integration variable, then evaluating the resulting integral.

SolutionIn Fig. 23-23, r x y y= + =( )2 2 1 2= =sin ,θ and cot θ = x y= . Differentiating the latter with respect to x, we get − =csc2θ θddx y= , which relates dx to dθ. (Extend the results in Appendix B, and recall that cosecant sine= −1.) The limits x = −∞ to∞ correspond to θ π= to 0 (since when x is negative so is cot ),θ thus the integral in Example 23-9 leading to

Equation 23-9 becomes E k y dxx y

k y y dy

ky

d=+

=−

= =−∞

∞z zzλ λθ θ

θ

λθ θ

π

π( ) (2 2 3 2

2

0

0

= = csc

sin )sin 3

( )[k y k yλ π λ= =− + =cos cos 0] 2 as before.

Ch23

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