Sm chapter19

Electric Forces and
Electric Fields
CHAPTER OUTLINE
19.1 Historical Overview
19.2 Properties of Electric
Charges ANSWERS TO QUESTIONS
19.3 Insulators and
Conductors
19.4 Coulomb’s Law Q19.1 A neutral atom is one that has no net charge. This means that it has
19.5 Electric Fields the same number of electrons orbiting the nucleus as it has protons
19.6 Electric Field Lines
19.7 Motion of Charged in the nucleus. A negatively charged atom has one or more excess
Particles in a Uniform electrons.
Electric Field
19.8 Electric Flux
19.9 Gauss’s Law Q19.2 The clothes dryer rubs dissimilar materials together as it tumbles the
19.10 Application of Gauss’s clothes. Electrons are transferred from one kind of molecule to
Law to Symmetric Charge
another. The charges on pieces of cloth, or on nearby objects charged
Distributions
19.11 Conductors in by induction, can produce strong electric fields that promote the
Electrostatic Equilibrium ionization process in the surrounding air that is necessary for a spark
19.12 Context ConnectionThe to occur. Then you hear or see the sparks.
Atmospheric Electric
Field

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.

519

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.
r r
Q19.8 Both figures are drawn correctly. E1 and E 2 are the electric fields separately created by the point
r r r
charges q and −q in Figure 19.11. The net electric field is the vector sum of E1 and E 2 , shown as E .
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.
z z b
When E is an unknown number, then we can say E cos θ dA = E cos θ dA . When E x , y , z is an
unknown function, then there is no such simplification.
g
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 19521
r
Q19.18 The electric fields outside are identical. The electric fields inside are very different. We have E = 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

e j
1.007 9 1.660 × 10 −27 kg − 9.11 × 10 −31 kg = 1.67 × 10 −27 kg .

Its charge, due to loss of one electron, is

e j
0 − 1 −1.60 × 10 −19 C = +1.60 × 10 −19 C .

(b) By similar logic, charge = +1.60 × 10 −19 C

e j
mass = 22.99 1.66 × 10 −27 kg − 9.11 × 10 −31 kg = 3.82 × 10 −26 kg

(c) charge of Cl − = −1.60 × 10 −19 C

e j
mass = 35.453 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 5.89 × 10 −26 kg

(d) e j
charge of Ca ++ = −2 −1.60 × 10 −19 C = +3.20 × 10 −19 C

e j e j
mass = 40.078 1.66 × 10 −27 kg − 2 9.11 × 10 −31 kg = 6.65 × 10 −26 kg

(e) e j
charge of N 3 − = 3 −1.60 × 10 −19 C = −4.80 × 10 −19 C

e
mass = 14.007 1.66 × 10 −27 kg j + 3e9.11 × 10 −31
j
kg = 2.33 × 10 −26 kg

continued on next page


(f) e
charge of N 4+ = 4 1.60 × 10 −19 C = +6.40 × 10 −19 C j
e j e
mass = 14.007 1.66 × 10 −27 kg − 4 9.11 × 10 −31 kg = 2.32 × 10 −26 kg j
(g) We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.

e
charge = 7 1.60 × 10 −19 C = 1.12 × 10 −18 C j
e j e
mass = 14.007 1.66 × 10 −27 kg − 7 9.11 × 10 −31 kg = 2.32 × 10 −26 kg j
(h) charge = −1.60 × 10 −19 C

b g
mass = 2 1.007 9 + 15.999 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 2.99 × 10 −26 kg

F 10.0 grams I FG 6.02 × 10 atoms IJ FG 47 electrons IJ =
P19.2 (a) N= GH 107.87 grams mol JK H 23
mol K H atom K 2.62 × 10 24

Q 1.00 × 10 −3 C
(b) # electrons added = = = 6.25 × 10 15
e 1.60 × 10 −19 C electron

or 2.38 electrons for every 10 9 already present .

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

F 70 000 grams I FG 6.02 × 10 molecules IJ FG 10 protons IJ = 2.3 × 10
N= GH 18 grams mol JK H 23
mol K H molecule K
28
protons .

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

e je
q = 0.01 1.6 × 10 −19 C 2.3 × 10 28 = 3.7 × 10 7 C . j
So F = ke
q1 q 2
= e9 × 10 j
e3.7 × 10 j
9
7 2
N = 4 × 10 25 N ~ 10 26 N .
r2 0.6 2
This force is almost enough to lift a weight equal to that of the Earth:

e
Mg = 6 × 10 24 kg 9.8 m s 2 = 6 × 10 25 N ~ 10 26 N .j

Chapter 19 523
r k q q
P19.4 The force on one proton is F = e 1 2 away from the other proton. Its magnitude is
r2

jFGH 12.6××10 I
−19 2
10 C
e 8.99 × 10 9 N ⋅ m C 2 −15 J
mK
= 57.5 N .

