Sm chapter19

519

Electric Forces andElectric Fields

CHAPTER OUTLINE 19.1 Historical Overview 19.2 Properties of Electric

Charges 19.3 Insulators and

Conductors 19.4 Coulomb’s Law 19.5 Electric Fields 19.6 Electric Field Lines 19.7 Motion of Charged

Particles in a Uniform Electric Field

19.8 Electric Flux 19.9 Gauss’s Law 19.10 Application of Gauss’s

Law to Symmetric Charge Distributions

19.11 Conductors in Electrostatic Equilibrium

19.12 Context ConnectionThe Atmospheric Electric Field

ANSWERS TO QUESTIONS Q19.1 A neutral atom is one that has no net charge. This means that it has

the same number of electrons orbiting the nucleus as it has protons in the nucleus. A negatively charged atom has one or more excess electrons.

Q19.2 The clothes dryer rubs dissimilar materials together as it tumbles the

clothes. Electrons are transferred from one kind of molecule to another. The charges on pieces of cloth, or on nearby objects charged by induction, can produce strong electric fields that promote the ionization process in the surrounding air that is necessary for a spark to occur. Then you hear or see the sparks.

Q19.3 To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and could

discharge with a spark, possibly causing an explosion of any flammable material in the oxygen-enriched atmosphere.

Q19.4 Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) of

two particles, and inversely proportional to the square of the separation distance. An electrical force exhibits the same proportionalities, with charge as the intrinsic property.

Differences: The electrical force can either attract or repel, while the gravitational force as described by Newton’s law can only attract. The electrical force between elementary particles is vastly stronger than the gravitational force.

Q19.5 No. The balloon induces polarization of the molecules in the wall, so that a layer of positive charge

exists near the balloon. This is just like the situation in Figure 19.5a, except that the signs of the charges are reversed. The attraction between these charges and the negative charges on the balloon is stronger than the repulsion between the negative charges on the balloon and the negative charges in the polarized molecules (because they are farther from the balloon), so that there is a net attractive force toward the wall. Ionization processes in the air surrounding the balloon provide ions to which excess electrons in the balloon can transfer, reducing the charge on the balloon and eventually causing the attractive force to be insufficient to support the weight of the balloon.

520 Electric Forces and Electric Fields

Q19.6 An electric field once established by a positive or negative charge extends in all directions from the charge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material at point A in Figure 19.18(a), so there is no charge, nor is there a force. There would be a force if a charge were present at point A, however. A field does exist at point A.

Q19.7 If a charge distribution is small compared to the distance of a field point from it, the charge

distribution can be modeled as a single particle with charge equal to the net charge of the distribution. Further, if a charge distribution is spherically symmetric, it will create a field at exterior points just as if all of its charge were a point charge at its center.

Q19.8 Both figures are drawn correctly.

rE1 and

rE2 are the electric fields separately created by the point

charges q and −q in Figure 19.11. The net electric field is the vector sum of rE1 and

rE2 , shown as

rE .

Figure 19.16 shows only one electric field line at each point away from the charge. At the point location of an object modeled as a point charge, the direction of the field is undefined, and so is its magnitude.

Q19.9 No. Life would be no different if electrons were + charged and protons were – charged. Opposite

charges would still attract, and like charges would repel. The naming of + and – charge is merely a convention.

Q19.10 At a point exactly midway between the two charges. Q19.11 In special orientations the force between two dipoles can be zero or a force of repulsion. In general

each dipole will exert a torque on the other, tending to align its axis with the field created by the first dipole. After this alignment, each dipole exerts a force of attraction on the other.

Q19.12 The negative charge will be drawn to the center of the positively charged ring. Since it will then

have velocity, it will continue on, to an equidistant point on the opposite side of the ring. It will then start moving back and arrive again at point P . This periodic motion will continue. If x is much less than a, the motion can be shown from the solution for the electric field to be simple harmonic.

Q19.13 The surface must enclose a positive total charge. Q19.14 The net flux through any gaussian surface is zero. We can argue it two ways. Any surface contains

zero charge so Gauss’s law says the total flux is zero. The field is uniform, so the field lines entering one side of the closed surface come out the other side and the net flux is zero.

Q19.15 Gauss’s law cannot tell the different values of the electric field at different points on the surface.

When E is an unknown number, then we can say E dA E dAcos cosθ θz z= . When E x y z, ,b g is an

unknown function, then there is no such simplification. Q19.16 Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels

every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surface of the conductor.

Q19.17 If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shell

carries no charge. The person is not harmed by touching this wall. If the person carries a (small) charge q, the electric field inside the sphere is no longer zero. Charge –q is induced on the inner wall of the sphere. The person will get a (small) shock when touching the sphere, as all the charge on his body jumps to the metal.

Chapter 19 521

Q19.18 The electric fields outside are identical. The electric fields inside are very different. We have rE = 0

everywhere inside the conducting sphere while E decreases gradually as you go below the surface of the sphere with uniform volume charge density.

Q19.19 There is zero force. The huge charged sheet creates a uniform field. The field can polarize the

neutral sheet, creating in effect a film of opposite charge on the near face and a film with an equal amount of like charge on the far face of the neutral sheet. Since the field is uniform, the films of charge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces add to zero.

SOLUTIONS TO PROBLEMS Section 19.1 Historical Overview No problems in this section

Section 19.2 Properties of Electric Charges

*P19.1 (a) The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its mass by a negligible amount, to

1 007 9 1 660 10 9 11 10 1 67 1027 31 27. . . .× − × = ×− − − kg kg kge j .

Its charge, due to loss of one electron, is

0 1 1 60 10 1 60 1019 19− − × = + ×− −. . C Ce j .

