CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS · 898 ELECTRIC FORCES AND ELECTRIC FIELDS CHAPTER...

CHAPTER 18 ELECTRIC FORCES AND

ELECTRIC FIELDS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

1. 1.9 × 1013

2. (b) Suppose that A is positive and B is negative. Since C and A also attract each other, C

must be negative. Thus, B and C repel each other, because they have like charges (both

negative). Suppose, however, that A is negative and B is positive. Since C and A also attract

each other, C must be positive. Again we conclude that B and C repeal each other, because

they have like charges (both positive).

3. (a) The ball is electrically neutral (net charge equals zero). However, it is made from a

conducting material, so it contains electrons that are free to move. The rod attracts some of

these (negative) electrons to the side of the ball nearest the rod, leaving the opposite side of

the ball positively charged. Since the negative side of the ball is closer to the positive rod

than the positive side, a net attractive force arises.

4. (d) The fact that the positive rod repels one object indicates that that object carries a net

positive charge. The fact that the rod repels the other object indicates that that object carries

a net negative charge. Since both objects are identical and made from conducting material,

they share the combined net charges equally after they are touched together. Since the rod

repels each object after they are touched, each object must then carry a net positive charge.

But the net electric charge of any isolated system is conserved, so the total net charge

initially must also have been positive. This means that the initial positive charge had the

greater magnitude.

5. (c) This distribution is not possible because of the law of conservation of electric charge.

The total charge on the three objects here is 98

q , whereas only q was present initially.

6. (c) This is an example of charging by induction. The negatively charged rod repels free

electrons in the metal. These electrons move through the point of contact and into the sphere

farthest away from the rod, giving it an induced charge of −q. The sphere nearest the rod

acquires an induced charge of +q. As long as the rod is kept in place while the spheres are

separated, these induced charges cannot recombine and remain on the spheres.

7. (b) Coulomb’s law states that the magnitude of the force is given by 1 2

2

q qF k

r . Doubling

the magnitude of each charge as in A would increase the numerator by a factor of four, but

this is offset by the change in separation, which increases the denominator by a factor of

22 = 4. Doubling the magnitude of only one charge as in D would increase the numerator by

896 ELECTRIC FORCES AND ELECTRIC FIELDS

a factor of two, but this is offset by the change in separation, which increases the

denominator by a factor of 2

2 2 .

8. (e) Coulomb’s law states that the magnitude of the force is given by 1 2

2

q qF k

r . The force

is directed along the line between the charges and is an attraction for unlike charges and a

repulsion for like charges. Charge B is attracted by charge A with a force of magnitude

2

q qk

d and repelled by charge C with a force of the same magnitude. Since both forces

point to the left, the net force acting on B has a magnitude of 2

2q q

kd

. Charge A is attracted

by charge B with a force of 2

q qk

dand also by charge C with a force of

2

2

q qk

d. Since both

forces point to the right, the net force acting on A has a magnitude of 2

1.25q q

kd

. Charge

C is pushed to the right by B with a force of 2

q qk

d and pulled to the left by A with a force

of

22

q qk

d. Since these two forces have different directions, the net force acting on C has a

magnitude of 2

0.75q q

kd

.

9. (b) According to Coulomb’s law, the magnitude of the force that any one of the point charges

exerts on another point charge is given by 2

q qF k

d , where d is the length of each side of

the triangle. The charge at B experiences a repulsive force from the charge at A and an

attractive force from the charge at C. Both forces have vertical components, but one points

in the +y direction and the other in the −y direction. These vertical components have equal

magnitudes and cancel, leaving a resultant that is parallel to the x axis.

10. 8.5 µC

11. (e) According to Equation 18.2, the force exerted on a charge by an electric field is

proportional to the magnitude of the charge. Since the negative charge has twice the

magnitude of the positive charge, the negative charge experiences twice the force.

Furthermore, the direction of the force on the positive charge is in the same direction as the

field, so that we can conclude that the field points due west. The force on the negative

charge points opposite to the field and, therefore, points due east.

Chapter 18 Answers to Focus on Concepts Questions 897

12. (c) The electric field created by a point charge has a magnitude 2

k qE

r and is inversely

proportional to the square of the distance r. If r doubles, the charge magnitude must increase

by a factor of 22 = 4 to keep the field the same.

13. (b) To the left of the positive charge the two contributions to the total field have opposite

directions. There is a spot in this region at which the field from the smaller, but closer,

positive charge exactly offsets the field from the greater, but more distant, negative charge.

14. (e) Consider the charges on opposite corners. In all of the arrangements these are like

charges. This means that the two field contributions created at the center of the square point

in opposite directions and, therefore, cancel. Thus, only the charge opposite the empty

corner determines the magnitude of the net field at the center of the square. Since the point

charges all have the same magnitude, the net field there has the same magnitude in each

arrangement.

15. 1.8 × 10−6 C/m2

16. (c) The tangent to the field line gives the direction of the electric field at a point. At A the

tangent points due south, at B southeast, and at C due east.

17. (a) The electric field has a greater magnitude where the field lines are closer together. They

are closest together at B and farthest apart at A. Therefore, the field has the greatest

magnitude at B and the smallest magnitude at A.

18. (d) In a conductor electric charges can readily move in response to an electric field. In A,

B, and C the electric charges experience an electric field and, hence, a force from

neighboring charges and will move outward, away from each other. They will rearrange

themselves so that the electric field within the metal is zero at equilibrium. This means that

they will reside on the outermost surface. Thus, only D could represent charges in

equilibrium.

19. 1.3 N·m2/C

20. 0.45 N·m2/C


CHAPTER 18 ELECTRIC FORCES AND

ELECTRIC FIELDS

PROBLEMS ______________________________________________________________________________

1. SSM REASONING The charge of a single proton is +e, and the charge of a single

electron is –e, where e = 1.60×10−19 C. The net charge of the ionized atom is the sum of the

charges of its constituent protons and electrons.

SOLUTION The ionized atom has 26 protons and 7 electrons, so its net electric charge q is

19 1826 7 19 19 1.60 10 C 3.04 10 Cq e e e

2. REASONING Since the object has a charge of 2.0 μC to begin with, it becomes neutral

when that charge is removed. To create a charge of +3.0 μC on the object, an additional

charge of 3.0 μC must be removed. Thus, the total charge that must be removed from the

object is 5.0 μC. The number of electrons that correspond to this charge can be obtained

by dividing the charge by the charge on a single electron, which is 191.60 10 Ce .

SOLUTION The number N of electrons corresponding to 5.0 μC is

6

1319

5.0 10 C3.1 10

1.60 10 CN

______________________________________________________________________________

3. REASONING

a. Since the objects are metallic and identical, the charges on each combine and produce a

net charge that is shared equally by each object. Thus, each object ends up with one-fourth

of the net charge.

b. The number of electrons (or protons) that make up the final charge on each object is

equal to the final charge divided by the charge of an electron (or proton).

SOLUTION a. The net charge is the algebraic sum of the individual charges. The charge q on each

object after contact and separation is one-fourth the net charge, or

14

1.6 C 6.2 C – 4.8 C – 9.4 C –1.6 Cq

Chapter 18 Problems 899

b. Since the charge on each object is negative, the charge is comprised of electrons. The

number of electrons on each object is the charge q divided by the charge e of a single

electron:

Number of electrons = 6

13

19

1.6 10 C1.0 10

1.60 10 C

q

e

____________________________________________________________________________________________

4. REASONING The conservation of electric charge states that, during any process, the net

electric charge of an isolated system remains constant (is conserved). Therefore, the net

charge (q1 + q2) on the two spheres before they touch is the same as the net charge after they

touch. When the two identical metal spheres touch, the net charge will spread out equally

over both of them. When the spheres are separated, the charge on each is the same.

SOLUTION

a. Since the final charge on each sphere is +5.0 C, the final net charge on both spheres is

2(+5.0 C) = +10.0 C. The initial net charge must also be +10.0 C. The only spheres

whose net charge is +10.0 C are

B D

B ( = 2.0 C) and D ( = +12.0 C)q q

b. Since the final charge on each sphere is +3.0 C, the final net charge on the three spheres

is 3(+3.0 C) = +9.0 C. The initial net charge must also be +9.0 C. The only spheres

whose net charge is +9.0 C are

A C DA ( = 8.0 C), C ( = +5.0 C) and D ( = +12.0 C)q q q

c. Since the final charge on a given sphere in part (b) is +3.0 C, we would have to add

3.0 C to make it electrically neutral. Since the charge on an electron is 1.6 1019 C,

the number of electrons that would have to be added is

613

19

3.0 10 CNumber of electrons 1.9 10

1.6 10 C

______________________________________________________________________________

5. SSM REASONING Identical conducting spheres equalize their charge upon touching.

When spheres A and B touch, an amount of charge +q, flows from A and instantaneously

neutralizes the –q charge on B leaving B momentarily neutral. Then, the remaining amount

of charge, equal to +4q, is equally split between A and B, leaving A and B each with equal

amounts of charge +2q. Sphere C is initially neutral, so when A and C touch, the +2q on A

splits equally to give +q on A and +q on C. When B and C touch, the +2q on B and the +q

on C combine to give a total charge of +3q, which is then equally divided between the

spheres B and C; thus, B and C are each left with an amount of charge +1.5q.


SOLUTION Taking note of the initial values given in the problem statement, and

summarizing the final results determined in the REASONING above, we conclude the

following:

a. Sphere C ends up with an amount of charge equal to +1.5q .

b. The charges on the three spheres before they were touched, are, according to the problem

statement, +5q on sphere A, –q on sphere B, and zero charge on sphere C. Thus, the total

charge on the spheres is 5q –q 0 4q .

c. The charges on the spheres after they are touched are +q on sphere A, +1.5q on sphere B,

and +1.5q on sphere C. Thus, the total charge on the spheres is q 1.5q 1.5q 4q .

______________________________________________________________________________

6. REASONING When N electrons, each carrying a charge −e = −1.6×10−19 C, are transferred

from the plate to the rod, the system consisting of the plate and the rod is isolated.

Therefore, the total charge q1i+ q2i of the system is unchanged by the process, where q1i is

the initial charge of the plate and q2i is the initial charge of the rod. At the end, the rod and

the plate each have the same final charge q1f = q2f. Therefore, each must have a charge equal

to half the total charge of the system: 11f 2f 1i 2i2

q q q q .

SOLUTION The final charge q2f on the rod is equal to its initial charge q2i plus the charge

transferred to it, which is equal to the product of the number N of electrons transferred and

the charge −e of each electron. Therefore,

2f 2i 2i

( )q q N e q Ne (1) Since the final charge on the rod is equal to half the total initial charge of the system, we can

substitute 12f 1i 2i2

q q q into Equation (1) and solve for N:

2i 1i1 1 1

1i 2i 2 2 1i 2i 2i 1i2 2 2 or or

2i i

q qq q q Ne Ne q q q q q N

e

Therefore, the number of electrons that must be transferred to the rod is

6 62i 1i 13

19

2.0 10 C 3.0 10 C1.6 10

2 2 1.6 10 C

q qN

e

7. REASONING

a. The number N of electrons is 10 times the number of water molecules in 1 liter of water.

The number of water molecules is equal to the number n of moles of water molecules times

Avogadro’s number NA: A

10N n N .

b. The net charge of all the electrons is equal to the number of electrons times the change on

one electron.


SOLUTION

a. The number N of water molecules is equal to A

10 n N , where n is the number of moles of

water molecules and NA is Avogadro’s number. The number of moles is equal to the mass

m of 1 liter of water divided by the mass per mole of water. The mass of water is equal to

its density times the volume, as expressed by Equation 11.1. Thus, the number of

electrons is

A A A

3 3 3

23 1

26

10 10 1018.0 g/mol 18.0 g/mol

1000 g1000 kg/m 1.00 10 m

10 6.022 10 mol1 kg

18.0 g/mol

3.35 10 electrons

VmN n N N N

b. The net charge Q of all the electrons is equal to the number of electrons times the change

on one electron: 26 19 73.35 10 1.60 10 C 5.36 10 CQ

.

______________________________________________________________________________

8. REASONING The magnitude of the electrostatic force that acts on particle 1 is given by

Coulomb’s law as 21 2

/F k q q r . This equation can be used to find the magnitude 2

q

of the charge.