P19.5 F1 = k e
q1 q 2
=
e8.99 × 10 9
je
N ⋅ m 2 C 2 7.00 × 10 −6 C 2.00 × 10 −6 C je j = 0.503 N
r 2
a0.500 mf 2

F2 = k e
q1 q 2
=
e8.99 × 10 9
N ⋅ m2 C je7.00 × 10 C je 4.00 × 10 C j
2 −6
= 1.01 N
−6

r2 a0.500 mf 2

r
Fx = 0.503 cos 60.0°+1.01 cos 60.0° = 0.755 N F1
Fy = 0.503 sin 60.0°−1.01 sin 60.0° = −0.436 N
r r
a f a f
F = 0.755 N $ − 0.436 N $ = 0.872 N at an angle of 330°
i j F2

FIG. P19.5

r k e q A qB $
*P19.6 In the first situation, FA on B ,1 = i . In the second situation, q A and qB are the same.
r12

r r k e q A qB $
FB on A ,2 = − FA on B =
r22
−i e j
F2 k e q A qB r12
=
F1 r22 k e q A qB
F1 r12 FG 13.7 mmIJ 2
F2 =
r22
= 2.62 µN
H 17.7 mmK = 1.57 µN

r
Then FB on A ,2 = 1.57 µN to the left .

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

F=
k e q1 q 2
e
= 8.99 × 10 N ⋅ m 9 2
C j
2 e12.0 × 10 Cje18.0 × 10 Cj =
−9 −9
2.16 × 10 −5 N .
r2 a0.300 mf 2

(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

F=
k e q1 q 2
e
= 8.99 × 10 N ⋅ m 9 2
C j
2 e3.00 × 10 Cje3.00 × 10 Cj =
−9 −9
8.99 × 10 −7 N .
r2 a0.300 mf 2


*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

F=
b g
r ke 3q Q bg e j
$ + ke q Q −$ .
i i
x 2
a f
d−x
2

3 1 x
The net force will be zero if = , or d − x = .
x2 a
d−x f 2
3

This gives an equilibrium position of the third bead of x = 0.634d .

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

j e0.529 × 10 j
−19 2
k e2 1.60 × 10 C
P19.9 (a) e
F = e 2 = 8.99 × 10 9 N ⋅ m 2 C 2 = 8.22 × 10 −8 N
r
e −10
m j 2

mv 2
(b) We have F = from which
r

v=
Fr
=
e
8.22 × 10 −8 N 0.529 × 10 −10 m j= 2.19 × 10 6 m s .
−31
m 9.11 × 10 kg

Section 19.5 Electric Fields
r r
P19.10 For equilibrium, Fe = − Fg
r
or e j
qE = − mg − $ .
j

r mg
Thus, E= $.
j
q

(a)
r mg
E= $=
j
e
9.11 × 10 −31 kg 9.80 m s 2je j
$ = − 5.58 × 10 −11 N C $
j j e j
q e
−1.60 × 10 −19
C j
(b)
r mg
E= $=
j
e
1.67 × 10 −27 kg 9.80 m s 2je
$=
j
j e1.02 × 10 −7
j
NC$
j
q e
1.60 × 10 −19 C j

Chapter 19 525
P19.11 The point is designated in the sketch. The magnitudes of the electric fields, r r
E2 E1
E1 , (due to the −2.50 × 10 −6 C charge) and E 2 (due to the 6.00 × 10 −6 C
charge) are

E1 =
ke q
=
e8.99 × 10 9
je
N ⋅ m 2 C 2 2.50 × 10 −6 C j (1)
FIG. P19.11
2 2
r d

E2 =
keq
=
e8.99 × 10 9
je
N ⋅ m 2 C 2 6.00 × 10 −6 C j (2)
r 2
ad + 1.00 mf 2

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

to get ad + 1.00 mf 2
= 2.40d 2

or d + 1.00 m = ±1.55d

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 m to the left of the − 2.50 µC charge .

k eQ +Q x = 0 q
* P19.12 The first charge creates at the origin field 2
to the right.
a x
Suppose the total field at the origin is to the right. Then q must
be negative: FIG. P19.12
k eQ $ k e q 2k Q
a 2
i+
3a
2
i
a fe j
−$ = e $
a2
i q = −9Q .

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

k eQ $ k e q $ 2 k eQ $
a2
i + 2 −i = 2 −i
9a a
e j e j q = +27Q .

e j e je j e− $jj = −e2.70 × 10
r k e q1 8.99 × 10 9 3.00 × 10 −9
P19.13 (a) E1 = 2 − $ =
r1
j
0.100
2
a f
3
j
NC$
j
r
E2
r
r E1
e j e je je j e
9 −9
r ke q2 8.99 × 10 6.00 × 10 E
E2 = 2 −$ =
r2
i
0.300
2
a − $ = − 5.99 × 10 2 N C $
i
f i j
r r r
e j e
E = E 2 + E 1 = − 5.99 × 10 2 N C $ − 2.70 × 10 3 N C $
i j j FIG. P19.13

r r
(b) e je
F = qE = 5.00 × 10 −9 C −599 $ − 2 700 $ N C
i j j
r
e
F = −3.00 × 10 −6 $ − 13.5 × 10 −6 $ N =
i j j e−3.00 $i − 13.5 $jj µN