(b) By similar logic, charge = + × −1 60 10 19. C

mass = × − × = ×− − −22 99 1 66 10 9 11 10 3 82 1027 31 26. . . . kg kg kge j

(c) charge of Cl C− −= − ×1 60 10 19.

mass = × + × = ×− − −35 453 1 66 10 9 11 10 5 89 1027 31 26. . . . kg kg kge j

(d) charge of Ca C C++ − −= − − × = + ×2 1 60 10 3 20 1019 19. .e j

mass = × − × = ×− − −40 078 1 66 10 2 9 11 10 6 65 1027 31 26. . . . kg kg kge j e j

(e) charge of N C C3 19 193 1 60 10 4 80 10− − −= − × = − ×. .e j

mass = × + × = ×− − −14 007 1 66 10 3 9 11 10 2 33 1027 31 26. . . . kg kg kge j e j

continued on next page


(f) charge of N C C4 19 194 1 60 10 6 40 10+ − −= × = + ×. .e j

mass = × − × = ×− − −14 007 1 66 10 4 9 11 10 2 32 1027 31 26. . . . kg kg kge j e j

(g) We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.

charge = × = ×− −7 1 60 10 1 12 1019 18. . C Ce j

mass = × − × = ×− − −14 007 1 66 10 7 9 11 10 2 32 1027 31 26. . . . kg kg kge j e j

(h) charge = − × −1 60 10 19. C

mass = + × + × = ×− − −2 1 007 9 15 999 1 66 10 9 11 10 2 99 1027 31 26. . . . .b g kg kg kg

P19.2 (a) N =FHG

IKJ ×FHG

IKJFHG

IKJ = ×

10 06 02 10 47 2 62 1023 24.. .

grams107.87 grams mol

atomsmol

electrons

atom

(b) #.

. electrons added C

1.60 10 C electron= = ×

×= ×

−

−Qe

1 00 106 25 10

3

1915

or 2 38. electrons for every 10 already present9 .

Section 19.3 Insulators and Conductors No problems in this section

Section 19.4 Coulomb’s Law P19.3 If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains

N =FHG

IKJ ×FHG

IKJFHG

IKJ = ×

70 0006 02 10 10 2 3 1023 28 grams

18 grams mol

moleculesmol

protons

molecule protons. . .

With an excess of 1% electrons over protons, each person has a charge

q = × × = ×−0 01 1 6 10 2 3 10 3 7 1019 28 7. . . . C Ce je j .

So F kq qre= = ×

×= ×1 2

29

7 2

2259 10

3 7 10

0 64 10e j e j.

. N N ~10 N26 .

This force is almost enough to lift a weight equal to that of the Earth:

Mg = × = ×6 10 6 1024 25 kg 9.8 m s N ~10 N2 26e j .

Chapter 19 523

P19.4 The force on one proton is rF =

k q qr

e 1 22 away from the other proton. Its magnitude is

8 99 101 6 102 10

57 5919

15

2

..

.× ⋅ ××FHG

IKJ =

−

− N m C C

m N2e j .

P19.5 F kq qre11 2

2

9 6 6

2

8 99 10 7 00 10 2 00 10

0 5000 503= =

× ⋅ × ×=

− −. . .

..

N m C C C

m N

2 2e je je ja f

F kq qre21 2

2

9 6 6

2

8 99 10 7 00 10 4 00 10

0 5001 01= =

× ⋅ × ×=

− −. . .

..

N m C C C

m N

2 2e je je ja f

FF

x

y

= °+ ° == °− ° = −

= − = °

0 503 60 0 1 01 60 0 0 7550 503 60 0 1 01 60 0 0 436

0 755 0 436 0 872

. cos . . cos . .

. sin . . sin . .

. $ . $ .

N N

N N N at an angle of 330rF i ja f a f

rF1

rF2

FIG. P19.5

*P19.6 In the first situation, rF iA B

e A Bk q q

r on ,1 =12

$ . In the second situation, qA and qB are the same.

r rF F iB A A B

e A B

e A B

e A B

k q q

r

FF

k q q

rr

k q q

FF rr

on ,2 on

N13.7 mm17.7 mm

N

= − = −

=

= = FHG

IKJ =

22

2

1 22

12

21 1

2

22

2

2 62 1 57

$

. .

e j

µ µ

Then

rFB A on ,2 N to the left= 1 57. µ .

P19.7 (a) The force is one of attraction . The distance r in Coulomb’s law is the distance between

centers. The magnitude of the force is

Fk q q

re= = × ⋅

× ×= ×

− −−1 2

29

9 9

258 99 10

12 0 10 18 0 10

0 3002 16 10.

. .

.. N m C

C C

m N2 2e j e je j

a f .

(b) The net charge of − × −6 00 10 9. C will be equally split between the two spheres, or

− × −3 00 10 9. C on each. The force is one of repulsion , and its magnitude is

Fk q q

re= = × ⋅

× ×= ×

− −−1 2

29

9 9

278 99 10

3 00 10 3 00 10

0 3008 99 10.

. .

.. N m C

C C

m N2 2e j e je j

a f .


*P19.8 Let the third bead have charge Q and be located distance x from the left end of the rod. This bead will experience a net force given by

rF i i= +

−−

k q Q

x

k q Q

d xe e3

2 2

b g b ga f e j

$ $ .

The net force will be zero if 3 12 2x d x

=−a f , or d x

x− =3

.

This gives an equilibrium position of the third bead of x d= 0 634. .

The equilibrium is stable if the third bead has positive charge .

P19.9 (a) Fk ere= = × ⋅

×

×= ×

−

−

−2

29

19 2

10 288 99 10

1 60 10

0 529 108 22 10.

.

.. N m C

C

m N2 2e j e j

e j

(b) We have Fmv

r=

2

from which

vFrm

= =× ×

×= ×

− −

−

8 22 10 10

9 11 102 19 10

8 10

316

.

..

N 0.529 m

kg m s

e j.

Section 19.5 Electric Fields P19.10 For equilibrium,

r rF Fe g= −

or q mgrE j= − −$e j .

Thus, rE j=

mgq$ .