SOLUTION Solving Coulomb’s law for the magnitude 2

q of the charge gives

226

2 9 2 2 61

3.4 N 0.26 m7.3 10 C

8.99 10 N m /C 3.5 10 C

F rq

k q

(18.1)

Since q1 is positive and experiences an attractive force, the charge q2 must be negative .

______________________________________________________________________________

9. SSM REASONING The number N of excess electrons on one of the objects is equal to

the charge q on it divided by the charge of an electron (e), or N = q/(e). Since the charge

on the object is negative, we can write q q , where q is the magnitude of the charge.

The magnitude of the charge can be found from Coulomb’s law (Equation 18.1), which

states that the magnitude F of the electrostatic force exerted on each object is given by 2/F k q q r , where r is the distance between them.


SOLUTION The number N of excess electrons on one of the objects is

q qqN

e e e

(1)

To find the magnitude of the charge, we solve Coulomb’s law, 2/F k q q r , for q :

2F rq

k

Substituting this result into Equation (1) gives

2

21 32

9 2 2

19

4.55 10 N 1.80 10 m

8.99 10 N m / C8

1.60 10 C

F r

q kN

e e

______________________________________________________________________________

10. REASONING

a. The magnitude of the electrostatic force that acts on each sphere is given by Coulomb’s

law as 21 2

/F k q q r , where 1 2

and q q are the magnitudes of the charges, and r is the

distance between the centers of the spheres.

b. When the spheres are brought into contact, the net charge after contact and separation

must be equal to the net charge before contact. Since the spheres are identical, the charge

on each after being separated is one-half the net charge. Coulomb’s law can be applied

again to determine the magnitude of the electrostatic force that each sphere experiences.

SOLUTION a. The magnitude of the force that each sphere experiences is given by Coulomb’s law as:

9 2 2 6 6

1 2 4

2 22

8.99 10 N m /C 20.0 10 C 50.0 10 C1.44 10 N

2.50 10 m

k q qF

r

Because the charges have opposite signs, the force is attractive . b. The net charge on the spheres is 20.0 C + 50.0 C = +30.0 C. When the spheres are

brought into contact, the net charge after contact and separation must be equal to the net

charge before contact, or +30.0 C. Since the spheres are identical, the charge on each after

being separated is one-half the net charge, so 1 2

15.0 Cq q . The electrostatic force

that acts on each sphere is now

9 2 2 6 6

1 2 3

2 22

8.99 10 N m /C 15.0 10 C 15.0 10 C3.24 10 N

2.50 10 m

k q qF

r

Since the charges now have the same signs, the force is repulsive .


______________________________________________________________________________

11. SSM REASONING Initially, the two spheres are neutral. Since negative charge is

removed from the sphere which loses electrons, it then carries a net positive charge.

Furthermore, the neutral sphere to which the electrons are added is then negatively charged.

Once the charge is transferred, there exists an electrostatic force on each of the two spheres,

the magnitude of which is given by Coulomb's law (Equation 18.1), 21 2

/F k q q r .

SOLUTION

a. Since each electron carries a charge of 191.60 10 C , the amount of negative charge

removed from the first sphere is

19

13 61.60 10 C3.0 10 electrons 4.8 10 C

1 electron

Thus, the first sphere carries a charge +4.8 10–6 C, while the second sphere carries a

charge 4.8 10–6 C. The magnitude of the electrostatic force that acts on each sphere is,

therefore,

29 2 2 6

1 2

2 2

8.99 10 N m /C 4.8 10 C0.83 N

0.50 m

k q qF

r

b. Since the spheres carry charges of opposite sign, the force is attractive .

______________________________________________________________________________

12. REASONING The magnitude of the force of attraction between the charges is given by

Coulomb's law: 21 2

/F k q q r (Equation 18.1), where 1 2

and q q are the magnitudes of

the charges and r is the separation of the charges. We will apply this equation twice, once

when the separation is rA and the magnitude of the force is A

1.5 NF and then again when

the separation is rB = rA/9 and the magnitude of the force is FB.

SOLUTION Applying Coulomb’s law when the separations are rA and rB, we obtain

1 2 1 2A B2 2

A B

and q q q q

F k F kr r

In order to eliminate the unknown charge magnitudes and the constant k, we divide the

equation on the right by the equation on the left:

2 221 2 BB A A

B A2A B B1 2 A

/ or

/

k q q rF r rF F

F r rk q q r

Using the fact that rB = rA/9, we find that the unknown force-magnitude is


2 2

A AB A

B A

1.5 N 120 N/ 9

r rF F

r r

______________________________________________________________________________

13. REASONING Let F2 and F1 represent the forces exerted on the

charge q at the origin by the point charges q1 and q2, respectively.

According to Equation 18.1, the magnitudes of these forces are

given by

1 21 22 2

1 2

and q qq q

F k F kr r

(1)

where r1 is the distance between q1 and q, r2 is the distance

between q2 and q, and k = 8.99×109 N·m2/C2. The directions of

the forces are determined by the signs of each charge-pair. The

sign of q1 is opposite that of q, so F1 is an attractive force,

pointing in the positive y direction. The signs of q2 and q are both

positive, so F2 is a repulsive force, pointing in the negative y direction (see the drawing).

Because the net force F = F1 + F2 acting on q points in the positive y direction, the force F1

must have a greater magnitude than the force F2. Therefore, the magnitude F of the net

electric force acting on q is equal to the magnitude of the attractive force F1 minus that of

the repulsive force F2:

1 2

F F F (2)

SOLUTION Substituting Equations (1) into Equation (2) yields

1 21 2 2 2

1 2

q qq qF F F k k

r r (3)

Solving Equation (3) for |q2|, we obtain

12 1 22 2 22 2

12 1

or qq qq q F

qk k F rk qrr r

Substituting the given values, we find that

6

2 52 2 9 2 2 6

25 10 C 27 N1.8 10 C0.34 m

8.99 10 N m /C 8.4 10 C0.22 mq

q2

q1 = −25 C

q = +8.4 C

F1

F2

y


14. REASONING The electrical force that each charge exerts on charge 2 is shown in the

following drawings. F21 is the force exerted on 2 by 1, and F23 is the force exerted on 2

by 3. Each force has the same magnitude, because the charges have the same magnitude

and the distances are equal.

The net electric force F that acts on charge 2 is shown in the following diagrams.

It can be seen from the diagrams that the largest electric force occurs in (a), followed by (c),

and then by (b). SOLUTION The magnitude F21 of the force exerted on 2 by 1 is the same as the

magnitude F23 of the force exerted on 2 by 3, since the magnitudes of the charges are the

same and the distances are the same. Coulomb’s law gives the magnitudes as

21 23 2

9 2 2 6 6

4

23

8.99 10 N m /C 8.6 10 C 8.6 10 C4.6 10 N

3.8 10 m

k q qF F

r

In part (a) of the drawing showing the net electric force acting on charge 2, both F21 and F23

point to the left, so the net force has a magnitude of

4 412

2 2 4.6 10 N 9.2 10 NF F

In part (b) of the drawing showing the net electric force acting on charge 2, F21 and F23

point in opposite directions, so the net force has a magnitude of 0 N .

2 3 1

+q +q +q

2

3

1

+q

+q

q

(b)

(c)

F23

2 3 1

q +q +q

(a)

F21

F23

F21 F21 F23

(a)

F21 F23

F

(b)

F21 F23

F = 0 N

(c)

F21

F23

F


In part (c) showing the net electric force acting on charge 2, the magnitude of the net force

can be obtained from the Pythagorean theorem:

2 2

2 2 4 4 421 23

4.6 10 N 4.6 10 N 6.5 10 NF F F

______________________________________________________________________________

15. SSM REASONING AND SOLUTION

a. Since the gravitational force between the spheres is one of attraction and the electrostatic

force must balance it, the electric force must be one of repulsion. Therefore, the charges

must have the same algebraic signs, both positive or both negative.

b. There are two forces that act on each sphere; they are the gravitational attraction FG of

one sphere for the other, and the repulsive electric force FE of one sphere on the other.

From the problem statement, we know that these two forces balance each other, so that

FG = FE. The magnitude of FG is given by Newton's law of gravitation (Equation 4.3:

2G 1 2

/F Gm m r ), while the magnitude of FE is given by Coulomb's law (Equation 18.1:

2E 1 2

/F k q q r ). Therefore, we have

21 2 21 22 2

or k q qGm m

Gm k qr r

since the spheres have the same mass m and carry charges of the same magnitude q .

Solving for q , we find

–11 2 2–6 –16

9 2 2

6.67 10 N m /kg(2.0 10 kg) 1.7 10 C

8.99 10 N m /C

Gq m

k

______________________________________________________________________________

16. REASONING The drawing at the right

shows the set-up. The force on the +q

charge at the origin due to the other +q

charge is given by Coulomb’s law

(Equation 18.1), as is the force due to the

+2q charge. These two forces point to the

left, since each is repulsive. The sum of

the two is twice the force on the +q charge at the origin due to the other +q charge alone.

SOLUTION Applying Coulomb’s law, we have


2 2 2

Force due to + Force due to +2 Twice the force due to charge at 0.50 m charge at + charge at 0.50 m

22

0.50 m 0.50 m

q qx x d q x

k q q k q q k q q

d

Rearranging this result and solving for d give

222 2

2or 2 0.50 m or 0.71 m

0.50 m

k q q k q qd d

d

We reject the negative root, because a negative value for d would locate the +2q charge to

the left of the origin. Then, the two forces acting on the charge at the origin would have

different directions, contrary to the statement of the problem. Therefore, the +2q charge is

located at a position of 0.71 mx .

______________________________________________________________________________

17. SSM REASONING Each particle will experience an electrostatic force due to the

presence of the other charge. According to Coulomb's law (Equation 18.1), the magnitude

of the force felt by each particle can be calculated from 21 2

/F k q q r , where

1 2 and q q are the respective charges on particles 1 and 2 and r is the distance between

them. According to Newton's second law, the magnitude of the force experienced by each

particle is given by F ma, where a is the acceleration of the particle and we have assumed

that the electrostatic force is the only force acting.

SOLUTION

a. Since the two particles have identical positive charges, 1 2

q q q , and we have, using

the data for particle 1,

2

1 12

k qm a

r

Solving for q , we find that

2 –6 3 2 –2 2

–81 19 2 2

(6.00 10 kg) (4.60 10 m/s ) (2.60 10 m)4.56 10 C

8.99 10 N m /C

m a rq

k

b. Since each particle experiences a force of the same magnitude (From Newton's third law),

we can write F1 = F2, or m1a1 = m2a2. Solving this expression for the mass m2 of particle 2,

we have

–6 3 2–61 1

2 3 22

(6.00 10 kg)(4.60 10 m/s )3.25 10 kg

8.50 10 m/s

m am

a

______________________________________________________________________________


18. REASONING AND SOLUTION Calculate the magnitude of each force acting on the

center charge. Using Coulomb’s law, we can write

9 2 2 6 64 3

43 2 243

8.99 10 N m / C 4.00 10 C 3.00 10 C

0.100 m

10.8 N (toward the south)

k q qF

r

9 2 2 6 65 3

53 2 253

8.99 10 N m / C 5.00 10 C 3.00 10 C

0.100 m

13.5 N (toward the east)

k q qF

r

Adding F43 and F53 as vectors, we have

2 22 2

43 53

–1 –143

53

= 10.8 N 13.5 N 17.3 N

10.8 Ntan tan 38.7 S of E

13.5 N

F F F

F

F

______________________________________________________________________________

19. REASONING According to Newton’s second law, the centripetal acceleration experienced

by the orbiting electron is equal to the centripetal force divided by the electron’s mass.

Recall from Section 5.3 that the centripetal force Fc is the name given to the net force

required to keep an object on a circular path of radius r. For an electron orbiting about two

protons, the centripetal force is provided almost exclusively by the electrostatic force of

attraction between the electron and the protons. This force points toward the center of the

circle and its magnitude is given by Coulomb’s law.