P19.14 (a) E=
keq
=
e8.99 × 10 je2.00 × 10 j = 14 400 N C
9 −6 r
E
r
E
r2 a1.12f 2

Ex = 0 and E = 2b14 400g sin 26.6° = 1.29 × 10
y
4
NC
r FIG. P19.14
so E = 1.29 × 10 4 $ N C .
j

r r
(b) e je
F = qE = −3.00 × 10 −6 1.29 × 10 4 $ = −3.86 × 10 −2 $ N
j j j
k b 2 q g $ k b3 q g $
ei cos 45.0°+ $j sin 45.0°j + k b4qg $j
r k q k q k q e e e
P19.15 (a) E = e 2 1 r1 + e 2 2 r2 + e 2 3 r3
$ $ $ = i+
r1 r2 r3 a2 2a 2 a2
r k q k q k q
E = 3.06 e2 $ + 5.06 e2 $ = 5.91 e2 at 58.8°
i j
a a a

r r k q2
(b) F = qE = 5.91 e 2 at 58.8°
a

P19.16 The electric field at any point x is

E=
ke q
−
a f ke q
=
k e q 4ax
.
ax − af cx − a− afh ex − a j
2 2 2 2 2

When x is much, much greater than a, we find E ≈
b g
4a k e q
.
x3

P19.17 E=
k e λl b g
=
e
ke Q l l
=
k eQ
=
8.99 × 10 9 22.0 × 10 −6 je j
a f a f a f a
d l+d d l+d d l+d 0.290 0.140 + 0.290 fa f
r
E = 1.59 × 10 6 N C , directed toward the rod.
FIG. P19.17

P19.18 E= z k e dq
x2
, where dq = λ 0 dx

z
∞
dx 1 FG IJ ∞
keλ 0
The direction is − $ or left for λ 0 > 0
E = keλ 0
x0 x
2
= keλ 0 −
x H K x0
=
x0
i

P19.19 E=
k e xQ
=
e8.99 × 10 je75.0 × 10 jx = 6.74 × 10 x
9 −6 5

ex 2
+ a2 j 32
ex + 0.100 j
2
ex + 0.010 0j
2 32 2 32

r
(a) At x = 0.010 0 m , E = 6.64 × 10 6 $ N C = 6.64$ MN C
i i

r
(b) At x = 0.050 0 m , E = 2.41 × 10 7 $ N C = 24.1$ MN C
i i

Chapter 19 527
r
(c) At x = 0.300 m , E = 6.40 × 10 6 $ N C = 6.40 $ MN C
i i

r
(d) At x = 1.00 m , E = 6.64 × 10 5 $ N C = 0.664$ MN C
i i

k e Qx
*P19.20 E=
ex 2
+ a2 j 32

dE
LM 1 3x 2
OP
For a maximum,
dx
= Qk e
MM ex j 32
−
ex + a j PQP
=0
2 52
N
2
+ a2 2

a
x 2 + a 2 − 3 x 2 = 0 or x = .
2
Substituting into the expression for E gives

k eQa k eQ 2 k eQ Q
E= = = = .
2 e aj
3
2
2 32 3 2
3
a 2 3 3a 2
6 3π ∈0 a 2

P19.21 Due to symmetry z
Ey = dE y = 0 , and Ex = dE sin θ = k e z z dq sin θ
r2
where dq = λds = λrdθ ,

so that, Ex =
keλ π
r 0 z k λ
sin θdθ = e − cos θ
r
a f π
0
=
2k e λ
r
FIG. P19.21
q L
where λ= and r = .
L π

Thus, Ex =
2 k e qπ
=
e je
2 8.99 × 10 9 N ⋅ m 2 C 2 7.50 × 10 −6 C π j .
L 2
a0.140 mf 2

Solving, Ex = 2.16 × 10 7 N C .
r
Since the rod has a negative charge, E = −2.16 × 10 7 $ N C = −21.6 $ MN C .
i i e j

r
P19.22 (a) The electric field at point P due to each element of length dx, is dE
k dq
dE = 2 e 2 and is directed along the line joining the element to
x +y
point P. By symmetry,

z
Ex = dE x = 0 and since dq = λdx ,

z z
E = Ey = dEy = dE cos θ where cos θ =
y
x2 + y2
.
FIG. P19.22

ze
l2
dx 2 k e λ sin θ 0
Therefore, E = 2 k e λy = .
0
2
x +y j
2 32 y

2k e λ
(b) For a bar of infinite length, θ 0 = 90° and Ey = .
y

P19.23 (a) The whole surface area of the cylinder is A = 2π r 2 + 2π rL = 2π r r + L . a f
e j b g
Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.025 0 m + 0.060 0 m = 2.00 × 10 −10 C

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

e j b ga
Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.060 0 m = 1.41 × 10 −10 C f
(c) e
Q = ρV = ρπ r 2 L = 500 × 10 −9 C m 3 π 0.025 0 m j b g b0.060 0 mg =
2
5.89 × 10 −11 C