(a) rE j j j= =

×

− ×= − ×

−

−−mg

q$

. .

.$ . $

9 11 10 9 80

1 60 105 58 10

31

1911

kg m s

C N C

2e je je j e j

(b) rE j j j= =

×

×= ×

−

−−mg

q$

. .

.$ . $

1 67 10 9 80

1 60 101 02 10

27

197

kg m s

C N C

2e je je j e j

Chapter 19 525

P19.11 The point is designated in the sketch. The magnitudes of the electric fields, E1 , (due to the − × −2 50 10 6. C charge) and E2 (due to the 6 00 10 6. × − C charge) are

Ek qr d

e1 2

9 6

2

8 99 10 2 50 10= =

× ⋅ × −. . N m C C2 2e je j (1)

Ek qr d

e2 2

9 6

2

8 99 10 6 00 10

1 00= =

× ⋅ ×

+

−. .

.

N m C C

m

2 2e je ja f (2)

rE1

rE2

FIG. P19.11

Equate the right sides of (1) and (2)

to get d d+ =1 00 2 402 2. . ma f

or d d+ = ±1 00 1 55. . m

which yields d = 1 82. m

or d = −0 392. m . The negative value for d is unsatisfactory because that locates a point between the charges where

both fields are in the same direction.

Thus, d = −1 82 2 50. . m to the left of the C chargeµ .

*P19.12 The first charge creates at the origin field k Qae2 to the right.

Suppose the total field at the origin is to the right. Then q must be negative:

k Qa

k q

a

k Qa

e e e2 2 23

2$ $ $i i i+ − =a f e j q Q= −9 .

In the alternative, the total field at the origin is to the left:

k Qa

k qa

k Qa

e e e2 2 29

2$ $ $i i i+ − = −e j e j q Q= +27 .

xq+Q x = 0

FIG. P19.12

P19.13 (a) rE j j j1

1

12

9 9

23

8 99 10 3 00 10

0 1002 70 10= − =

× ×− = − ×

−k q

re $

. .

.$ . $e j e je j

a f e j e j N C

r

r r r

E i i i

E E E i j

22

22

9 9

22

2 12 3

8 99 10 6 00 10

0 3005 99 10

5 99 10 2 70 10

= − =× ×

− = − ×

= + = − × − ×

−k q

re $

. .

.$ . $

. $ . $

e j e je ja f e j e j

e j e j

N C

N C N C

rE

rE1

rE2

FIG. P19.13

(b) r rF E i j= = × − −−q 5 00 10 599 2 7009. $ $ C N Ce je j

rF i j i j= − × − × = − −− −3 00 10 13 5 10 3 00 13 56 6. $ . $ . $ . $e j e j N Nµ


P19.14 (a) Ek qr

e= =× ×

=−

2

9 6

2

8 99 10 2 00 10

1 1214 400

. .

.

e je ja f N C

Ex = 0 and Ey = ° = ×2 14 400 26 6 1 29 104b gsin . . N C

so rE j= ×1 29 104. $ N C .

rE

rE

FIG. P19.14

(b) r rF E j j= = − × × = − ×− −q 3 00 10 1 29 10 3 86 106 4 2. . $ . $e je j N

P19.15 (a) rE r r r i i j j= + + = + °+ ° +

k qr

k qr

k qr

k q

a

k q

a

k q

ae e e e e e1

12 1

2

22 2

3

32 3 2 2 2

2 3

245 0 45 0

4$ $ $ $ $ cos . $ sin . $b g b g e j b g

rE i j= + = °3 06 5 06 5 912 2 2. $ . $ .

k qa

k qa

k qa

e e e at 58.8

(b) r rF E= = °q

k qae5 91

2

2. at 58.8

P19.16 The electric field at any point x is

Ek q

x a

k q

x a

k q ax

x a

e e e=−

−− −

=−a f a fc ha f

e j2 2 2 2 2

4.

When x is much, much greater than a, we find Ea k q

xe≈

43

b g.

P19.17 Ek

d d

k Q

d dk Q

d de e e=+

=+

=+

=× ×

+

−λll

l l

l la fb ga f a f

e je ja fa f8 99 10 22 0 10

0 290 0 140 0 290

9 6. .

. . .

rE = ×1 59 106. , N C directed toward the rod.

FIG. P19.17

P19.18 Ek dqxe= z 2 , where dq dx= λ 0

E kdxx

kx

kxe

xe

x

e= = −FHGIKJ =

∞ ∞

zλ λ λ0 2 0

0

00 0

1 The direction is or left for − >$i λ 0 0

P19.19 Ek xQ

x a

x

x

x

x

e=+

=× ×

+= ×

+

−

2 2 3 2

9 6

2 2 3 2

5

2 3 2

8 99 10 75 0 10

0 100

6 74 10

0 010 0e je je je j e j

. .

.

.

.

(a) At x = 0 010 0. m , rE i i= × =6 64 10 6 646. $ . $ N C MN C

(b) At x = 0 050 0. m , rE i i= × =2 41 10 24 17. $ . $ N C MN C


Chapter 19 527

(c) At x = 0 300. m , rE i i= × =6 40 10 6 406. $ . $ N C MN C

(d) At x = 1 00. m , rE i i= × =6 64 10 0 6645. $ . $ N C MN C

*P19.20 Ek Qx

x a

e=+2 2 3 2e j

For a maximum, dEdx

Qkx a

x

x ae=

+−

+

L

NMMM

O

QPPP

=1 30

2 2 3 2

2

2 2 5 2e j e j

x a x2 2 23 0+ − = or xa=2

.

Substituting into the expression for E gives

Ek Qa

a

k Q

a

k Qa

Qa

e e e= = = =∈2 3

23 3 6 33

22 3 2 3

22 2

02e j π

.