SOLUTION The magnitude ac of the centripetal acceleration is equal to the magnitude Fc

of the centripetal force divided by the electron’s mass: c c

/a F m (Equation 5.3). The

centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where F is

the magnitude of the electrostatic force of attraction between the electron and the two

protons, Thus, c

/a F m . The magnitude of the electrostatic force is given by Coulomb’s

law, 21 2

/F k q q r (Equation 18.1), where 1

q e and 2

2q e are the magnitudes

of the charges, r is the radius of the orbit, and 9 2 28.99 10 N m / Ck . Substituting this

expression for F into c/a F m , and using m = 9.11 10

31 kg for the mass of the electron,

we find that


2

c 2

9 2 2 19 1923 2

231 11

2

2

8.99 10 N m / C 1.60 10 C 2 1.60 10 C7.19 10 m/s

9.11 10 kg 2.65 10 m

e ek

F k e eram m mr

____________________________________________________________________________________________

20. REASONING The unknown charges can

be determined using Coulomb’s law to

express the electrostatic force that each

unknown charge exerts on the 4.00 μC

charge. In applying this law, we will use

the fact that the net force points downward

in the drawing. This tells us that the

unknown charges are both negative and

have the same magnitude, as can be

understood with the help of the free-body

diagram for the 4.00 μC charge that is

shown at the right. The diagram shows

the attractive force F from each negative charge directed along the lines between the charges.

Only when each force has the same magnitude (which is the case when both unknown

charges have the same magnitude) will the resultant force point vertically downward. This

occurs because the horizontal components of the forces cancel, one pointing to the right and

the other to the left (see the diagram). The vertical components reinforce to give the

observed downward net force.

SOLUTION Since we know from the REASONING that the unknown charges have the

same magnitude, we can write Coulomb’s law as follows:

6 6A B

2 2

4.00 10 C 4.00 10 Cq qF k k

r r

The magnitude of the net force acting on the 4.00 μC charge, then, is the sum of the

magnitudes of the two vertical components F cos 30.0º shown in the free-body diagram:

6 6A B

2 2

6A

2

4.00 10 C 4.00 10 Ccos30.0 cos30.0

4.00 10 C2 cos30.0

q qF k k

r r

qk

r

Solving for the magnitude of the charge gives

qA qB

+4.00 μC

30.0º F cos 30.0º

F sin 30.0º

F


2

A 6

2

6

9 2 2 6

2 4.00 10 C cos30.0

405 N 0.0200 m2.60 10 C

2 8.99 10 N m / C 4.00 10 C cos30.0

F rq

k

Thus, we have 6A B

2.60 10 Cq q .

21. REASONING

a. There are two electrostatic forces that act on q1; that due to q2 and that due to q3. The

magnitudes of these forces can be found by using Coulomb’s law. The magnitude and

direction of the net force that acts on q1 can be determined by using the method of vector

components. b. According to Newton’s second law, Equation 4.2b, the acceleration of q1 is equal to the

net force divided by its mass. However, there is only one force acting on it, so this force is

the net force.

SOLUTION

a. The magnitude F12 of the force exerted on q1

by q2 is given by Coulomb’s law, Equation 18.1,

where the distance is specified in the drawing:

9 2 2 6 6

1 212 2 2

12

8.99 10 N m /C 8.00 10 C 5.00 10 C0.213 N

1.30 m

k q qF

r

Since the magnitudes of the charges and the distances are the same, the magnitude of F13 is

the same as the magnitude of F12, or F13 = 0.213 N. From the drawing it can be seen that the

x-components of the two forces cancel, so we need only to calculate the y components of the

forces.

Force y component

F12 +F12 sin 23.0 = +(0.213 N) sin 23.0 = +0.0832 N

F13 +F13 sin 23.0 = +(0.213 N) sin 23.0 = +0.0832 N

F Fy = +0.166 N

Thus, the net force is 0.166 N (directed along the + axis) .y F

q1

q3

q2

F13

+x

+y

F12

1.30 m

1.30 m 23.0 23.0


b. According to Newton’s second law, Equation 4.2b, the acceleration of q1 is equal to the

net force divided by its mass. However, there is only one force acting on it, so this force is

the net force:

2

3

0.166 N111m/s

1.50 10 kgm

Fa

where the plus sign indicates that the acceleration is along the +y axis .

______________________________________________________________________________

22. REASONING When the airplane and the other end of the guideline carry point charges +q

and −q, the airplane is subject to an attractive electric force of magnitude 2

1 2

2 2 2

q q q q qF k k k

r r r

(Equation 18.1), where 9 2 28.99 10 N m / Ck and r is the

length of the guideline. This electric force provides part of the centripetal force

22

c2

mvF

r

(Equation 5.3) necessary to keep the airplane (mass = m) flying along its circular path at the

higher speed v2 (which is associated with the greater kinetic energy KE2). The remainder of

the centripetal force is provided by the maximum tension Tmax in the guideline, so we have

that

22

max c2

mvF T F

r (1)

When the airplane is neutral and flying at the slower speed v1 (which is associated with the

smaller kinetic energy KE1), there is no electrical force, so the centripetal force

21

c1

mvF

r

acting on the airplane is due solely to the maximum tension Tmax in the guideline:

21

max c1

mvT F

r (2)

We note that the kinetic energy of the airplane is given by 212

KE mv (Equation 6.2), so

that the quantities in the numerators of Equations (1) and (2) are proportional to the kinetic

energies KE2 and KE1 of the airplane:

2 22 2 1 1

2 KE and 2 KE mv mv (3)

SOLUTION Solving 2

2

qF k

r (Equation 18.1) for q2 and taking the square root of both

sides, we obtain

2 2

2 or Fr Fr

q qk k

(4)

Substituting Equations (3) into Equations (1) and (2) yields


2 1

max max

2 KE 2 KE and F T T

r r (5)

Substituting the second of Equations (5) into the first of Equations (5), we obtain

1 2 2 1 2 1

2 KE 2 KE 2 KE 2 KE 2 KE KE or F F

r r r r r

(6)

Substituting Equation (6) into Equation (4), we find that

2

2 12 KE KEFr

qk r

2r 2 12 KE KE r

k k

Therefore, the magnitude q of the charge on the airplane is

5

9 2 2

2 51.8 J 50.0 J 3.0 m3.5 10 C

8.99 10 N m /Cq

23. REASONING The kinetic energy of the orbiting electron is 212

KE mv (Equation 6.2),

where m and v are its mass and speed, respectively. We can obtain the speed by noting that

the electron experiences a centripetal force whose magnitude Fc is given by 2c

/F mv r

(Equation 5.3), where r is the radius of the orbit. The centripetal force is provided almost

exclusively by the electrostatic force of attraction F between the electron and the protons, so

Fc = F. The electrostatic force points toward the center of the circle and its magnitude is

given by Coulomb’s law as 21 2

/F k q q r (Equation 18.1), where 1

q and 2

q are the

magnitudes of the charges.

SOLUTION The kinetic energy of the electron is

212

KE mv (6.2)

Solving the centripetal-force expression, 2c

/F mv r (Equation 5.3), for the speed v, and

substituting the result into Equation 6.2 gives

21 12 2

KE mv m crF

m

1c2

rF

(1)

The centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where

F is the magnitude of the electrostatic force of attraction between the electron and the three

protons. This force is given by Coulomb’s law, 21 2

/F k q q r (Equation 18.1).

Substituting Coulomb’s law into Equation 1 yields

1 2 1 21 1 1c2 2 2 2

KE2

k q q k q qrF rF r

rr


Setting 1

q e and 2

3q e , we have

9 2 2 19 1917

11

3KE

2

8.99 10 N m / C 1.60 10 C 3 1.60 10 C1.96 10 J

2 1.76 10 m

k e e

r

______________________________________________________________________________

24. REASONING Since the spring is stretched because of the charges, the charges must be

exerting a repulsive force on each other. Only like charges repel each other. The magnitude

of the repulsive force between the charges is given by Coulomb's law: 21 2

/F k q q r

(Equation 18.1), where 1 2

and q q are the magnitudes of the charges and r is the separation

of the charges. This repulsive force provides the applied force needed to stretch the spring,

which is Appliedspringx

F k x (Equation 10.1), where kspring is the spring constant and x is the

displacement of the spring from its unstrained length.

SOLUTION a. Since only like charges repel each other, the charges must be

either both positive or both negative

b. Since the charges have equal magnitudes, we know that 1 2

q q q , so that

Coulomb’s law can be used to determine q :

2 2

1 2

2 2 or

q q q FrF k k q

kr r

In this result for q we know that the electrostatic force F provides the applied force needed

to stretch the spring, so that Appliedspringx

F F k x (Equation 10.1). With this substitution

for F, our expression for q becomes

22spring

2

69 2 2

220 N/m 0.020 m 0.32 m 0.020 m7.5 10 C

8.99 10 N m /C

k xrFrq

k k

______________________________________________________________________________


25. SSM REASONING Consider the drawing

at the right. It is given that the charges qA,

q1, and q2 are each positive. Therefore, the

charges q1 and q2 each exert a repulsive

force on the charge qA. As the drawing

shows, these forces have magnitudes FA1

(vertically downward) and FA2 (horizontally

to the left). The unknown charge placed at the empty corner of the rectangle is qU, and it exerts a force on qA that has a magnitude FAU.

In order that the net force acting on qA point in the vertical direction, the horizontal

component of FAU must cancel out the horizontal force FA2. Therefore, FAU must point as

shown in the drawing, which means that it is an attractive force and qU must be negative,

since qA is positive.

SOLUTION The basis for our solution is the fact that the horizontal component of FAU

must cancel out the horizontal force FA2. The magnitudes of these forces can be expressed

using Coulomb’s law 2/F k q q r , where r is the distance between the charges q and q .

Thus, we have

A U A 2AU A22 22

and 4 4

k q q k q qF F

d d d

where we have used the fact that the distance between the charges qA and qU is the diagonal

of the rectangle, which is 2 24d d according to the Pythagorean theorem, and the fact

that the distance between the charges qA and q2 is 4d. The horizontal component of FAU is

AUcosF , which must be equal to FA2, so that we have

A U UA 2 2

2 22cos or cos

17 164 4

k q q qk q q q

d d d

The drawing in the REASONING, reveals that 2 2cos 4d / 4 4 / 17d d .

Therefore, we find that

U 2 6 6U 2

4 17 17 17 17 or 3.0 10 C 3.3 10 C

17 16 64 6417

q qq q

As discussed in the REASONING, the algebraic sign of the charge qU is negative .

qU q1

qA q2 = +3.0 C

FA2 FAU

d

4d

q

FA1


26. REASONING AND SOLUTION In order for the net force on any charge to be directed

inward toward the center of the square, the charges must be placed with alternate + and –

signs on each successive corner. The magnitude of the force on any charge due to an

adjacent charge located at a distance r is

22 9 2 2 6

2 2

8.99 10 N m / C 2.0 10 C0.40 N

0.30 m

k qF

r

The forces due to two adjacent charges are perpendicular to one another and produce a

resultant force that has a magnitude of

22

adjacent2 2 0.40 N 0.57 NF F

The magnitude of the force due to the diagonal charge that is located at a distance of 2r is

diagonal

2 2

2 2

0.40 N0.20 N

222

k q k qF

rr

since the diagonal distance is 2r . The force Fdiagonal is directed opposite to Fadjacent (since

the diagonal charges are of the same sign). Therefore, the net force acting on any of the

charges is directed inward and has a magnitude

Fnet = Fadjacent – Fdiagonal = 0.57 N – 0.20 N = 0.37 N

______________________________________________________________________________

27. SSM REASONING The charged insulator experiences an electric force due to the

presence of the charged sphere shown in the drawing in the text. The forces acting on the

insulator are the downward force of gravity (i.e., its weight, W mg), the electrostatic force 2

1 2/F k q q r (see Coulomb's law, Equation 18.1) pulling to the right, and the tension T

in the thread pulling up and to the left at an angle with respect to the vertical, as shown in

the drawing in the problem statement. We can analyze the forces to determine the desired

quantities q and T.