Section 19.6 Electric Field Lines

q1 −6 1
*P19.24 (a) = = −
q 2 18 3

(b) q1 is negative, q 2 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
r r r
vectors due to each of the two lower point charges: E = E1 + E 2 .
r q
The electric field from a point charge is E = k e 2 r .
$ r r
r E2 E1
As shown in the solution figure at right,
r q
E1 = k e 2 to the right and upward at 60°
a
r q
E 2 = k e 2 to the left and upward at 60°
a FIG. P19.26
r r r q q
a
ei j j e i j
a
j
E = E1 + E 2 = k e 2 cos 60° $ + sin 60° $ + − cos 60° $ + sin 60° $ = k e 2 2 sin 60° $
j e j
q
= 1.73 k e 2 $ j
a

Section 19.7 Motion of Charged Particles in a Uniform Electric Field

P19.27 (a) a=
qE 1.60 × 10 −19 640
=
a f
= 6.13 × 10 10 m s 2
−27
m 1.67 × 10

(b) v f = vi + at e
1.20 × 10 6 = 6.13 × 10 10 t j t = 1.95 × 10 −5 s

1 1
(c) x f − xi =
2
d i
vi + v f t xf =
2
e je j
1.20 × 10 6 1.95 × 10 −5 = 11.7 m

1 1
(d) K=
2 2
e je
mv 2 = 1.67 × 10 −27 kg 1.20 × 10 6 m s j 2
= 1.20 × 10 −15 J


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

Work done = ∆K
1
so, − Fd = − mvi2 (since the final velocity = 0 )
2
which becomes eEd = K

K
and E= .
ed

x 0.050 0
P19.29 (a) t= = = 1.11 × 10 −7 s = 111 ns
v x 4.50 × 10 5

(b) ay =
qE 1.602 × 10
=
e
−19
9.60 × 10 3 je j
= 9.21 × 10 11 m s 2
m 1.67 × 10 −27e j
1 1
y f − yi = v yi t +
2
ayt 2 : yf =
2
e je
9.21 × 10 11 1.11 × 10 −7 j 2
= 5.68 × 10 −3 m = 5.68 mm

(c) v x = 4.50 × 10 5 m s e je j
v yf = v yi + a y t = 9.21 × 10 11 1.11 × 10 −7 = 1.02 × 10 5 m s

Section 19.8 Electric Flux

P19.30 e je j
Φ E = EA cos θ = 2.00 × 10 4 N C 18.0 m 2 cos 10.0° = 355 kN ⋅ m 2 C

P19.31 Φ E = EA cos θ a
A = π r 2 = π 0.200 f 2
= 0.126 m 2

a
5.20 × 10 5 = E 0.126 cos 0° f E = 4.14 × 10 6 N C = 4.14 MN C

Section 19.9 Gauss’s Law

P19.32 (a) E=
k eQ
: 8.90 × 10 2 =
e8.99 × 10 jQ 9

r2 a0.750f 2

r
But Q is negative since E points inward. Q = −5.57 × 10 −8 C = −55.7 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 Q
kQ δ →0
curved surface is e 2 radially outward (normal to the
R
surface). Therefore, the flux is this field strength times the
area of half a sphere:

z
r r
Φ curved = E ⋅ dA = Elocal A hemisphere
FG Q IJ FG 1 4π R IJ = 1 Qa2π f =
2 +Q
Φ curved = k e
H R2 K H 2 K 4π ∈ 0 2 ∈0 FIG. P19.33

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

−Q
Φ curved + Φ flat = 0 or Φ flat = − Φ curved = .
2 ∈0

qin 170 × 10 −6 C
P19.34 ΦE = = = 1.92 × 10 7 N ⋅ m 2 C
∈0 8.85 × 10 −12 C 2 N ⋅ m 2

1.92 × 10 7 N ⋅ m 2 C
(a) bΦ g
E one face =
1
6
ΦE =
6
bΦ gE one face = 3.20 MN ⋅ m 2 C

(b) Φ E = 19.2 MN ⋅ m 2 C

(c) The answer to (a) would change because the flux through each face of the cube would
not be equal with an asymmetric charge distribution. The sides of the cube nearer the
charge would have more flux and the ones further away would have less. The answer
to (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
k eQr
P19.35 (a) E= = 0
a3

(b) E=
k eQr
=
e8.99 × 10 je26.0 × 10 ja0.100f =
9 −6
365 kN C
a3 a0.400f 3

(c) E=
k eQ
=
e8.99 × 10 je26.0 × 10 j =
9 −6
1. 46 MN C
r2 a0.400f 2

(d) E=
k eQ
=
e8.99 × 10 je26.0 × 10 j =
9 −6
649 kN C
r2 a0.600f 2

The direction for each electric field is radially outward .