P19.21 Due to symmetry E dEy y= =z 0 , and E dE kdq

rx e= =z zsinsinθ θ

2

where dq ds rd= =λ λ θ ,

so that, Ekr

dkr

krx

e e e= = − =zλ θ θ λ θ λππ

sin cos0

0

2a f

where λ =qL

and rL=π

.

FIG. P19.21

Thus, Ek qLxe= =

× ⋅ × −2 2 8 99 10 7 50 10

0 1402

9 6

2π π. .

.

N m C C

m

2 2e je ja f .

Solving, Ex = ×2 16 107. N C .

Since the rod has a negative charge, rE i i= − × = −2 16 10 21 67. $ . $e j N C MN C .


P19.22 (a) The electric field at point P due to each element of length dx, is

dEk dq

x ye=+2 2 and is directed along the line joining the element to

point P. By symmetry,

E dEx x= =z 0 and since dq dx= λ ,

E E dE dEy y= = =z z cosθ where cosθ =+

y

x y2 2.

Therefore, E k ydx

x y

kye

e=+

=z22

2 2 3 20

20λ λ θ

e jl sin

.

drE

FIG. P19.22

(b) For a bar of infinite length, θ 0 90= ° and Ekyye=

2 λ.

P19.23 (a) The whole surface area of the cylinder is A r rL r r L= + = +2 2 22π π π a f .

Q A= = × + = ×− −σ π15 0 10 2 0 025 0 0 025 0 0 060 0 2 00 109 10. . . . . C m m m m C2e j b g

(b) For the curved lateral surface only, A rL= 2π .

Q A= = × = ×− −σ π15 0 10 2 0 025 0 0 060 1 41 109 10. . . . C m m 0 m C2e j b ga f

(c) Q V r L= = = × = ×− −ρ ρπ π2 9 2 11500 10 0 025 0 0 060 0 5 89 10 C m m m C3e j b g b g. . .

Section 19.6 Electric Field Lines

*P19.24 (a) qq

1

2

618

13

= − = −

(b) q q1 2 is negative, is positive

P19.25

FIG. P19.25

Chapter 19 529

P19.26 (a) The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the

three charges individually produce fields that cancel out. In addition to the center of the triangle, the electric field

lines in the second figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically.

(b) You may need to review vector addition in Chapter One. The

electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges:

r r rE E E= +1 2 .

The electric field from a point charge is rE r= k

qre 2

$ .

As shown in the solution figure at right,

rE1 2= k

qae to the right and upward at 60°

rE2 2= k

qae to the left and upward at 60°

rE1

rE2

FIG. P19.26

r r rE E E i j i j j

j

= + = ° + ° + − ° + ° = °

=

1 2 2 2

2

60 60 60 60 2 60

1 73

kq

ak

qa

kq

a

e e

e

cos $ sin $ cos $ sin $ sin $

. $

e j e j e j

Section 19.7 Motion of Charged Particles in a Uniform Electric Field

P19.27 (a) aqEm

= =×

×= ×

−

−1 60 10 640

1 67 106 13 10

19

2710.

..

a f m s2

(b) v v atf i= + 1 20 10 6 13 106 10. .× = ×e jt t = × −1 95 10 5. s

(c) x x v v tf i i f− = +12d i x f = × × =−1

21 20 10 1 95 10 11 76 5. . .e je j m

(d) K mv= = × × = ×− −12

12

1 67 10 1 20 10 1 20 102 27 6 2 15. . . kg m s Je je j


P19.28 The required electric field will be in the direction of motion .

Work done = ∆K

so, − = −Fd mvi12

2 (since the final velocity = 0 )

which becomes eEd K=

and EKed

= .

P19.29 (a) tx

vx= =

×= × =−0 050 0

4 50 101 11 10 1115

7..

. s ns

(b) aqEmy = =

× ×

×= ×

−

−

1 602 10 9 60 10

1 67 109 21 10

19 3

2711

. .

..

e je je j

m s2

y y v t a tf i yi y− = + 12

2: y f = × × = × =− −12

9 21 10 1 11 10 5 68 10 5 6811 7 2 3. . . .e je j m mm

(c) vx = ×4 50 105. m s v v a tyf yi y= + = × × = ×−9 21 10 1 11 10 1 02 1011 7 5. . .e je j m s

Section 19.8 Electric Flux

P19.30 ΦE EA= = × ° = ⋅cos . . cos .θ 2 00 10 18 0 10 0 3554 N C m kN m C2 2e je j

P19.31 ΦE EA= cosθ A r= = =π π2 20 200 0 126. .a f m2

5 20 10 0 126 05. . cos× = °Ea f E = × =4 14 10 4 146. . N C MN C

Section 19.9 Gauss’s Law

P19.32 (a) Ek Qre= 2 : 8 90 10

8 99 10

0 7502

9

2..

.× =

×e ja f

Q

But Q is negative since rE points inward. Q = − × = −−5 57 10 55 78. . C nC

(b) The negative charge has a spherically symmetric charge distribution, concentric with

the spherical shell.

Chapter 19 531

P19.33 (a) With δ very small, all points on the hemisphere are nearly at a distance R from the charge, so the field everywhere on the

curved surface is k QRe

2 radially outward (normal to the

surface). Therefore, the flux is this field strength times the area of half a sphere:

Φ

Φ

curved local hemisphere

curved

= ⋅ =

= FHGIKJFHG

IKJ = ∈

= +∈

z r rE Ad E A

kQR

R QQ

e 22

0 0

12

41

42

2π

ππa f

Qδ → 0

FIG. P19.33

(b) The closed surface encloses zero charge so Gauss’s law gives

Φ Φcurved flat+ = 0 or Φ Φflat curved= − = −∈Q

2 0.

P19.34 ΦEq

=∈

= ×× ⋅

= × ⋅−

−in

2 22 C

8.85 10 C N m N m C

0

6

127170 10

1 92 10.