SOLUTION.

a. We can see from the diagram given with the problem statement that

xT F which gives 2

1 2sin /T k q q r

and

yT W which gives cosT mg

Dividing the first equation by the second yields

21 2

/sintan

cos

k q q rT

T mg


Solving for q, we find that

1 2–1

2

9 2 2 –6 –6–1

–2 2 2

tan

(8.99 10 N m /C )(0.600 10 C)(0.900 10 C) tan 15.4

(8.00 10 kg)(9.80 m/s )(0.150 m)

k q q

mgr

b. Since cosT mg , the tension can be obtained as follows:

2 2 (8.00 10 kg) (9.80 m/s )

0.813 Ncos cos 15.4

mgT

______________________________________________________________________________

28. REASONING AND SOLUTION Since the objects attract each other initially, one object

has a negative charge, and the other object has a positive charge. Assume that the negative

charge has a magnitude of 1

q and that the positive charge has a magnitude of 2

q .

Assume also that 2

q is greater than 1

q . The magnitude F of the initial attractive force

between the objects is

1 2

2

k q qF

r (1)

After the objects are brought into contact and returned to their initial positions, the charge

on each object is the same and has a magnitude of 2 1/ 2q q . The magnitude F of the

force between the objects is now

2

2 1

2

/ 2k q qF

r

(2)

It is given that F is the same in Equations (1) and (2). Therefore, we can equate

Equations (1) and (2) and rearrange the result to find that

2 2

1 2 1 26 0q q q q (3)

The solutions to this quadratic equation are

1 2 1 25.828 and 0.1716q q q q (4)

Since we have assumed that 2

q is greater than 1

q , we must choose the solution

1 20.1716q q . Substituting this result into Equation (1) and using the given values of

F = 1.20 N and r = 0.200 m, we find that


1 20.957 C and 5.58 Cq q (5)

Note that we need not have assumed that 2

q is greater than 1

q . We could have assumed

that 2

q is less than 1

q . Had we done so, we would have found that

1 25.58 C and 0.957 Cq q (6) Considering Equations (5) and (6) and remembering that q1 is the negative charge, we

conclude that the two possible solutions to this problem are

1 2 1 20.957 C and 5.58 C or 5.58 C and 0.957 Cq q q q

______________________________________________________________________________

29. SSM REASONING The electric field created by a point charge is inversely proportional

to the square of the distance from the charge, according to Equation 18.3. Therefore, we

expect the distance r2 to be greater than the distance r1, since the field is smaller at r2 than it

is at r1. The ratio r2/r1, then, should be greater than one.

SOLUTION Applying Equation 18.3 to each position relative to the charge, we have

1 22 21 2

and k q k q

E Er r

Dividing the expression for E1 by the expression for E2 gives

2 21 1 2

2 22 2 1

/

/

E k q r r

E k q r r

Solving for the ratio r2/r1 gives

2 1

1 2

248 N/C1.37

132 N/C

r E

r E

As expected, this ratio is greater than one.

30. REASONING

a. The magnitude of the electric field is obtained by dividing the magnitude of the force

(obtained from the meter) by the magnitude of the charge. Since the charge is positive, the

direction of the electric field is the same as the direction of the force.

b. As in part (a), the magnitude of the electric field is obtained by dividing the magnitude of

the force by the magnitude of the charge. Since the charge is negative, however, the

direction of the force (as indicated by the meter) is opposite to the direction of the electric

field. Thus, the direction of the electric field is opposite to that of the force.


SOLUTION a. According to Equation 18.2, the magnitude of the electric field is

40.0 N

2.0 N /C20.0 C

FE

q

As mentioned in the REASONING, the direction of the electric field is the same as the

direction of the force, or due east .

b. The magnitude of the electric field is

20.0 N2.0 N /C

10.0 C

FE

q

Since the charge is negative, the direction of the electric field is opposite to the direction of

the force, or due east . Thus, the electric fields in parts (a) and (b) are the same.

______________________________________________________________________________

31. SOLUTION Knowing the electric field at a spot allows us to calculate the force that acts on

a charge placed at that spot, without knowing the nature of the object producing the field.

This is possible because the electric field is defined as E = F/q0, according to Equation 18.2.

This equation can be solved directly for the force F, if the field E and charge q0 are known.

SOLUTION Using Equation 18.2, we find that the force has a magnitude of

–60

260 000 N/C 7.0 10 C 1.8 NF E q

If the charge were positive, the direction of the force would be due west, the same as the

direction of the field. But the charge is negative, so the force points in the opposite

direction or due east. Thus, the force on the charge is 1.8 N due east .

______________________________________________________________________________

32. REASONING AND SOLUTION The electric

field lines must originate on the positive charges

and terminate on the negative charge. They

cannot cross one another. Furthermore, the

number of field lines beginning or terminating on

any charge must be proportional to the magnitude

of the charge. Thus, for every field line that

leaves the charge +q, two field lines must leave

the charge +2q. These three lines must terminate

on the 3q charge. If the sketch is to have six

field lines, two of them must originate on +q, and

four of them must originate on the charge +2q.

______________________________________________________________________________

+q +2q

-3q


33. REASONING Each charge creates an electric field at the center of the square, and the four

fields must be added as vectors to obtain the net field. Since the charges all have the same

magnitude and since each corner is equidistant from the center of the square, the magnitudes

of the four individual fields are identical. Each is given by Equation 18.3 as 2

k qE

r . The

directions of the various contributions are not the same, however. The field created by a

positive charge points away from the charge, while the field created by a negative charge

points toward the charge.

SOLUTION The drawing at the right shows

each of the field contributions at the center of

the square (see black dot). Each is directed

along a diagonal of the square. Note that ED

and EB point in opposite directions and,

therefore, cancel, since they have the same

magnitude. In contrast EA and EC point in

the same direction toward corner A and,

therefore, combine to give a net field that is

twice the magnitude of EA or EC. In other

words, the net field at the center of the square

is given by the following vector equation:

2

A B C D A B C B A C AE E E E E E E E E E E E

Using Equation 18.3, we find that the magnitude of the net field is

A 22 2

k qE E

r

In this result r is the distance from a corner to the center of the square, which is one half of

the diagonal distance d. Using L for the length of a side of the square and taking advantage

of the Pythagorean theorem, we have 2 21 12 2

r d L L . With this substitution for r, the

magnitude of the net field becomes

9 2 2 12

2 2 22 21

2

4 8.99 10 N m / C 2.4 10 C42 54 N/C

0.040 m

k q k qE

LL L

34. REASONING

Part (a) of the drawing given in the text. The electric field produced by a charge points

away from a positive charge and toward a negative charge. Therefore, the electric field E+2

produced by the +2.0 C charge points away from it, and the electric fields E3 and E5

produced by the 3.0 C and 5.0 C charges point toward them (see the left-hand side of

the following drawing). The magnitude of the electric field produced by a point charge is

A

B C

D

+ +

+

EA

EC EB

ED


given by Equation 18.3 as E = k q /r2. Since the distance from each charge to the origin is

the same, the magnitude of the electric field is proportional only to the magnitude q of the

charge. Thus, the x component Ex of the net electric field is proportional to

5.0 C (2.0 C + 3.0 C). Since only one of the charges produces an electric field in the y

direction, the y component Ey of the net electric field is proportional to the magnitude of this

charge, or 5.0 C. Thus, the x and y components are equal, as indicated at the right-hand

side of the following drawing, where the net electric field E is also shown.

Part (b) of the drawing given in the text. Using the same arguments as earlier, we find

that the electric fields produced by the four charges are shown at the left-hand side of the

following drawing. These fields also produce the same net electric field E as before, as

indicated at the right-hand side of the following drawing.

SOLUTION Part (a) of the drawing given in the text. The net electric field in the x direction is

9 2 2 6 9 2 2 6

2 2

7

8.99 10 N m /C 2.0 10 C 8.99 10 N m /C 3.0 10 C

0.061 m 0.061 m

1.2 10 N /C

xE

The net electric field in the y direction is

5.0 C

+2.0 C 3.0 C

E5

E+2

E3

E E

y

Ex

+6.0 C

1.0 C +4.0 C

+1.0 C E

+6 E E

y

Ex E

+4

E1

E+1

O


9 2 2 6

7

2

8.99 10 N m /C 5.0 10 C1.2 10 N /C

0.061 my

E

The magnitude of the net electric field is

2 2

2 2 7 7 71.2 10 N/C 1.2 10 N/C 1.7 10 N/Cx y

E E E

Part (b) of the drawing given in the text. The magnitude of the net electric field is the

same as determined for part (a); E = 7 1.7 10 N/C . ______________________________________________________________________________

35. REASONING At every position in space, the

net electric field E is the vector sum of the

external electric field Eext and the electric

field Epoint created by the point charge at the

origin: E = Eext + Epoint. The external electric

field is uniform, which means that it has the

same magnitude Eext = 4500 N/C at all

locations and that it always points in the

positive x direction (see the drawing). The

magnitude Epoint of the electric field due to

the point charge at the origin is given by point 2

k qE

r (Equation 18.3), where r is the

distance between the origin and the location where the electric field is to be evaluated, q is

the charge at the origin, and k = 8.99×109 N·m2/C2. All three locations given in the problem

are a distance r = 0.15 m from the origin, so we have that

9 2 2 9

point 2 2

8.99 10 N m /C 8.0 10 C3200 N/C

0.15 m

k qE

r

The direction of the electric field Epoint varies from location to location, but because the

charge q is negative, Epoint is always directed towards the origin (see the drawing).

SOLUTION a. At x = −0.15 m, the electric field of the point charge and the external electric field both

point in the positive x direction (see the drawing), so the magnitude E of the net electric

field is the sum of the magnitudes of the individual electric fields:

ext point4500 N/C 3200 N/C 7700 N/CE E E

b. At x = +0.15 m, the electric field of the point charge is opposite the external electric field

(see the drawing). Therefore, the magnitude E of the net electric field is the difference

between the magnitudes of the individual electric fields:

Epoint

y

Epoint Epoint

Eext

Eext

Eext

x


ext point4500 N/C 3200 N/C 1300 N/CE E E

c. At y = +0.15 m, the electric fields Epoint and Eext are perpendicular (see the drawing).

This makes them, in effect, the x-component (Eext) and y-component (Epoint) of the net

electric field E. The magnitude E of the net electric field, then, is given by the Pythagorean

theorem (Equation 1.7):

2 22 2ext point

5500 N/C4500 N/C 3200 N/CE E E

36. REASONING

a. The magnitude E of the electric field is given by 0

/E (Equation 18.4), where is

the charge density (or charge per unit area) and 0 is the permittivity of free space.

b. The magnitude F of the electric force that would be exerted on a K+ ion placed inside the

membrane is the product of the magnitude 0

q of the charge and the magnitude E of the

electric field (see Equation 18.2), or 0

F q E . SOLUTION

a. The magnitude of the electric field is

6 25

12 2 20

7.1 10 C/m8.0 10 N/C

8.85 10 C / N mE

b. The magnitude F of the force exerted on a K+ ion (q0 = +e) is

19 5 130

1.60 10 C 8.0 10 N/C 1.3 10 NF q E e E ______________________________________________________________________________

37. SSM REASONING The drawing at the right shows

the set-up. Here, the electric field E points along the +y

axis and applies a force of +F to the +q charge and a

force of –F to the –q charge, where q = 8.0 C denotes

the magnitude of each charge. Each force has the same

magnitude of F = E q , according to Equation 18.2.

The torque is measured as discussed in Section 9.1.

According to Equation 9.1, the torque produced by each

force has a magnitude given by the magnitude of the

force times the lever arm, which is the perpendicular

distance between the point of application of the force

and the axis of rotation. In the drawing the z axis is the axis of rotation and is midway

between the ends of the rod. Thus, the lever arm for each force is half the length L of the rod

or L/2, and the magnitude of the torque produced by each force is (E q )(L/2).

+x

+y

+z

+q

–q

E

+F

–F


SOLUTION The +F and the –F force each cause the rod to rotate in the same sense about

the z axis. Therefore, the torques from these forces reinforce one another. Using the

expression (E q )(L/2) for the magnitude of each torque, we find that the magnitude of the

net torque is

3 –6

Magnitude ofnet torque 2 2

5.0 10 N/C 8.0 10 C 4.0 m 0.16 N m

L LE q E q E q L

______________________________________________________________________________

38. REASONING Before the 3.0-C point charge q is introduced into the region, the region

contains a uniform electric field E of magnitude 1.6 104

N/C. After the 3.0-C charge is

introduced into the region, the net electric field changes. In addition to the uniform electric

field E , the region will also contain the electric field Eq due to the point charge q. The field

at any point in the region is the vector sum of E and Eq. The field Eq is radial as discussed

in the text, and its magnitude at any distance r from the charge q is given by 2/q

E k q r

(Equation 18.3). There will be one point P in the region where the net electric field Enet is

zero. This point is located where the field E has the same magnitude and points in the

direction opposite to the field Eq. We will use this reasoning to find the distance r0 from the

charge q to the point P.