*P19.36
F σ IJ = qFG Q A IJ
mg = qE = qG
Q 2 ∈0 mg 2 8.85 × 10
= =
e
−12
0.01 9.8 ja fa f
= −2.48 µC m 2
H2∈ K H 2∈ K
0 0 A q −0.7 × 10 −6

P19.37 (a)
2k λ
E= e 4
3.60 × 10 =
e jb
2 8.99 × 10 9 Q 2.40 g
r 0.190

Q = +9.13 × 10 −7 C = +913 nC

r
(b) E= 0

r
P19.38 (a) E= 0

(b) E=
k eQ
=
e8.99 × 10 je32.0 × 10 j = 7.19 MN C
9 −6
r
E = 7.19 MN C radially outward
r2 a0.200f
2

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 2 L and it encloses charge ρπ r 2 L .
Because the charge distribution is long, no electric flux passes
r r
through the r rcircular end caps; E ⋅ dA = EdA cos 90.0° = 0 . The curved
surface has E ⋅ dA = EdA cos 0° , and E must be the same strength
everywhere over the curved surface. FIG. P19.39

r r
z
Gauss’s law, E ⋅ dA =
q
∈0
, becomes E z
Curved
dA =
ρπ r 2 L
∈0
.
Surface

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

ρπ r 2 L r ρr
b g
E 2π r L =
∈0
. Thus, E=
2 ∈0
radially away from the cylinder axis .

4 3 3Q
*P19.40 The charge density is determined by Q = πa ρ ρ=
3 4π a 3

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

qin 4π r 3 ρ 4π r 3 3Q Qr 3
ΦE = = = =
∈0 3 ∈0 3 ∈0 4π a 3 ∈0 a 3

Q
(b) ΦE = . Note that the answers to parts (a) and (b) agree at r = a .
∈0


Chapter 19 533
(c) ΦE

Q
∈0

0 r
0 a

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.
.

k e q1 q 2 a46f e1.60 × 10 Cj
2 −19 2

F= e
= 8.99 × 10 N ⋅ m9 2
C j
2
= 3.50 × 10 3 N = 3.50 kN
r2
e2 × 5.90 × 10 mj −15 2

Section 19.11 Conductors in Electrostatic Equilibrium

z
qin
qin λ
P19.42 b g
EdA = E 2π rl =
∈0
E= l
2π ∈0 r
=
2π ∈0 r
r
(a) r = 3.00 cm E= 0

r 30.0 × 10 −9
(b) r = 10.0 cm E= 5 400 N C , outward
e
2π 8.85 × 10 −12 0.100 ja f=
r 30.0 × 10 −9
(c) r = 100 cm E= = 540 N C , outward
e
2π 8.85 × 10 −12 1.00 ja f
σ conductor
P19.43 The fields are equal. The Equation 19.25 E = for the field outside the aluminum looks
∈0
σ insulator
different from Equation 19.24 E = for the field around glass. But its charge will spread out
2 ∈0
Q
to cover both sides of the aluminum plate, so the density is σ conductor = . The glass carries charge
2A
Q Q
only on area A, with σ insulator = . The two fields are the same in magnitude, and both are
A 2 A ∈0
perpendicular to the plates, vertically upward if Q is positive.


σ
P19.44 (a) E=
∈0
e je j
σ = 8.00 × 10 4 8.85 × 10 −12 = 7.08 × 10 −7 C m 2

σ = 708 nC m 2 , positive on one face and negative on the other.

(b) σ=
Q
A
e
Q = σA = 7.08 × 10 −7 0.500 ja f 2
C

Q = 1.77 × 10 −7 C = 177 nC , positive on one face and negative on the other.

r
P19.45 (a) E= 0

(b)
k eQ
E= 2 =
e
8.99 × 10 9 8.00 × 10 −6 je
= 7.99 × 10 7 N C
j r
r 0.030 0
2
b g
r
(c) E= 0

(d)
k eQ
E= 2 =
e
8.99 × 10 9 4.00 × 10 −6 je
= 7.34 × 10 6 N C
j r
r 0.070 0
2
b g
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
σ
E= or σ =∈0 E .
∈0

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

e je
σ =∈0 Emin = 8.85 × 10 −12 C 2 N ⋅ m 2 2.80 × 10 4 N C = 248 nC m 2 . j
(b) Where the radius of curvature is the smallest,

e je
σ =∈0 Emax = 8.85 × 10 −12 C 2 N ⋅ m 2 5.60 × 10 4 N C = 496 nC m 2 . 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 .
qin
0 = λl + qin so = −λ
l
Outside surface: The total charge on the metal cylinder is 2λl = qin + qout

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

(b) E=
b g = 6k λ =
2 k e 3λ e 3λ
radially outward
r r 2π ∈0 r

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=
1 q FG IJ e
−8
1 4.00 × 10 C j
= 8.00 × 10 −8 C m 2 = 80.0 nC m 2 .
2 A
=
H K
2 0.500 m 2 a f
r FG IJ F
σ $ 8.00 × 10 −8 C m 2 I b9.04 kN Cgk
H K GH JK
(b) E= k= $
k= $
∈0 8.85 × 10 −12 C 2 N ⋅ m 2

r
(c) E= b−9.04 kN Cgk
$


Section 19.12 Context ConnectionThe Atmospheric Electric Field
r σ
P19.50 (a) E= toward a negative charge.
∈0

b ge j
σ = E ∈0 = 120 N C 8.85 × 10 −12 C 2 N ⋅ m 2 = 1.06 × 10 −9 C m 2

(b) e j e
Q = σ A = σ 4π R 2 = −1.06 × 10 −9 C m 2 4π 6.37 × 10 6 m jLNM e j OQP =
2
−5.42 × 10 5 C