(a) Φ ΦE Eb gone face

2 N m C= =

× ⋅16

1 92 106

7. ΦEb gone face

2 MN m C= ⋅3 20.

(b) ΦE = ⋅19 2. MN m C2

(c)

The answer to (a) would change because the flux through each face of the cube wouldnot be equal with an asymmetric charge distribution. The sides of the cube nearer thecharge would have more flux and the ones further away would have less. The answerto (b) would remain the same, since the overall flux would remain the same.

Section 19.10 Application of Gauss’s Law to Symmetric Charge Distributions

P19.35 (a) Ek Qr

ae= =3 0

(b) Ek Qr

ae= =

× ×=

−

3

9 6

3

8 99 10 26 0 10 0 100

0 400365

. . .

.

e je ja fa f kN C

(c) Ek Qre= =

× ×=

−

2

9 6

2

8 99 10 26 0 10

0 4001 46

. .

..

e je ja f MN C

(d) Ek Qre= =

× ×=

−

2

9 6

2

8 99 10 26 0 10

0 600649

. .

.

e je ja f kN C

The direction for each electric field is radially outward .


*P19.36 mg qE q qQ A

= =∈FHGIKJ = ∈FHGIKJ

σ2 20 0

QA

mgq

=∈

=×

− ×= −

−

−2 2 8 85 10 0 01 9 8

0 7 102 480

12

6

. . .

..

e ja fa f C m2µ

P19.37 (a) Ekre=

2 λ 3 60 10

2 8 99 10 2 40

0 1904

9

.. .

.× =

×e jb gQ

Q = + × = +−9 13 10 9137. C nC

(b)

rE = 0

P19.38 (a)

rE = 0

(b) Ek Qre= =

× ×=

−

2

9 6

2

8 99 10 32 0 10

0 2007 19

. .

..

e je ja f MN C

rE = 7 19. MN C radially outward

P19.39 If ρ is positive, the field must be radially outward. Choose as the

gaussian surface a cylinder of length L and radius r, contained inside the charged rod. Its volume is π r L2 and it encloses charge ρπ r L2 . Because the charge distribution is long, no electric flux passes through the circular end caps;

r rE A⋅ = ° =d EdA cos .90 0 0 . The curved

surface has r rE A⋅ = °d EdA cos0 , and E must be the same strength

everywhere over the curved surface.

FIG. P19.39

Gauss’s law, r rE A⋅ =

∈z dq

0, becomes E dA

r L

CurvedSurface

z =∈

ρπ 2

0.

Now the lateral surface area of the cylinder is 2π rL :

E r Lr L

22

0π

ρπb g =∈

. Thus, rE =

∈ρ r

2 0 radially away from the cylinder axis .

*P19.40 The charge density is determined by Q a= 43

3π ρ ρπ

= 34 3

Qa

(a) The flux is that created by the enclosed charge within radius r:

ΦEq r r Q

aQr

a=

∈=

∈=

∈=

∈in

0

3

0

3

03

3

03

43

4 33 4

π ρ ππ

(b) ΦEQ=∈0

. Note that the answers to parts (a) and (b) agree at r a= .


Chapter 19 533

(c)

ar

ΦE

Q∈0

00

FIG. P19.40(c)

P19.41 The distance between centers is 2 5 90 10 15× × −. m . Each produces a field as if it were a point charge

at its center, and each feels a force as if all its charge were a point at its center. .

Fk q q

re= = × ⋅

×

× ×= × =

−

−

1 22

92 19 2

15 238 99 10

46 1 60 10

2 5 90 103 50 10 3 50.

.

.. . N m C

C

m N kN2 2e j

a f e je j

Section 19.11 Conductors in Electrostatic Equilibrium

P19.42 EdA E rlqz = =∈

20

πb g in Er r

ql=∈

=∈

in

2 20 0πλ

π

(a) r = 3 00. cm

rE = 0

(b) r = 10 0. cm rE = ×

×=

−

−30 0 10

2 8 85 10 0 1005 400

9

12

.

. .πe ja f N C , outward

(c) r = 100 cm rE = ×

×=

−

−30 0 10

2 8 85 10 1 00540

9

12

.

. .πe ja f N C , outward

P19.43 The fields are equal. The Equation 19.25 E =∈

σ conductor

0 for the field outside the aluminum looks

different from Equation 19.24 E =∈

σ insulator

2 0 for the field around glass. But its charge will spread out

to cover both sides of the aluminum plate, so the density is σ conductor = QA2

. The glass carries charge

only on area A, with σ insulator = QA

. The two fields are Q

A2 0∈ the same in magnitude, and both are

perpendicular to the plates, vertically upward if Q is positive.


P19.44 (a) E =∈σ

0 σ = × × = ×− −8 00 10 8 85 10 7 08 104 12 7. . .e je j C m2

σ = 708 nC m2 , positive on one face and negative on the other.

(b) σ = QA

Q A= = × −σ 7 08 10 0 5007 2. .e ja f C

Q = × =−1 77 10 1777. C nC , positive on one face and negative on the other.

P19.45 (a)

rE = 0

(b) Ek Qre= =

× ×= ×

−

2

9 6

27

8 99 10 8 00 10

0 030 07 99 10

. .

..

e je jb g

N C rE = 79 9. MN C radially outward

(c)

rE = 0

(d) Ek Qre= =

× ×= ×

−

2

9 6

26

8 99 10 4 00 10

0 070 07 34 10

. .

..

e je jb g

N C rE = 7 34. MN C radially outward

P19.46 The electric field on the surface of a conductor varies inversely with the radius of curvature of the

surface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa. The local charge density and the electric field intensity are related by

E =∈σ

0 or σ =∈0 E .

(a) Where the radius of curvature is the greatest,

σ =∈ = × ⋅ × =−0

12 48 85 10 2 80 10 248Emin2 2 2 C N m N C nC m. .e je j .

(b) Where the radius of curvature is the smallest,

σ =∈ = × ⋅ × =−0

12 48 85 10 5 60 10 496Emax . . C N m N C nC m2 2 2e je j .