SOLUTION Let us assume that the field E points to the right and that the charge q is

negative (the problem is done the same way if q is positive, although then the relative

positions of P and q will be reversed). Since q is negative, its electric field is radially

inward (i.e., toward q); therefore, in order for the field Eq to point in the opposite direction

to E , the charge q will have to be to the left of the point P where Enet is zero, as shown in

the drawing at the right. Using 20

/q

E k q r (Equation 18.3) and

solving for the distance r0, we find

0/

qr k q E . Since the magnitude

Eq must be equal to the magnitude of E

at the point P, we have

9 2 2 –6

0 4

(8.99 10 Nm /C ) (3.0 10 C)1.3 m

1.6 10 N/C

k qr

E

______________________________________________________________________________

39. SSM REASONING Two forces act on the charged ball (charge q); they are the

downward force of gravity mg and the electric force F due to the presence of the charge q in

the electric field E. In order for the ball to float, these two forces must be equal in

magnitude and opposite in direction, so that the net force on the ball is zero (Newton's


second law). Therefore, F must point upward, which we will take as the positive direction.

According to Equation 18.2, F = qE. Since the charge q is negative, the electric field E

must point downward, as the product qE in the expression F = qE must be positive, since

the force F points upward. The magnitudes of the two forces must be equal, so that

mg q E . This expression can be solved for E.

SOLUTION The magnitude of the electric field E is

23

–6

(0.012 kg)(9.80 m/s )6.5 10 N/C

18 10 C

mgE

q

As discussed in the reasoning, this electric field points downward . ______________________________________________________________________________

40. REASONING The proton and the electron have the same charge magnitude e, so the

electric force that each experiences has the same magnitude. The directions are different,

however. The proton, being positive, experiences a force in the same direction as the

electric field (due east). The electron, being negative, experiences a force in the opposite

direction (due west).

Newton’s second law indicates that the direction of the acceleration is the same as the

direction of the net force, which, in this case, is the electric force. The proton’s acceleration

is in the same direction (due east) as the electric field. The electron’s acceleration is in the

opposite direction (due west) as the electric field.

Newton’s second law indicates that the magnitude of the acceleration is equal to the

magnitude of the electric force divided by the mass. Although the proton and electron

experience the same force magnitude, they have different masses. Thus, they have

accelerations of different magnitudes.

SOLUTION According to Newton’s second law, Equation 4.2, the acceleration a of an

object is equal to the net force divided by the object’s mass m. Here there is only one force,

the electric force F, so it is the net force. According to Equation 18.2, the magnitude of the

electric force is equal to the product of the magnitude of the charge and the magnitude of the

electric field, or F = 0

q E. Thus, the magnitude of the acceleration can be written as

0q EF

am m

The magnitude of the acceleration of the proton is

19 4

0 12 2

27

1.60 10 C 8.0 10 N /C7.7 10 m/s

1.67 10 kg

q Ea

m

The magnitude of the acceleration of the electron is

19 4

0 16 2

31

1.60 10 C 8.0 10 N /C1.4 10 m/s

9.11 10 kg

q Ea

m

______________________________________________________________________________


41. REASONING AND SOLUTION Figure 1 at the right shows

the configuration given in text Figure 18.21a. The electric field

at the center of the rectangle is the resultant of the electric fields

at the center due to each of the four charges. As discussed in

Conceptual Example 11, the magnitudes of the electric field at

the center due to each of the four charges are equal. However,

the fields produced by the charges in corners 1 and 3 are in

opposite directions. Since they have the same magnitudes, they

combine to give zero resultant.

The fields produced by the charges in corners 2 and 4 point in

the same direction (toward corner 2). Thus, EC = EC2 + EC4,

Figure 1

where EC is the magnitude of the electric field at the center of the rectangle, and EC2 and

EC4 are the magnitudes of the electric field at the center due to the charges in corners 2 and

4 respectively. Since both EC2 and EC4 have the same magnitude, we have EC = 2 EC2.

The distance r, from any of the charges to the center

of the rectangle, can be found using the Pythagorean

theorem (see Figure 2 at the right):

2 2(3.00 cm) +(5.00 cm) 5.83 cmd

2Therefore, 2.92 cm 2.92 10 m2

dr

Figure 2 The electric field at the center has a magnitude of

9 2 2 12

2 22 2 2 2

2 2(8.99 10 N m /C )(8.60 10 C)2 1.81 10 N/C

(2.92 10 m)C C

k qE E

r

Figure 3 at the right shows the configuration given in text

Figure 18.21b. All four charges contribute a non-zero

component to the electric field at the center of the rectangle.

As discussed in Conceptual Example 11, the contribution

from the charges in corners 2 and 4 point toward corner 2 and

the contribution from the charges in corners 1 and 3 point

toward corner 1.

Notice also, the magnitudes of E24 and E13 are equal, and,

from the first part of this problem, we know that

E24 = E13 = 1.81 102 N/C

Figure 3

1 2

34

1

4

+ q

+ q + q

- q

1 2

34

1

4

d

3.00 cm

5.00 cm

1 2

34

1

4

- q

+ q + q

- q

C

E13

E24


The electric field at the center of the rectangle is the vector sum of E24 and E13. The x

components of E24 and E13 are equal in magnitude and opposite in direction; hence

(E13)x – (E24)x = 0

Therefore,

C 13 24 13 13( ) ( ) 2( ) 2( )sin

y y yE E E E E

From Figure 2, we have that

5.00 cm 5.00 cmsin 0.858

5.83 cmd

and

2 2C 13

2 sin 2 1.81 10 N/C 0.858 3.11 10 N/CE E

______________________________________________________________________________

42. REASONING AND SOLUTION The magnitude of the force on q1 due to q2 is given by

Coulomb's law:

1 212 2

12

k q qF

r

(1)

The magnitude of the force on q1 due to the electric field of the capacitor is given by

1C 1 C 10

F q E q

(2)

Equating the right hand sides of Equations (1) and (2) above gives

1 212

012

k q qq

r

Solving for r12 gives

0 212

12 2 2 9 2 2 6–2

2

=

[8.85×10 C /(N m )](8.99 ×10 N m /C )(5.00 ×10 C)= = 5.53×10 m

(1.30 ×10 C/m )

k qr

______________________________________________________________________________

43. REASONING The electric field is given by Equation 18.2 as the force F that acts on a test

charge q0, divided by q0. Although the force is not known, the acceleration and mass of the

charged object are given. Therefore, we can use Newton’s second law to determine the force

as the mass times the acceleration and then determine the magnitude of the field directly


from Equation 18.2. The force has the same direction as the acceleration. The direction of

the field, however, is in the direction opposite to that of the acceleration and force. This is

because the object carries a negative charge, while the field has the same direction as the

force acting on a positive test charge.

SOLUTION According to Equation 18.2, the magnitude of the electric field is

0

FE

q

According to Newton’s second law, the net force acting on an object of mass m and

acceleration a is F = ma. Here, the net force is the electrostatic force F, since that force

alone acts on the object. Thus, the magnitude of the electric field is

3 3 2

5

60 0

3.0 10 kg 2.5 10 m/s2.2 10 N/C

34 10 C

F maE

q q

The direction of this field is opposite to the direction of the acceleration. Thus, the field

points along the x axis .

44. REASONING The external electric field E exerts a force FE = qE

(Equation 18.2) on the sphere, where q = +6.6 C is the net charge

of the sphere. The external electric field E is directed upward, so

the force FE it exerts on the positively charged sphere is also

directed upward (see the free-body diagram). Balancing this

upward force are two downward forces: the weight mg of the

sphere (where m is the mass of the sphere and g is the acceleration

due to gravity) and the force Fs exerted on the sphere by the spring

(see the free-body diagram). We know that the spring exerts a

downward force on the sphere because the equilibrium length

L = 0.059 m of the spring is shorter than its unstrained length

L0 = 0.074 m. The magnitude Fs of the spring force is given by s

F kx (Equation 10.2,

without the minus sign), where k is the spring constant of the spring and

x = L0 − L = 0.074 m – 0.059 m = 0.015 m is the distance by which the spring has been

compressed.

SOLUTION Solving FE = qE (Equation 18.2) for the magnitude E of the external electric

field, we find that

EF

Eq

(1)

The sphere is in equilibrium, so the upward force (FE) must exactly balance the two

downward forces (mg and Fs). Therefore, the magnitudes of the three forces are related by

q = +6.6 C

mg Fs

FE

Free-body diagram


E s

F mg F (2) Substituting Equation (2) into Equation (1) yields

smg F

Eq

(3)

Lastly, substituting

sF kx (Equation 10.2, without the minus sign) into Equation (3), we

obtain the desired electric field magnitude:

3 2

4

6

5.1 10 kg 9.80 m/s 2.4 N/m 0.015 m1.3 10 N/C

6.6 10 C

mg kxE

q

45. SSM REASONING The two charges lying on the x axis produce no net electric field at

the coordinate origin. This is because they have identical charges, are located the same

distance from the origin, and produce electric fields that point in opposite directions. The

electric field produced by q3 at the origin points away from the charge, or along the y

direction. The electric field produced by q4 at the origin points toward the charge, or along

the +y direction. The net electric field is, then, E = –E3 + E4, where E3 and E4 can be

determined by using Equation 18.3.

SOLUTION The net electric field at the origin is

3 43 4 2 2

3 4

9 2 2 6 9 2 2 6

2 22 2

6

8.99 10 N m /C 3.0 10 C 8.99 10 N m /C 8.0 10 C

5.0 10 m 7.0 10 m

3.9 10 N/C

k q k qE E E

r r

The plus sign indicates that the net electric field points along the + directiony .

______________________________________________________________________________

46. REASONING The electric field is a vector. Therefore, the total field E is the vector sum of

its two parts, or 1 2 E E E . We will carry out this vector addition by using the method of

components (see Section 1.8).


SOLUTION The drawing at the right shows

the two vectors E1 and E2, together with their x

and y components. In the following table, we

calculate the components of each vector. We

also show the x component Ex of the total field

as the sum of the individual x components of

E1 and E2 and the y component Ey of the total

field as the sum of the individual y components

of E1 and E2. Note that the calculations in the

table carry additional significant figures. Rounding off to the correct number of significant

figures will be done when we calculate the final answers.

Vector x component y component

1E 1 1 1

cos 1200 N/C cos35

983 N/C

xE E

1 1 1sin 1200 N/C sin35

688 N/C

yE E

2E 2 2 2

cos 1700 N/C cos55

975 N/C

xE E

2 2 2sin 1700 N/C sin55

1393 N/C

yE E

1 2 E E E

1 2983 N/C 975 N/C

1958 N/C

x x xE E E

1 2 688 N/C 1393 N/C

2081 N/C

y y yE E E

Since the components Ex and Ey of the total field are perpendicular, we can use the

Pythagorean theorem to calculate the magnitude E of the total field and trigonometry to

calculate the directional angle θ:

2 22 2

1 1

1958 N/C 2081 N/C 2900 N/C

2081 N/Ctan tan 47

1958 N/C

x y

y

x

E E E

E

E

47. REASONING Since we know the initial velocity and time, we can determine the particle’s

displacement from an equation of kinematics, provided its acceleration can be determined.

The acceleration is given by Newton’s second law as the net force acting on the particle

divided by its mass. The net force is the electrostatic force, since the particle is moving in an

electric field. The electrostatic force depends on the particle’s charge and the electric field,

both of which are known.