−5.42 × 10 5 C
FG 1 electron IJ = 3.38 × 10 24 excess electrons
H −1.6 × 10 C K
−19

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

z r r
E ⋅ dA =
q
∈0
: b+120 N CgA + b−100 N CgA = ρAa100 mf
∈ 0

ρ=
b20 N Cge8.85 × 10 −12
C 2 N ⋅ m2 j= 1.77 × 10 −12 C m3
100 m

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

F e−5 × 10 Cje−1.37 × 10
5 5
j IJ =
jGG
k e q1 q 2 C
F= e
= 8.99 × 10 9 N ⋅ m 2 C 2
JK 4.18 × 10 3 N .
r2
H e3.84 × 10 mj 8 2

(b) The gravitational force is

F=
Gm1 m 2
=
e6.67 × 10 −11
je je
N ⋅ m 2 kg 2 5.98 × 10 24 kg 7.36 × 10 22 kg j
r 2
e3.84 × 10 mj
8 2

F = 1.99 × 10 20 N toward Earth.

Thus, the electric force is weaker by

1.99 × 10 20 N
= 4.77 × 10 16 times and in the opposite direction .
4.18 × 10 3 N

Additional Problems

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

b
k e q 12 µ C g = k qb45 µ Cg
e FG 15 cm + d IJ 2
45
d 2
a15 cm + df 2 H d K =
12
= 3.75

15 cm + d = 1.94d d = 16.0 cm .

The third charge is at x = −16.0 cm . The equilibrium of the −12 µ C requires

b g = k b45 µ Cg12 µ C
k e q 12 µ C e
q = 51.3 µ C .
a16.0 cmf a15 cmf
2 2

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 . r
T
So T = 2.03 × 10 −2 N . r
qE
From ∑ Fx = 0 , we have qE = T sin 15.0°
r
T sin 15.0° e 2.03 × 10 j
−2 Fg = 0.019 6 N
N sin 15.0°
or q= = = 5.25 × 10 −6 C = 5.25 µC .
E 1.00 × 10 3 N C
FIG. P19.54

k e q1 q 2 15.0
P19.55 F= 2
: tan θ = θ = 14.0°
r 60.0

F =
e8.99 × 10 je10.0 × 10 j
9 −6 2
= 40.0 N
r
F3
1
a0.150f 2 r
F2

F =
e8.99 × 10 je10.0 × 10 j
9 −6 2

= 2.50 N
r r
Fnet F1
3
a0.600f 2

F =
e8.99 × 10 je10.0 × 10 j
9 −6 2
= 2.35 N
FIG. P19.55
2
a0.619f 2

Fx = − F3 − F2 cos 14.0° = −2.50 − 2.35 cos 14.0° = −4.78 N
Fy = − F1 − F2 sin 14.0° = −40.0 − 2.35 sin 14.0° = −40.6 N

Fnet = Fx2 + Fy2 = a4.78f + a40.6f
2 2
= 40.9 N
Fy −40.6
tan φ = =
Fx −4.78
φ = 263°


P19.56 FIG. P19.56(a): d cos 30.0° = 15.0 cm,
15.0 cm
or d=
cos 30.0°

θ = sin −1
FG d IJ
FIG. P19.56(b):
H 50.0 cm K
F 15.0 cm I = 20.3°
θ = sin −1 GH 50.0 cmacos 30.0°f JK FIG. P19.56(a)

Fq
= tan θ
mg

or Fq = mg tan 20.3° (1) r
Fq
r r
FIG. P19.56(c): Fq = 2 F cos 30.0° Fg = mg

Fq = 2
LM k q e
2 OP cos 30.0° (2) FIG. P19.56(b)
MN a0.300 mf 2
PQ
Combining equations (1) and (2),

2
LM k q OP cos 30.0° = mg tan 20.3°
e
2

NM a0.300 mf QP 2
r
F

q2 =
a
mg 0.300 m tan 20.3° f 2
r
Fq
r
F
2 k e cos 30.0°

q2 =
e2.00 × 10 kg je9.80 m s ja0.300 mf tan 20.3°
−3 2 2
FIG. P19.56(c)
2e8.99 × 10 N ⋅ m C j cos 30.0°
9 2 2

q = 4.20 × 10 −14 C 2 = 2.05 × 10 −7 C = 0.205 µC

Q
P19.57 Charge resides on each block, which repel as point charges:
2

F=
e je j = kbL − L g
ke
Q
2
Q
2
2 i
L

Q = 2L
k bL − L g
= 2a0.400 mf
i b100 N mga0.100 mf = 26.7 µC
k e e8.99 × 10 N ⋅ m C j 9 2 2

Chapter 19 539

P19.58 Charge
Q
resides on each block, which repel as point charges: F=
ke e je j = kbL − L g .
Q
2
Q
2
2 i
2 L

Solving for Q, Q = 2L
b
k L − Li g .
ke

Observing the spring extension gives a way of measuring the charge. If L is large compared to Li , the
3
charge is proportional to L to the power. The charge is proportional to the spring constant to the
2
1
power. We demonstrate dimensional correctness by evaluating the units of the right hand side:
2

F N ⋅ m⋅C I
mG
2 12

H m ⋅ N ⋅ m JK 2
= C.