P19.47 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting

shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length = −λ .

0 = +λl qin so qin

l= −λ

Outside surface: The total charge on the metal cylinder is 2λl = +q qin out

qout = +2λ λl l so the outside charge/length is 3λ .

(b) Ek

rkr r

e e= = =∈

2 3 6 32 0

λ λ λπ

b g radially outward

Chapter 19 535

*P19.48 (a) Consider a gaussian surface in the shape of a rectangular box with two faces perpendicular to the direction of the field. It encloses some charge, so the net flux out of the box is nonzero. The field must be stronger on one side than on the other. It cannot be uniform in magnitude.

(b) Now the volume contains no charge. The net flux out of the

box is zero. The flux entering is equal to the flux exiting. The field magnitude is uniform.

+ + +

+ + +

+

FIG. P19.48(a) P19.49 (a) The charge density on each of the surfaces (upper and lower) of the plate is:

σ = FHGIKJ =

×= × =

−−1

212

4 00 10

0 5008 00 10 80 0

8

28q

A

.

.. .

C

m C m nC m2 2e j

a f .

(b) rE k k k=

∈FHGIKJ =

×× ⋅

FHG

IKJ =

−

−σ

0

8

128 00 10

8 85 109 04$ .

.$ . $ C m

C N m kN C

2

2 2 b g

(c) rE k= −9 04. $ kN Cb g

Section 19.12 Context ConnectionThe Atmospheric Electric Field

P19.50 (a) rE =

∈σ

0 toward a negative charge.

σ = ∈ = × ⋅ = ×− −E 012 9120 1 06 10 N C 8.85 10 C N m C m2 2 2b ge j .

(b) Q A R= = = − × ×LNM

OQP = − ×−σ σ π π4 1 06 10 4 6 37 10 5 42 102 9 6 2 5e j e j e j. . . C m m C2

− ×− ×FHG

IKJ = ×−5 42 10

11 6 10

3 38 10519

24..

. C electron

C excess electrons

P19.51 Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.

r rE A⋅ =

∈z dq

0: + + − =

∈120 100

100

0 N C N C

mb g b g a fA A

Aρ

ρ =× ⋅

= ×−

−20 8 85 10

1001 77 10

1212

N C C N m

m C m

2 23b ge j.

.

The charge is positive , to produce the net outward flux of electric field.


P19.52 (a) The Moon would feel a force away from Earth of magnitude

Fk q q

re= = × ⋅

− × − ×

×

F

HGG

I

KJJ = ×1 2

29 2

5 5

8 238 99 10

5 10 1 37 10

3 84 104 18 10.

.

.. N m C

C C

m N2e j e je j

e j.

(b) The gravitational force is

F

Gm mr

F

= =× ⋅ × ×

×

= ×

−1 22

11 24 22

8 2

20

6 67 10 5 98 10 7 36 10

3 84 10

1 99 10

. . .

.

.

N m kg kg kg

m

N toward Earth.

2 2e je je je j

Thus, the electric force is weaker by

1 99 104 18 10

4 77 1020

316.

..

××

= × N N

times and in the opposite direction .

Additional Problems

*P19.53 The two given charges exert equal-size forces of attraction on each other. If a third charge, positive or negative, were placed between them they could not be in equilibrium. If the third charge were at a point x > 15 cm , it would exert a stronger force on the 45 Cµ than on the −12 Cµ , and could not produce equilibrium for both. Thus the third charge must be at x d= − < 0 . Its equilibrium requires

k q

d

k q

de e12 45

152 2

C C

cm

µ µb g b ga f=

+

15 4512

3 752 cm +F

HGIKJ = =d

d.

15 1 94 cm + =d d. d = 16 0. cm .

d x15 cm

q

x = 0+–

–12 Cµ 45 Cµ

FIG. P19.53

The third charge is at x = −16 0. cm . The equilibrium of the −12 Cµ requires

k q ke e12

16 0

45 122 2

C

cm

C C

15 cm

µ µ µb ga f

b ga f.

= q = 51 3. Cµ .

All six individual forces are now equal in magnitude, so we have equilibrium as required, and this is the only solution.

Chapter 19 537

P19.54 From the free-body diagram shown,

Fy∑ = 0 : T cos . .15 0 1 96 10 2° = × − N .

So T = × −2 03 10 2. N .

From Fx∑ = 0 , we have qE T= °sin .15 0

or qT

E= ° =

× °

×= × =

−−sin . . sin .

.. .

15 0 2 03 10 15 0

1 00 105 25 10 5 25

2

36

N

N C C C

e jµ .

rT

qrE

rFg = 0 019 6. N

FIG. P19.54

P19.55 Fk q q

re= 1 2

2 : tan..

θ = 15 060 0

θ = °14 0.

F

F

F

F F FF F F

x

y

1

9 6 2

2

3

9 6 2

2

2

9 6 2

2

3 2

1 2

8 99 10 10 0 10

0 15040 0

8 99 10 10 0 10

0 6002 50

8 99 10 10 0 10

0 6192 35

14 0 2 50 2 35 14 0 4 7814 0 40 0 2 35 14 0 40 6

=× ×

=

=× ×

=

=× ×

=

= − − ° = − − ° = −= − − ° = − − ° = −

−

−

−

. .

..

. .

..

. .

..

cos . . . cos . .sin . . . sin . .

e je ja f

e je ja f

e je ja f

N

N

N

N N

rF1

rF3rF2

rFnet

FIG. P19.55

F F F

F

F

x y

y

x

net N= + = + =

= = −−

= °

2 2 2 24 78 40 6 40 9

40 64 78

263

. . .

tan.

.

a f a f

φ

φ


P19.56 FIG. P19.56(a): d cos . .30 0 15 0° = cm,

or d =°

15 030 0

.cos .

cm

FIG. P19.56(b): θ = FHG

IKJ

−sin.