SOLUTION To obtain the displacement x of the particle we employ Equation 3.5a from the

equations of kinematics: 210 2x x

x v t a t . We use this equation because two of the

variables, the initial velocity v0x and the time t, are known. The initial velocity is zero, since

+y

+x

E2

35º

55º E1

E2x

E1x E1y

E2y


the particle is released from rest. The acceleration ax can be found from Newton’s second

law, as given by Equation 4.2a, as the net force Fx acting on the particle divided by its

mass m: /x x

a F m . Only the electrostatic force Fx acts on the proton, so it is the net

force. Setting x x

F F in Newton’s second law gives /x x

a F m . Substituting this result

into Equation 3.5a, we have that

2 21 10 02 2

xx x x

Fx v t a t v t t

m

(1)

Since the particle is moving in a uniform electric field Ex, it experiences an electrostatic

force Fx given by 0x x

F q E (Equation 18.2), where q0 is the charge. Substituting this

expression for Fx into Equation (1) gives

2 201 10 02 2

6 22 2 21

2 5

12 10 C 480 N/C0 m/s 1.6 10 s 1.6 10 s 1.9 10 m

3.8 10 kg

x xx x

F q Ex v t t v t t

m m

______________________________________________________________________________

48. REASONING

a. The drawing at the right shows the

electric fields at point P due to the two

charges in the case that the second

charge is positive. The presence of the

second charge causes the magnitude of the net field at P to be twice as great as it is when

only the first charge is present. Since both fields have the same direction, the magnitude of

E2 must, then, be the same as the magnitude of E1. But the second charge is further away

from point P than is the first charge, and more distant charges create weaker fields. To offset

the weakness that comes from the greater distance, the second charge must have a greater

magnitude than that of the first charge.

b. The drawing at the right shows the

electric fields at point P due to the two

charges in the case that the second

charge is negative. The presence of the

second charge causes the magnitude of the net field at P to be twice as great as it is when

only the first charge is present. Since the fields now have opposite directions, the magnitude

of E2 must be greater than the magnitude of E1. This is necessary so that E2 can offset E1

and still lead to a net field with twice the magnitude as E1. To create this greater field E2, the

second charge must now have a greater magnitude than it did in question (a).

+q1 q2 P

d d

E1 E2

+q1 q2 P

d d +

E2

E1


SOLUTION a. The magnitudes of the field contributions of each charge are given according to

Equation 18.3 as 2

k qE

r . With q2 present, the magnitude of the net field at P is twice what

it is when only q1 is present. Using Equation 18.3, we can express this fact as follows:

1 2 1 2 1

2 2 2 2 22 or =

2 2

k q k q k q k q k q

d d dd d

Solving for 2

q gives

2 14 4 0.50 C 2.0 Cq q

Thus, the second charge is q2 = +2.0 μC .

b. Now that the second charge is negative, we have

2 1 1 2 1

2 2 2 2 22 or =3

2 2

k q k q k q k q k q

d d dd d

Solving for 2

q gives

2 112 12 0.50 C 6.0 Cq q

Thus, the second charge is q2 = 6.0 μC .

49. REASONING AND SOLUTION From kinematics, vy2 = v0y

2 + 2ayy. Since the electron

starts from rest, v0y = 0 m/s. The acceleration of the electron is given by

y

F eEa

m m

where e and m are the electron's charge magnitude and mass, respectively, and E is the

magnitude of the electric field. The magnitude of the electric field between the plates of a

parallel plate capacitor is E = s/0, where s is the magnitude of the charge per unit area on

each plate. Thus, ay = es/(m0). Combining this expression for a with the kinematics

equation we have

2

0

2y

ev y

m

Solving for vy gives


19 7 2 2

7

31 12 2 20

2 1.60 10 C 1.8 10 C/m 1.5 10 m21.0 10 m/s

9.11 10 kg 8.85 10 C / N my

e yv

m

______________________________________________________________________________

50. REASONING The following drawing shows the two particles in the electric field Ex. They

are separated by a distance d. If the particles are to stay at the same distance from each other

after being released, they must have the same acceleration, so ax, 1 = ax, 2. According to

Newton’s second law (Equation 4.2a), the acceleration ax of each particle is equal to the net

force Fx acting on it divided by its mass m, or /x x

a F m .

SOLUTION The net force acting on each particle and its resulting acceleration are:

q1: The charge q1 experiences a force 1 x

q E due to the electric field (see Equation 18.2).

The charge also experiences an attractive force in the +x direction due to the presence

of q2. This force is given by Coulomb’s law as + 21 2

/k q q d (see Equation 18.1). The

net force acting on q1 is

1 2, 1 1 2x x

q qF q E k

d

The acceleration of q1 is

1 21 2, 1

,11 1

xx

x

q qq E kF

dam m

q2: The charge q2 experiences a force 2 x

q E due to the electric field. It also experiences an

attractive force in the x direction due to the presence of q1. This force is given by

Coulomb’s law as 21 2

/k q q d . The net force acting on q2 is

1 2, 2 2 2x x

q qF q E k

d

The acceleration of q2 is

Ex

d

+x

q1 = 7.0 C

m1 = 1.4 105

kg

q2 = +18 C

m2 = 2.6 105

kg


1 22 2, 2

, 22 2

xx

x

q qq E kF

dam m

Setting ax, 1 = ax, 2 gives

1 2 1 21 22 2

1 2

x x

q q q qq E k q E k

d d

m m

Solving this expression for d, we find that

1 21 2

2 1

2 1

29 6 6

2 5 5

6 6

5 5

1 1

N m 1 18.99 10 7.0 10 C 18 10 C

C 1.4 10 kg 2.6 10 kg6.5 m

18 10 C 7.0 10 C2500 N/C

2.6 10 kg 1.4 10 kg

x

k q qm m

dq q

Em m

______________________________________________________________________________

51. SSM REASONING AND SOLUTION The net electric field at point P in Figure 1 is the

vector sum of the fields E+ and E–, which are due, respectively, to the charges +q and –q.

These fields are shown in Figure 2.

According to Equation 18.3, the magnitudes of the fields E+ and E– are the same, since the

triangle is an isosceles triangle with equal sides of length . Therefore, 2–

/E E k q .

ℓ ℓ


The vertical components of these two fields cancel, while the horizontal components

reinforce, leading to a total field at point P that is horizontal and has a magnitude of

– 2cos cos 2 cos

P

k qE E E

At point M in Figure 1, both E+ and E– are horizontal and point to the right. Again using

Equation 18.3, we find

– 2 2 2

2M

k q k q k qE E E

d d d

Since EM

/EP = 9.0, we have

2

2 2 2

2 / 19.0

2 cos / cos /

M

P

E k q d

E k q d

But from Figure 1, we can see that d/ = cos . Thus, it follows that

3

3

19.0 or cos 1/ 9.0 0.48

cos

The value for is, then, –1cos 0.48 61 .

______________________________________________________________________________

52. REASONING In the drawing that accompanies the problem statement, we assume that the

electron is initially moving in the +x direction and then begins moving upward in the

+y direction as it moves through the capacitor. The upward part of the motion occurs

because the electric field E of the capacitor, which points downward from the positive plate

toward the negative plate, exerts a force F on the electron. The electric field is given by

0/E F q (Equation 18.2), where

0q e is the charge on the electron. Newton’s second

law applies to the force (assumed to be the only force present), so that yF ma

(Equation 4.1), where m is the mass of the electron and ay is the electron’s acceleration in

the y direction. The kinematics equations apply to the motion in both the x and y directions,

and with their aid we can determine the acceleration ay. Knowing the acceleration, we will

be able to determine the force and, hence, the electric field.

SOLUTION According to 0

/E F q (Equation 18.2) and yF ma (Equation 4.1), we

have

0 0

ymaF

Eq q

(1)

From kinematics we know that the electron’s displacement in the upward direction at the

time t that it exits the capacitor is 20.150 10 my and is given by 1 2

2 yy a t

(Equation 3.5b with 0 0 m / syv since the electron is initially not moving in the


y direction). We can solve this equation to obtain 22 /y

a y t and substitute this result into

Equation (1):

20 0

2yma m y

Eq q t

(2)

We also know that the electron’s displacement in the horizontal direction at the time t that it

exits the capacitor is 22.00 10 mx and is given by 0xx v t , since the horizontal

speed of the electron is 60 7.00 10 m / sxv and remains constant during the motion. We

can solve this equation to obtain 0

/ xt x v and substitute this result into Equation (2):

20

2 2 20 00 0

22 2

/

x

x

m yvm y m yE

q t q xq x v (3)

With 190 1.60 10 Cq e as the electron charge and 319.11 10 kgm as the

electron mass, Equation (3) gives

231 2 62

02 2

19 20

9.11 10 kg 2 0.150 10 m 7.00 10 m / s22090 N/C

1.60 10 C 2.00 10 m

xm yvE

q x

This result is negative, because the electric field points downward in the drawing that

accompanies the problem, which is the direction that we assumed to be the negative

y direction. The magnitude of the electric field, then, is 2090 N/C .

______________________________________________________________________________

53. REASONING AND SOLUTION Since the thread makes an angle of 30.0° with the

vertical, it can be seen that the electric force on the ball, Fe, and the gravitational force, mg,

are related by

Fe = mg tan 30.0° The force Fe is due to the charged ball being in the electric field of the parallel plate

capacitor. That is,

e ballF E q (1)

where ball

q is the magnitude of the ball's charge and E is the magnitude of the field due to

the plates. According to Equation 18.4 E is

0

qE

A (18.4)

where q is the magnitude of the charge on each plate and A is the area of each plate.

Substituting Equation 18.4 into Equation (1) gives


balle

0

tan 30.0q q

F mgA

Solving for q yields

0

ball

–12 2 2 2 –3 2

–6

–8

tan 30.0

8.85 10 C / N m 0.0150 m 6.50 10 kg 9.80 m/s tan 30.0

0.150 10 C

3.25 10 C

Amgq

q

______________________________________________________________________________

54. REASONING AND SOLUTION Gauss' Law is given by text Equation 18.7: E = 0

Q

,

where Q is the net charge enclosed by the Gaussian surface.

a. E 3.5 10– 6 C

8.851012 C 2 /(N m 2 ) 4.0 105 N m 2 /C

b. E 2.3 10– 6 C

8.851012 C 2 /(N m 2 ) –2.6105 N m 2 /C

c. E (3.5 10– 6 C) (2.3 10– 6 C)

8.851012 C2 /(N m 2 ) 1.4 105 N m 2 /C

______________________________________________________________________________

55. SSM REASONING As discussed in Section 18.9, the magnitude of the electric flux E

through a surface is equal to the magnitude of the component of the electric field that is

normal to the surface multiplied by the area of the surface, EE A

, where E is the

magnitude of the component of E that is normal to the surface of area A. We can use this

expression and the figure in the text to determine the desired quantities.

SOLUTION

a. The magnitude of the flux through surface 1 is

2 2E 11

( cos 35 ) (250 N/C)(cos 35 )(1.7 m ) 350 N m /CE A

b. Similarly, the magnitude of the flux through surface 2 is

2 2E 22

( cos 55 ) (250 N/C)(cos 55 )(3.2 m ) 460 N m /CE A

______________________________________________________________________________


56. REASONING In each case, Gauss’ law can be used to determine the electric flux E

. For

a Gaussian surface that encloses a net charge Q, this law is E

0

Q

(Equation 18.7), where

12 2 20

8.85 10 C / N m .

SOLUTION

a. Since the net charge surrounded by the surface is 62.0 10 CQ , Gauss’ law shows

that

65 2

E 12 2 20

2.0 10 C2.3 10 N m /C

8.85 10 C / N m

Q

b. Since the net charge surrounded by the surface is the same as in part a and since Gauss’

law holds for a closed surface of any shape, the flux is the same as in part a;

5 2E

2.3 10 N m /C .

c. Since the net charge surrounded by the surface is the same as in part a and since Gauss’

law holds for a closed surface of any shape, the flux is the same as in part a;

5 2E

2.3 10 N m /C .

______________________________________________________________________________

57. REASONING The magnitude of the electric flux E through the circular surface is

determined by the angle (less than 90º) between the electric field and the normal to the

surface, as well as the magnitude E of the electric field and the area A of the surface:

EcosE A (18.6)

We will use Equation 18.6 to determine the angle .

SOLUTION Solving Equation 18.6 for the angle , we obtain

1E Ecos or cosEA EA

(1)

The surface is circular, with a radius r, so its area is 2A r . Making this substitution in

Equation (1) yields

21 1 1EE

242

78 N m /Ccos cos cos 58

1.44 10 N/C 0.057 mE rEA


58. REASONING The charge Q inside the rectangular box is related to the electric flux E that

passes through the surfaces of the box by Gauss’ law, 0 E

Q (Equation 18.7), where 0

is the permittivity of free space. The electric flux is the algebraic sum of the flux through

each of the six surfaces.