P19.59 (a) From the 2Q charge we have Fe − T2 sin θ 2 = 0 and mg − T2 cos θ 2 = 0 .

Fe T sin θ 2
Combining these we find = 2 = tan θ 2 .
mg T2 cos θ 2

From the Q charge we have Fe = T1 sin θ 1 = 0 and mg − T1 cos θ 1 = 0 .

Fe T sin θ 1
Combining these we find = 1 = tan θ 1 or θ 2 = θ 1 . FIG. P19.59
mg T1 cos θ 1

k e 2QQ 2 k eQ 2
(b) Fe = =
r2 r2
r 2
If we assume θ is small then tan θ ≈ .
l
Substitute expressions for Fe and tan θ into either equation found in part (a) and solve for r.

Fe 2k Q 2 1 r F I 4k eQ 2 l F I 13

mg
= tan θ then e 2
r mg
≈
2l GH JK
and solving for r we find r ≈
mg GH JK .

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.
1
In terms of lengths s, a 3 , and r, shown in Figure P19.60, the charge at the origin exerts an
2
attractive force

k eQq

es + 1
2 a 3 j 2

k eQq
The other two charges exert equal repulsive forces of magnitude . The horizontal components
r2
of the two repulsive forces add, balancing the attractive force,


LM 2 cos θ 1
OP
Fnet = k Qq Me − P=0
NM r
es +
2
1
2 a 3j P
2
Q
1 a
2 1
From Figure P19.60 r= s= a cot θ
sin θ 2

FG 4 IJ k QqFG 2 cos θ sin θ − 2 1
I
JJ = 0 .
The equilibrium condition, in terms of θ, is Fnet =
H a K GH 2 e
e 3 + cot θ j K
2

Thus the equilibrium value of θ satisfies 2 cos θ sin 2 θ e 3 + cot θ j 2
=1.

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
1
s= a cot 81.7° = 0.072 9 a .
2

θ 2 cos θ sin 2 θ e 3 + cot θ j 2

60° 4
70° 2.654
80° 1.226
90° 0
81° 1.091
FIG. P19.60
81.5° 1.024
81.7° 0.997

A second zero-field point is on the negative side of the x-axis, where θ = −9.16° and s = −3.10 a .

P19.61 (a) Zero contribution from the same face due to symmetry,
opposite face contributes

FG k q sin φIJ where
e FG s IJ + FG s IJ
2 2
+ s 2 = 1.5 s = 1.22s
4
Hr K2
r=
H 2K H 2K
s k e qs 4 keq keq
sin φ = E=4 = = 2.18
r r 3
a1.22f 3
s 2
s2 FIG. P19.61

(b) $
The direction is the k direction.

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:
σ
E+ = E− = .
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
r
region is E = 0 .

(b) In the region between the sheets, E+ and E− are both r r r
directed toward the right and the net field is
r σ
E= to the right .
∈0
FIG. P19.62
r
(c) To the right of the negative sheet, E+ and E− are again oppositely directed and E = 0 .

P19.63 The magnitude of the field due to the each sheet given by
Equation 19.24 is
σ
E= directed perpendicular to the sheet. r r r
2 ∈0

(a) In the region to the left of the pair of sheets, both fields are
directed toward the left and the net field is FIG. P19.63
r σ
E= to the left .
∈0

(b) In the region between the sheets, the fields due to the individual sheets are oppositely
directed and the net field is
r
E= 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
r σ
E= to the right .
∈0


F − xi + 0.150 m$
$ I k λe− xi$ + 0.150 m$jjdx r
dE
P19.64
r
dE =
k dq
e G j
J= e

+ a0.150 mf G
H x + a0.150 mf J K x + a0.150 mf
2 2 2 32
x 2 2 2

e− x$i + 0.150 m$jjdx
z z
r r 0. 400 m
E= dE = k e λ
x + a0.150 mf
2 2 3 2
all charge x =0
FIG. P19.64

r
LM $
+i
0. 400 m
a0.150 mf$jx OP 0. 400 m

E = keλ MM
x + a0.150 mf
2
+
a0.150 mf x + a0.150 mf
2 PP 2 2 2

r
N 0 Q 0

E = e8.99 × 10 N ⋅ m C je35.0 × 10 C mj i
9 2 2 $a 2.34 − 6.67 f m + $a6.24 − 0f m
−9
j −1 −1

r
E = e −1.36 i + 1.96 $j × 10 N C = e −1.36 i + 1.96 $ j kN C
$ j $3
j

P19.65 (a) z r r
e q
E ⋅ dA = E 4π r 2 = in
∈0
j
FG 4 π r IJ
3
For r < a , qin = ρ
H3 K
ρr
so E= .
3 ∈0