1

50 0d cm

θ =°

FHG

IKJ = °−sin

.cos .

.1 15 030 0

20 3 cm

50.0 cma f

F

mgq = tanθ

or F mgq = °tan .20 3 (1)

FIG. P19.56(c): F Fq = °2 30 0cos .

Fk q

qe=

LNMM

OQPP °2

0 30030 0

2

2.cos .

ma f (2)

Combining equations (1) and (2),

20 300

30 0 20 32

2

k qmge

.cos . tan .

ma fLNMM

OQPP ° = °

qmg

k

q

q

e

22

23 2

9

14 7

0 300 20 32 30 0

2 00 10 9 80 0 300 20 3

2 8 99 10 30 0

4 20 10 2 05 10 0 205

=°

°

=× °

× ⋅ °

= × = × =

−

− −

. tan .cos .

. . . tan .

. cos .

. . .

m

kg m s m

N m C

C C C

2

2 2

2

a f

e je ja fe j

µ

FIG. P19.56(a)

rFq

r rF gg m=

FIG. P19.56(b)

rF

rF

rFq

FIG. P19.56(c)

P19.57 Charge Q2

resides on each block, which repel as point charges:

Fk

Lk L L

Q Lk L L

k

eQ Q

i

i

e

= = −

=−

=× ⋅

=

2 22

92 2 0 400

100 0 100

8 99 1026 7

e je j b gb g a f b ga f

e j.

.

.. m

N m m

N m C C

2 2µ

Chapter 19 539

P19.58 Charge Q2

resides on each block, which repel as point charges: Fk

Lk L L

eQ Q

i= = −2 2

2

e je j b g.

Solving for Q, Q Lk L L

ki

e=

−2b g

.

Observing the spring extension gives a way of measuring the charge. If L is large compared to Li , the

charge is proportional to L to the 32

power. The charge is proportional to the spring constant to the

12

power. We demonstrate dimensional correctness by evaluating the units of the right hand side:

mN m Cm N m

C2⋅ ⋅

⋅ ⋅FHG

IKJ =2

1 2

.

P19.59 (a) From the 2Q charge we have F Te − =2 2 0sinθ and mg T− =2 2 0cosθ .

Combining these we find Fmg

TT

e = =2 2

2 22

sincos

tanθθ

θ .

From the Q charge we have F Te = =1 1 0sinθ and mg T− =1 1 0cosθ .

Combining these we find Fmg

TT

e = =1 1

1 11

sincos

tanθθ

θ or θ θ2 1= .

(b) Fk QQ

rk Qre

e e= =2 2

2

2

2

FIG. P19.59

If we assume θ is small then tanθ ≈r 2l

.

Substitute expressions for Fe and tanθ into either equation found in part (a) and solve for r.

Fmg

e = tanθ then 2 1

2

2

2k Qr mg

re FHGIKJ ≈ l

and solving for r we find rk Qmge≈

FHG

IKJ

4 2 1 3l

.

P19.60 At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be

located by determining the angle θ corresponding to equilibrium.

In terms of lengths s, 12

3a , and r, shown in Figure P19.60, the charge at the origin exerts an

attractive force

k Qq

s a

e

+ 12

23e j

The other two charges exert equal repulsive forces of magnitude k Qqre

2 . The horizontal components

of the two repulsive forces add, balancing the attractive force,



F k Qqr s a

enet = −+

L

NMMM

O

QPPP

=2 1

302 1

2

2cosθ

e j

From Figure P19.60 ra

=12

sinθ s a= 1

2cotθ

The equilibrium condition, in terms of θ, is Fa

k Qqenet = FHGIKJ −

+

F

HGG

I

KJJ =

42

1

302

22cos sin

cotθ θ

θe j.

Thus the equilibrium value of θ satisfies 2 3 12 2cos sin cotθ θ θ+ =e j .

One method for solving for θ is to tabulate the left side. To three significant figures a value of θ corresponding to equilibrium is 81.7°.

The distance from the vertical side of the triangle to the equilibrium position is

s a a= ° =12

81 7 0 072 9cot . . .

FIG. P19.60

θ θ θ θ2 3

6070809081

81 581 7

2 2cos sin cot

.

.

+

°°°°°°°

e j4

2.6541.226

01.0911.0240.997

A second zero-field point is on the negative side of the x-axis, where θ = − °9 16. and s a= −3 10. . P19.61 (a) Zero contribution from the same face due to symmetry,

opposite face contributes

4 2

k qr

e sinφFHG

IKJ where r

s ss s s= FHG

IKJ + FHGIKJ + = =

2 21 5 1 22

2 22 . .

sinφ = sr

Ek qsr

k qs

k qs

e e e= = =44

1 222 183 3 2 2..a f

(b) The direction is the direction.$k

FIG. P19.61

Chapter 19 541

P19.62 Consider the field due to a single sheet and let E+ and E− represent the fields due to the positive and negative sheets. The field at any distance from each sheet has a magnitude given by Equation 19.24:

E E+ −= =∈σ

2 0.

(a) To the left of the positive sheet, E+ is directed toward the

left and E− toward the right and the net field over this

region is rE = 0 .

(b) In the region between the sheets, E+ and E− are both

directed toward the right and the net field is

rE =

∈σ

0 to the right .

r r r

FIG. P19.62 (c) To the right of the negative sheet, E+ and E− are again oppositely directed and

rE = 0 .

P19.63 The magnitude of the field due to the each sheet given by

Equation 19.24 is

E =∈σ

2 0 directed perpendicular to the sheet.

(a) In the region to the left of the pair of sheets, both fields are

directed toward the left and the net field is

r r r

FIG. P19.63

rE =

∈σ

0 to the left .

(b) In the region between the sheets, the fields due to the individual sheets are oppositely

directed and the net field is

rE = 0 .

(c) In the region to the right of the pair of sheets, both fields are directed toward the right and

the net field is

rE =

∈σ

0 to the right .