SOLUTION The charge inside the box is

0 E 0 1 2 3 4 5 6

2 2 2 212

2

2 2 2

8

C N m N m N m8.85 10 +1500 2200 4600

C C CN m

N m N m N m 1800 3500 5400

C C C

2.1 10 C

Q

______________________________________________________________________________

59. REASONING We use a Gaussian surface that is a sphere centered within the solid sphere.

The radius r of this surface is smaller than the radius R of the solid sphere. Equation 18.7

gives Gauss’ law as follows:

E

0Electric flux,

cosQ

E A

(18.7)

We will deal first with the left side of this equation and evaluate the electric flux ФE. Then

we will evaluate the net charge Q within the Gaussian surface.

SOLUTION The positive charge is

spread uniformly throughout the solid

sphere and, therefore, is spherically

symmetric. Consequently, the electric

field is directed radially outward, and for

each element of area A is perpendicular

to the surface. This means that the

angle between the normal to the surface and the field is 0º, as the drawing shows.

Furthermore, the electric field has the same magnitude everywhere on the Gaussian surface.

Because of these considerations, we can write the electric flux as follows:

cos cos0E A E A E A

The term A is the entire area of the spherical Gaussian surface or 4πr2. With this

substitution, the electric flux becomes

2cos 4E A E A E r (1)

Normal

E Angle between E

and the normal is 0º


Now consider the net charge Q within the Gaussian surface. This charge is the charge

density times the volume 343

r encompassed by that surface:

3

3433 34

3

Volume ofChargeGaussian density

surface

q qrQ r

R R

(2)

Substituting Equations (1) and (2) into Equation 18.7 gives

3 3

2

0

/4

qr RE r

Rearranging this result shows that

3 3

3200

/

44

qr R qrE

Rr

60. REASONING Gauss’ Law, 0

cosQ

E A

(Equation 18.7), relates the electric field

magnitude E on a Gaussian surface to the net charge Q enclosed by that surface;

EcosE A (Equation 18.6) is the electric flux through the Gaussian surface

(divided into many tiny sections of area ΔA) and 0 is the permittivity of free space. We are

to determine the magnitude E of the electric field due to electric charges that are spread

uniformly over the surfaces of two concentric spherical shells. The electric field due to these

charges possesses spherical symmetry, so we will choose Gaussian surfaces in the shape of

spheres concentric with the shells. The radius r of each Gaussian surface will be equal to the

distance from the common center of the shells and will be the distance at which we are to

evaluate the electric field.

Because the electric field has spherical symmetry, the magnitude E of the electric field is

constant at all points on any such spherical Gaussian surface. Furthermore, the electric field

is directed either radially outward (if the net charge within the Gaussian surface is positive)

or radially inward (if the net charge within the Gaussian surface is negative). This means

that the angle between the electric field and the normal to any such spherical Gaussian

surface is either 0.0° or 180°. The quantity cosE , therefore, is constant, and may be

factored out of the summation in Equation 18.6:

Ecos cosE A E A (1)

The sum ΔA of all the tiny sections of area ΔA that compose a spherical Gaussian surface

is the total surface area ΔA = A = 4 r 2 of a sphere of radius r. Thus, Equation (1) becomes

2E

cos 4 cosE A r E (2)


SOLUTION

a. The outer shell has a radius r2 = 0.15 m, and we are to determine the electric field at a

distance r = 0.20 m from the common center of the shells. Therefore, we will choose a

spherical Gaussian surface (radius r = 0.20 m) that encloses both shells and shares their

common center. According to Gauss’ law and Equation (2), we have that the net electric

flux E through this sphere is

2

0

cos 4 cosQ

E A r E

(3)

Solving Equation (3) for E yields

2

04 cos

QE

r (4)

Because the chosen Gaussian surface encloses both shells, the net charge Q enclosed by the

surface is Q = q1 + q2. The positive charge q2 on the outer shell has a larger magnitude than

the negative charge q1 on the inner shell, so that Q is a positive net charge. Therefore, the

electric field is directed radially outward , and the angle between the electric field and the

normal to the surface of the spherical Gaussian surface is = 0.0°. Therefore, Equation (4)

gives the electric field magnitude as

6 61 2

2 2 212 2 20 0

5

1.6 10 C 5.1 10 C

4 cos 4 cos 4 8.85 10 C / N m 0.20 m cos0.0

7.9 10 N/C

q qQE

r r

b. We again choose a spherical Gaussian surface concentric with the shells, this time of

radius r = 0.10 m. The radius of this sphere is greater than the radius (r1 = 0.050 m) of the

inner shell but less than the radius (r2 = 0.15 m) of the outer shell. Therefore, this Gaussian

surface is located between the two shells and encloses only the charge on the inner shell:

Q = q1. This is a negative charge, so that the electric field is directed radially inward , and

the angle between the electric field and the normal to the surface of the Gaussian sphere is

= 180°. From Equation (4), then, we have that

61

2 2 212 2 20 0

6

1.6 10 C

4 cos 4 cos 4 8.85 10 C / N m 0.10 m cos180

1.4 10 N/C

qQE

r r

c. Choosing a spherical Gaussian surface with a radius of r = 0.025 m, we see that it is

entirely inside the inner shell (r1 = 0.050 m). Therefore, the enclosed charge is zero:

Q = 0 C. Equation (4) shows that the electric field at this distance from the common center

is zero:

2 20 0

0 C0 N/C

4 cos 4 cos

QE

r r


61. SSM REASONING The electric flux through each

face of the cube is given by E

( cos )E A (see

Section 18.9) where E is the magnitude of the electric

field at the face, A is the area of the face, and is the

angle between the electric field and the outward

normal of that face. We can use this expression to

calculate the electric flux E

through each of the six

faces of the cube.

SOLUTION a. On the bottom face of the cube, the outward normal points parallel to the –y axis, in

the opposite direction to the electric field, and = 180°. Therefore,

2 1 2

E bottom1500 N/C cos180° 0.20 m 6.0 10 N m /C

On the top face of the cube, the outward normal points parallel to the +y axis, and = 0.0°.

The electric flux is, therefore,

2 1 2E top

( ) (1500 N/C)(cos 0.0 )(0.20 m) +6.0 10 N m /C

On each of the other four faces, the outward normals are perpendicular to the direction of

the electric field, so = 90°. So for each of the four side faces,

2 2E sides

( ) (1500 N/C)(cos 90 )(0.20 m) 0N m /C

b. The total flux through the cube is

E total E top E E E E Ebottom side 1 side 2 side 3 side 4( ) ( )

Therefore,

1 2 1 2 2E total

( ) +6.0 10 N m /C 6.0 10 N m /C 0 0 0 0 0N m /C

______________________________________________________________________________

62. REASONING Because the charge is distributed uniformly along the straight wire, the

electric field is directed radially outward, as the following end view of the wire illustrates.


And because of symmetry, the magnitude of the electric field is the same at all points

equidistant from the wire. In this situation we will use a Gaussian surface that is a cylinder

concentric with the wire. The drawing shows that this cylinder is composed of three parts,

the two flat ends (1 and 3) and the curved wall (2). We will evaluate the electric flux for

this three-part surface and then set it equal to Q/0 (Gauss’ law) to find the magnitude of the

electric field.

SOLUTION Surfaces 1 and 3 – the flat ends of the cylinder – are parallel to the electric

field, so cos = cos 90° = 0. Thus, there is no flux through these two surfaces:

1 = 3 = 20 N m /C .

Surface 2 – the curved wall – is everywhere perpendicular to the electric field E, so

cos = cos 0° = 1. Furthermore, the magnitude E of the electric field is the same for all

points on this surface, so it can be factored outside the summation in Equation 18.6:

2 E cos0 A EA

The area A of this surface is just the circumference 2 r of the cylinder times its length L:

A = (2 r)L. The electric flux through the entire cylinder is, then,

E 1 2 3 0 E 2rL 0 E 2rL

Following Gauss’ law, we set E equal to Q/0, where Q is the net charge inside the

Gaussian cylinder: E(2 rL) = Q/0. The ratio Q/L is the charge per unit length of the wire

and is known as the linear charge density : = Q/L. Solving for E, we find that

0 0

/

2 2

Q LE

r r

______________________________________________________________________________


63. REASONING The electric field lines

must originate on the positive charges and

terminate on the negative charges. They

cannot cross one another. Furthermore, the

number of field lines beginning or ending

on any charge must be proportional to the

magnitude of the charge.

SOLUTION If 10 electric field lines leave

the +5q charge, then six lines must

originate from the +3q charge, and eight

lines must end on each –4q charge. The

drawing shows the electric field lines that

meet these criteria. ______________________________________________________________________________

64. REASONING The gravitational force is an attractive force. To neutralize this force, the

electrical force must be a repulsive force. Therefore, the charges must both be positive or

both negative. Newton’s law of gravitation, Equation 4.3, states that the gravitational force

depends inversely on the square of the distance between the earth and the moon. Coulomb’s

law, Equation 18.1 states that the electrical force also depends inversely on the square of the

distance. When these two forces are added together to give a zero net force, the distance

can be algebraically eliminated. Thus, we do not need to know the distance between the

two bodies.

SOLUTION Since the repulsive electrical force neutralizes the attractive gravitational

force, the magnitudes of the two forces are equal:

e m2 2

Electrical Gravitationalforce, force,

Equation 18.1 Equation 4.3

GM Mk q q

r r

Solving this equation for the magnitude q of the charge on either body, we find

2

11 24 222

13e m2

92

N m6.67 10 5.98 10 kg 7.35 10 kg

kg5.71 10 C

N m8.99 10

C

GM Mq

k

_____________________________________________________________________________

65. REASONING AND SOLUTION The +2q of charge initially on the sphere lies entirely on

the outer surface. When the +q charge is placed inside of the sphere, then a q charge will

still be induced on the interior of the sphere. An additional +q will appear on the outer

surface, giving a net charge of +3q . _____________________________________________________________________________


66. REASONING Since the charged droplet (charge = q) is suspended motionless in the

electric field E, the net force on the droplet must be zero. There are two forces that act on

the droplet, the force of gravity W mg , and the electric force F = qE due to the electric

field. Since the net force on the droplet is zero, we conclude that mg q E . We can use

this reasoning to determine the sign and the magnitude of the charge on the droplet.

SOLUTION

a. Since the net force on the droplet is zero, and the weight of magnitude W

points downward, the electric force of magnitude F q E must point

upward. Since the electric field points upward, the excess charge on the

droplet must be positive in order for the force F to point upward.

b. Using the expression mg q E , we find that the magnitude of the excess charge on the

droplet is

–9 2–12(3.50 10 kg)(9.80 m/s )

4.04 10 C8480 N/C

mgq

E

The charge on a proton is 1.60 10–19 C, so the excess number of protons is

–12 7

–19

1 proton4.04 10 C 2.53 10 protons

1.60 10 C

______________________________________________________________________________

67. SSM REASONING

a. The drawing shows the two point charges q1 and q2. Point A is located at x = 0 cm, and

point B is at x = +6.0 cm.

Since q1 is positive, the electric field points away from it. At point A, the electric field E1

points to the left, in the x direction. Since q2 is negative, the electric field points toward it.

At point A, the electric field E2 points to the right, in the +x direction. The net electric field

is E = E1 + E2. We can use Equation 18.3, 2/E k q r , to find the magnitude of the

electric field due to each point charge.

q2

A

q1

B 3.0 cm 3.0 cm 3.0 cm E1

E2


b. The drawing shows the electric fields produced by the charges q1 and q2 at point B,

which is located at x = +6.0 cm.

Since q1 is positive, the electric field points away from it. At point B, the electric field

points to the right, in the +x direction. Since q2 is negative, the electric field points toward

it. At point B, the electric field points to the right, in the +x direction. The net electric field

is E = +E1 + E2.