For a < r < b and c < r , qin = Q . FIG. P19.65

Q
So E= .
4π r 2 ∈0

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.

q1 −Q
Therefore, q1 + Q = 0 and σ1 = 2
= .
4π b 4π b 2

Let q 2 = induced charge on the outside surface of the hollow sphere. Since the hollow
sphere is uncharged, we require

q1 Q
q1 + q 2 = 0 and σ2 = 2
= .
4π c 4π c 2

Chapter 19 543
k e xQ
P19.66 The field on the axis of the ring is calculated in Example 19.5, E = Ex =
ex + a j 2 2 3 2

LM x OP
The force experienced by a charge −q placed along the axis of the ring is F = − k Qq
e MM ex + a j 2 3 2 PP
N Q
2

F k Qq IJ x
F = −G e
and when x << a , this becomes
Ha K 3

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

k eQq
effective spring constant of k=
a3

k 1 k eQq
Since ω = 2π f = , we have f= .
m 2π ma 3

P19.67 The resultant field within the cavity is the superposition of two
r
fields, one E + due to a uniform sphere of positive charge of r
r r1
radius 2a, and the other E − due to a sphere of negative charge of
radius a centered within the cavity. r r
a r

F
4 π r 3ρ I r ρr ρr
r

3 ∈0GH JK
= 4π r 2 E+ so E+ =
3 ∈0
r=
$
3 ∈0

F
4 π r13 ρ I r ρ r1
b g −ρ r
−
3 ∈0 GH JK
= 4π r12 E− so E− =
3 ∈0
− r1 =
$
3 ∈0
r1 .
FIG. P19.67
r r r
Since r = a + r1 ,
r
E− =
b g
r r
−ρ r − a
3 ∈0
r r r r r r r
ρr ρr ρa ρa ρa $
E = E+ + E− = − + = = 0$ +
i j.
3 ∈0 3 ∈0 3 ∈0 3 ∈0 3 ∈0

Thus, Ex = 0

ρa
and Ey = at all points within the cavity.
3 ∈0

ANSWERS TO EVEN PROBLEMS
P19.2 (a) 2.62 × 10 24 electrons; P19.6 1.57 µN to the left
(b) 2.38 electrons for every
10 9 already present P19.8 0.634d , stable if the third bead has positive
charge
P19.4 57.5 N away from the other proton


P19.10 e
(a) − 5.58 × 10 −11 N C $ ;
j j P19.42 (a) 0; (b) 5 400 N/C, outward;
(c) 540 N/C, outward
e
(b) 1.02 × 10 −7 N C $
j j
P19.44 (a) 708 nC m 2 and −708 nC m 2 ;
P19.12 −9Q and +27Q (b) 177 nC and −177 nC
r
P19.14 (a) E = 1.29 × 10 4 $ N C ;
j P19.46 (a) 248 nC m 2 ; (b) 496 nC m 2
−2 $
(b) −3.86 × 10 j N
P19.48 (a) If the volume charge density is
P19.16 see the solution nonzero, the field cannot be uniform in
magnitude.
keλ 0 (b) The field must be uniform in
P19.18 , The direction is − $ or left for λ 0 > 0
i magnitude.
x0

P19.20 see the solution P19.50 (a) negative, 1.06 × 10 −9 C m 2 ;
(b) −5.42 × 10 5 C , 3.38 × 10 24 excess
P19.22 (a) see the solution; (b) see the solution electrons

1 P19.52 (a) away from Earth, 4.18 × 10 3 N ;
P19.24 (a) − ; (b) q1 is negative and q 2 is positive
3 (b) the electric force is weaker by
4.77 × 10 16 times and in the opposite
q direction
P19.26 (a) at the center; (b) 1.73 k e 2 $
j
a
P19.54 5. 25 µC
K
P19.28 in the direction of motion,
ed P19.56 0. 205 µC

P19.30 355 kN ⋅ m 2 C
P19.58 2L
b
k L − Li g
P19.32 (a) −55.7 nC ; (b) negative, spherically ke
symmetric, and concentric with the shell
P19.60 0.072 9 a
2 2
P19.34 (a) 3.20 MN ⋅ m C ; (b) 19.2 MN ⋅ m C ;
(c) see the solution σ
P19.62 (a) 0; (b) to the right; (c) 0
∈0
P19.36 −2.48 µC m 2

P19.38 (a) 0; (b) 7.19 MN C radially outward
P19.64 e−1.36$i + 1.96$jj kN C
Qr 3 Q P19.66 see the solution
P19.40 (a) ; (b) ; (c) see the solution
∈0 a 3 ∈0

Sm chapter19

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