P19.64 dk dq

x

x

x

k x dx

x

e erE

i j i j=

+− +

+

F

HGG

I

KJJ =

− +

+2 2 2 2 2 2 3 20 150

0 150

0 150

0 150

0 150.

$ . $

.

$ . $

. m

m

m

m

ma f a fe ja f

λ

r rE E

i j= =

− +

+z z

=

d kx dx

xe

xall charge

m m

mλ

$ . $

.

. 0 150

0 1502 2 3 20

0 400 e ja f

drE

FIG. P19.64

r

r

r

Ei j

E i j

E i j i j

= +

++

+

L

NMMM

O

QPPP

= × ⋅ × − + −

= − + × = − +

− − −

kx

x

xeλ

$

.

. $

. .

. . $ . . $ .

. $ . $ . $ . $

. .

2 2

0

0 400

2 2 2

0

0 400

9 9 1 1

3

0 150

0 150

0 150 0 150

8 99 10 35 0 10 2 34 6 67 6 24 0

1 36 1 96 10 1 36 1 96

m

m

m m

N m C C m m m

N C kN C

m m

2 2

a fa f

a f a f

e je j a f a fe j e j

P19.65 (a) r rE A⋅ = =

∈z d E rq

4 2

0πe j in

For r a< , q rin = FHGIKJρ π4

33

so Er

=∈

ρ3 0

.

For a r b< < and c r< , q Qin = .

So EQr

=∈4 2

0π.

FIG. P19.65

For b r c≤ ≤ , E = 0 , since E = 0 inside a conductor.

(b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the

conductor, the total charge enclosed by a spherical surface of radius b r c≤ ≤ must be zero.

Therefore, q Q1 0+ = and σπ π1

12 24 4

= = −qb

Qb

.

Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphere is uncharged, we require

q q1 2 0+ = and σπ π2

12 24 4

= =q

cQ

c.

Chapter 19 543

P19.66 The field on the axis of the ring is calculated in Example 19.5, E Ek xQ

x ax

e= =+2 2 3 2e j

The force experienced by a charge −q placed along the axis of the ring is F k Qqx

x ae= −

+

L

NMMM

O

QPPP2 2 3 2e j

and when x a<< , this becomes Fk Qq

axe= −FHGIKJ3

This expression for the force is in the form of Hooke’s law, with an

effective spring constant of kk Qq

ae= 3

Since ω π= =2 fkm

, we have fk Qqma

e= 12 3π

.

P19.67 The resultant field within the cavity is the superposition of two

fields, one rE+ due to a uniform sphere of positive charge of

radius 2a, and the other rE− due to a sphere of negative charge of

radius a centered within the cavity.

43

43

0

2π ρπ

rr E

∈FHGIKJ = + so

r r

E rr

+ =∈

=∈

ρ ρr3 30 0

$

−∈FHGIKJ = −

43

413

012π ρ

πr

r E so r rE r r− =

∈− = −

∈ρ ρr1

01

013 3

$b g .

Since r r rr a r= + 1 ,

rr r

Er a

− =− −

∈ρb g3 0

ra

rr

rr1

FIG. P19.67

r r r r r r r

E E Er r a a

i j= + =∈

−∈

+∈

=∈

= +∈+ −

ρ ρ ρ ρ ρ3 3 3 3

030 0 0 0 0

$ $a.

Thus, Ex = 0

and Ea

y =∈

ρ3 0

at all points within the cavity.

ANSWERS TO EVEN PROBLEMS P19.2 (a) 2 62 1024. × electrons; (b) 2.38 electrons for every

10 already present9 P19.4 57 5. N away from the other proton

P19.6 1 57. Nµ to the left P19.8 0 634. d , stable if the third bead has positive

charge


P19.10 (a) − × −5 58 10 11. $ N Ce jj ;

(b) 1 02 10 7. $× − N Ce jj

P19.12 −9Q and +27Q P19.14 (a)

rE j= ×1 29 104. $ N C ;

(b) − × −3 86 10 2. $j N P19.16 see the solution

P19.18 kxeλ 0

0, The direction is or left for − >$i λ 0 0

P19.20 see the solution P19.22 (a) see the solution; (b) see the solution

P19.24 (a) − 13

; (b) q1 is negative and q2 is positive

P19.26 (a) at the center; (b) 1 73 2. $kq

ae j

P19.28 in the direction of motion, Ked

P19.30 355 kN m C2⋅ P19.32 (a) −55 7. nC ; (b) negative, spherically

symmetric, and concentric with the shell P19.34 (a) 3 20. MN m C2⋅ ; (b) 19.2 MN m C2⋅ ;

(c) see the solution P19.36 −2 48. C m2µ P19.38 (a) 0; (b) 7 19. MN C radially outward

P19.40 (a) Qr

a

3

03∈

; (b) Q∈0

; (c) see the solution

P19.42 (a) 0; (b) 5 400 N/C, outward; (c) 540 N/C, outward P19.44 (a) 708 nC m2 and −708 nC m2 ; (b) 177 nC and −177 nC P19.46 (a) 248 nC m2 ; (b) 496 nC m2 P19.48 (a) If the volume charge density is

nonzero, the field cannot be uniform in magnitude.

(b) The field must be uniform in magnitude.

P19.50 (a) negative, 1 06 10 9. × − C m2 ; (b) − ×5 42 105. C , 3 38 1024. × excess

electrons P19.52 (a) away from Earth, 4 18 103. × N ; (b) the electric force is weaker by

4 77 1016. × times and in the opposite direction

P19.54 5 25. Cµ P19.56 0 205. Cµ

P19.58 2Lk L L

ki

e

−b g

P19.60 0 072 9. a

P19.62 (a) 0; (b) σ∈0

to the right; (c) 0

P19.64 − +1 36 1 96. $ . $i je j kN C

P19.66 see the solution

Education

Sm chapter19