SOLUTION

a. The net electric field at the origin (point A) is E = E1 + E2:

1 21 2 2 2

1 2

9 2 2 6 9 2 2 6

2 22 2

7

8.99 10 N m /C 8.5 10 C 8.99 10 N m /C 21 10 C

3.0 10 m 9.0 10 m

6.2 10 N/C

k q k qE E E

r r

The minus sign tells us that the net electric field points along the x axis.

b. The net electric field at x = +6.0 cm (point B) is E = E1 + E2:

1 21 2 2 2

1 2

9 2 2 6 9 2 2 6

2 22 2

8

8.99 10 N m /C 8.5 10 C 8.99 10 N m /C 21 10 C

3.0 10 m 3.0 10 m

2.9 10 N/C

k q k qE E E

r r

The plus sign tells us that the net electric field points along the +x axis.

______________________________________________________________________________

q2 A

q1

B 3.0 cm 3.0 cm 3.0 cm

E1

E2


68. REASONING The magnitude F of the forces that point charges q1 and q2 exert on each

other varies with the distance r separating them according to 1 2

2

q qF k

r (Equation 18.1),

where k = 8.99×109 N·m2/C2. We note that both charges are given in units of

microcoulombs (C), rather than the base SI units of coulombs (C). We will replace the

prefix with 10−6 when calculating the distance r from Equation 18.1.

SOLUTION Solving 1 2

2

q qF k

r (Equation 18.1) for the distance r, we obtain

1 2 1 22 or q q q q

r k r kF F

Therefore, when the force magnitude F is 0.66 N, the distance between the charges must be

6 6

1 2 9 2 28.4 10 C 5.6 10 C

8.99 10 N m /C 0.80 m0.66 N

q qr k

F

69. SSM REASONING The electrons transferred increase the magnitudes of the positive and

negative charges from 2.00 μC to a greater value. We can calculate the number N of electrons

by dividing the change in the magnitude of the charges by the magnitude e of the charge on

an electron. The greater charge that exists after the transfer can be obtained from Coulomb’s

law and the value given for the magnitude of the electrostatic force.

SOLUTION The number N of electrons transferred is

after beforeq q

Ne

where after

q and before

q are the magnitudes of the charges after and before the transfer of

electrons occurs. To obtain after

q , we apply Coulomb’s law with a value of 68.0 N for the

electrostatic force: 2

2after

after2 or

q FrF k q

kr

Using this result in the expression for N, we find that

22

6before 9 2 2

12

19

68.0 N 0.0300 m2.00 10 C

8.99 10 N m / C3.8 10

1.60 10 C

Frq

kN

e


70. REASONING The fact that the net electric

field points upward along the vertical axis

holds the key to this problem. The drawing

at the right shows the fields from each

charge, together with the horizontal

components of each. The reason that the

net field points upward is that these

horizontal components point in opposite

directions and cancel. Since they cancel,

they must have equal magnitudes, a fact

that will quickly lead us to a solution.

SOLUTION Setting the magnitudes of the horizontal components of the fields equal gives

2 1sin 60.0 sin30.0E E

The magnitude of the electric field created by a point charge is given by Equation 18.3.

Using this expression for E1 and E2 and noting that each point charge is the same distance r

from the center of the circle, we obtain

2 12 12 2

sin 60.0 sin30.0 or sin 60.0 sin30.0k q k q

q qr r

Solving for the ratio of the charge magnitudes gives

2

1

sin 30.00.577

sin 60.0

q

q

71. SSM REASONING The drawing

shows the arrangement of the three

charges. Let Eq represent the electric

field at the empty corner due to the –q

charge. Furthermore, let E1 and E2 be the

electric fields at the empty corner due to

charges +q1 and +q2, respectively.

According to the Pythagorean theorem, the distance from the charge –q to the empty corner

along the diagonal is given by (2d )2 d 2 5d 2 d 5 . The magnitude of each

electric field is given by Equation 18.3, 2/E k q r . Thus, the magnitudes of each of the

electric fields at the empty corner are given as follows:

–q

+q1

+q2

d 5 d

2d

E2E1

Eq

q1

E2 sin 60.0º E

1 sin 30.0º

E2

E1

q2

30.0º 60.0º

60.0º 30.0º


2 2 255

q

k q k q k qE

r dd

1 1 21 22 2 2

and 42

k q k q k qE E

d dd

The angle q that the diagonal makes with the horizontal is 1tan ( / 2 ) 26.57d d . Since

the net electric field Enet at the empty corner is zero, the horizontal component of the net

field must be zero, and we have

11 2 2

cos 26.57– cos 26.57 0 or – 0

4 5q

k q k qE E

d d

Similarly, the vertical component of the net field must be zero, and we have

22 2 2

sin 26.57– sin 26.57 0 or – 0

5q

k q k qE E

d d

These last two expressions can be solved for the charge magnitudes 1 2

and q q .

SOLUTION Solving the last two expressions for 1 2

and q q , we find that

4

1 5

1

2 5

cos 26.57 0.716

sin 26.57 0.0895

q q q

q q q

______________________________________________________________________________

72. REASONING

The drawing at the right shows the forces that act

on the charges at each corner. For example, FAB is

the force exerted on the charge at corner A by the

charge at corner B. The directions of the forces

are consistent with the fact that like charges repel

and unlike charges attract. Coulomb’s law

indicates that all of the forces shown have the

same magnitude, namely, 2 2/F k q L , where q

is the magnitude of each of the charges and L is the

length of each side of the equilateral triangle. The

magnitude is the same for each force, because q

and L are the same for each pair of charges.

+

+ A

B

C

FAB

FBA

FCB

FBC

FCA

FAC


The net force acting at each corner is the sum of the two force vectors shown in the drawing,

and the net force is greatest at corner A. This is because the angle between the two vectors at

A is 60º. With the angle less than 90º, the two vectors partially reinforce one another. In

comparison, the angles between the vectors at corners B and C are both 120º, which means

that the vectors at those corners partially offset one another.

The net forces acting at corners B and C have the same magnitude, since the magnitudes of

the individual vectors are the same and the angles between the vectors at both B and C are

the same (120º). Thus, vector addition by either the tail-to-head method (see Section 1.6) or

the component method (see Section 1.8) will give resultant vectors that have different

directions but the same magnitude. The magnitude of the net force is the smallest at these

two corners.

SOLUTION As pointed out in the

REASONING, the magnitude of any

individual force vector is 2 2/F k q L .

With this in mind, we apply the component

method for vector addition to the forces at

corner A, which are shown in the drawing at

the right, together with the appropriate

components. The x component x

F and the

y component y

F of the net force are

AB ACA

ABA

cos60.0 cos60.0 1

sin 60.0 sin 60.0

x

y

F F F F

F F F

where we have used the fact that AB AC

F F F . The Pythagorean theorem indicates that

the magnitude of the net force at corner A is

22 2 22A A A

22 2 2 2

2

26

2 29 2 2

2

cos60.0 1 sin 60.0

cos60.0 1 sin 60.0 cos60.0 1 sin 60.0

5.0 10 C8.99 10 N m / C cos60.0 1 sin 60.0

0.030 m

430 N

x yF F F F F

qF k

L

60.0

º

FAB

FAC

+y

+x

FAB

cos 60.0

º

FAB

sin 60.0º

A


We now apply the component method

for vector addition to the forces at

corner B. These forces, together with

the appropriate components are shown

in the drawing at the right. We note

immediately that the two vertical

components cancel, since they have

opposite directions. The two horizontal

components, in contrast, reinforce since

they have the same direction. Thus, we

have the following components for the

net force at corner B:

BC BAB

B

cos60.0 cos60.0 2 cos60.0

0

x

y

F F F F

F

where we have used the fact that BC BA

F F F . The Pythagorean theorem indicates that

the magnitude of the net force at corner B is

22 2 2

B B B

262

9 2 2

2 2

2 cos60.0 0 2 cos60.0

5.0 10 C2 cos60.0 2 8.99 10 N m / C cos60.0

0.030 m

250 N

x yF F F F F

qk

L

As discussed in the REASONING, the magnitude of the net force acting on the charge at

corner C is the same as that acting on the charge at corner B, so C

250 NF .

73. REASONING The magnitude E of the electric field is the magnitude F of the electric force

exerted on a small test charge divided by the magnitude of the charge: E = F/ q . According

to Newton’s second law, Equation 4.2, the net force acting on an object is equal to its mass

m times its acceleration a. Since there is only one force acting on the object, it is the net

force. Thus, the magnitude of the electric field can be written as

F maE

q q

FBA

FBC

+y

+x

FBA

cos 60.0

º

FBC

cos 60.0º 60.0º

B


The acceleration is related to the initial and final velocities, v0 and v, and the time t through

Equation 2.4, as 0v v

at

. Substituting this expression for a into the one above for E gives

0

0

v vm

m v vtmaE

q q q t

SOLUTION The magnitude E of the electric field is

5 3

0 4

6

9.0 10 kg 2.0 10 m/s 0 m/s2.5 10 N /C

7.5 10 C 0.96 s

m v vE

q t

______________________________________________________________________________

74. REASONING We will use Coulomb’s law to calculate the force that any one charge exerts

on another charge. Note that in such calculations there are three separations to consider.

Some of the charges are a distance d apart, some a distance 2d, and some a distance 3d. The

greater the distance, the smaller the force. The net force acting on any one charge is the

vector sum of three forces. In the following drawing we represent each of those forces by an

arrow. These arrows are not drawn to scale and are meant only to “symbolize” the three

different force magnitudes that result from the three different distances used in Coulomb’s

law. In the drawing the directions are determined by the facts that like charges repel and

unlike charges attract. By examining the drawing we will be able to identify the greatest and

the smallest net force.

The greatest net force occurs for charge C, because all three force contributions point in the

same direction and two of the three have the greatest magnitude, while the third has the next

greatest magnitude. The smallest net force occurs for charge B, because two of the three

force contributions cancel.

SOLUTION Using Coulomb’s law for each contribution to the net force, we calculate the

ratio of the greatest to the smallest net force as follows:

2 2 2

2 2 2 1C 4

2 2 2 14B

2 2 2

1 129.0

2

q q qk k k

d dF d

F q q qk k k

d d d

d +

C

+

B

+

A

d d

D


75. REASONING The magnitude of the electric field between the plates of a parallel plate

capacitor is given by Equation 18.4 as 0

E

, where σ is the charge density for each plate

and ε0 is the permittivity of free space. It is the charge density that contains information

about the radii of the circular plates, for charge density is the charge per unit area. The area

of a circle is πr2. The second capacitor has a greater electric field, so its plates must have the

greater charge density. Since the charge on the plates is the same in each case, the plate area

and, hence, the plate radius, must be smaller for the second capacitor. As a result, we expect

that the ratio r2/r1 is less than one.

SOLUTION Using q to denote the magnitude of the charge on the capacitor plates and

A = πr2 for the area of a circle, we can use Equation 18.4 to express the magnitude of the

field between the plates of a parallel plate capacitor as follows:

20 0

qE

r

Applying this result to each capacitor gives

1 22 20 1 0 2

First capacitor Second capacitor

and q q

E Er r

Dividing the expression for E1 by the expression for E2 gives

2 20 11 2

222 10 2

/

/

q rE r

E rq r

Solving for the ratio r2/r1 gives

52 1

51 2

2.2 10 N/C0.76

3.8 10 N/C

r E

r E

As expected, this ratio is less than one.

76. REASONING AND SOLUTION

a. To find the charge on each ball we first need to determine the electric force acting on

each ball. This can be done by noting that each thread makes an angle of 18° with respect to

the vertical.

Fe = mg tan 18° = (8.0 10–4 kg)(9.80 m/s2) tan 18° = 2.547 10–3 N


We also know that 2

1 2e 2 2

k q q k qF

r r

where r = 2(0.25 m) sin 18° = 0.1545 m. Now

–3

–8e9 2 2

2.547 10 N0.1545 m 8.2 10 C

8.99 10 N m / C

Fq r

k

b. The tension is due to the combination of the weight of the ball and the electric force, the

two being perpendicular to one another. The tension is therefore,

2 22 2 –4 2 –3 –3

e8.0 10 kg 9.80 m/s 2.547 10 N 8.2 10 NT mg F

______________________________________________________________________________

CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS · 898 ELECTRIC FORCES AND ELECTRIC FIELDS CHAPTER...

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Transcript of CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS · 898 ELECTRIC FORCES AND ELECTRIC FIELDS CHAPTER...