Chapter 22
22.1 Sketch qualitatively the electric field lines both between and outside two concentric conducting
spherical shells when uniform positive charge q1 is on the inner shell and a uniform negative
charge −q 2 is on the outer. Consider the cases q1 > q2 , q1 = q2 , and q1 < q2 .
q1=
q2
q1>
q2
q1<
q2
22.4 What is the magnitude of a point charge that would create and electric field of 1.00N/C at
points 1.00m away.
E=
q
4 π ε0 r 2
q = 4 π ε0 r 2 E
= 1.112 × 10−10C
22.6 Two particles are fixed to an x axis: particle 1 of charge −2.00 × 10−7 C at x = 6.00 cm and
paticle 2 of charge +2.00 × 10−7 C at x = 21.00 cm . Midway between the particles, what is their
net electric field in unit vector notation.
We begin with the picture below.
+q
E+
-q
x
E7.5cm
7.5cm
Both of the electric fields point to the length. They both have the same magnitude as well. We
can calculate the total magnitude
E=
q
q
q
+
=2
2
2
4π ε 0 r
4 π ε 0r
4π ε 0 r 2
=2
−7
+2.00 × 10 C
−2
2
4π ε 0 (7.5 × 10 m )
= 6.4 × 10 5 N / C
The field points to the left in the x direction, so we can write the electric field in vector notation:

E = −6.4 × 105 N / C î
22.10 In Fig. 22-31, what is the magnitude of the electric field at point P due to the four point
charges shown?
The fields due to the two +5q charges cancel exactly. We only need consider the fields due to the
+3q and -12q charges.
-12q
+3q
+5q
E12
P
E3
+5q
+3q
−12q
+
2
4 π ε0 d
4 π ε 0 (2d)2
= 0!
E=
22.12. Calculate the direction and magnitude of the electric field at the point p.
The fields due to the +q cancel exactly, We only need compute the field due to the +2q.
+q
E2
P
+2q
+q
E=
+2q
at 45°
4 π ε0 (a2 / 2)
22.13 Figure 22-34 shows two charged particles on an x axis: −q = −3.2 × 10 −19 C at x = −3.0m
and q = +3.2 × 10 −19 C at x = +3.0m . What are the (a) magnitude and (b) direction (relative to the
positive direction of the x axis) of the net electric field produced at point P at y = +4.0m
x
r-
-q
(-3.0, 0.0)
(0.0, 4.0)
r+
+q
(+3.0, 0.0)
We can do this problem most easily by using the full vector form for the electric field due to a point
charge.

r+ = (0 − 3.0) iˆ + (4.0 − 0) ˆj

r− = (0 − (−3)) ˆi + (4 − 0) ˆj
r+ = 32 + 4 2 = 5m

ˆ
ˆ
ˆr+ = r+ = −3 i + 4 j
r+
5
r− = (−3) 2 + 4 2 = 5m

ˆ
ˆ
ˆr− = r− = 3i + 4 j
r−
5
 

E = E+ + E −
3.2 × 10 −19C −3 ˆi + 4 ˆj 3.2 × 10 −19C 3 ˆi + 4 ˆj
=
⋅
−
⋅
4 π ε 0 (5m) 2
5
4 π ε 0 (5m) 2
5
−19
3.2 × 10 C −6 iˆ + 0 ˆj
=
⋅
4 π ε 0 (5m) 2
5
= −4.39 × 10 −10 N / C
22.17 Calculate the electric dipole moment of an electron and a proton 4.3 nm apart
p = q d = 1.6 ×10−19 C ⋅ 4.3 ×10−9 m = 6.88 ×10−28 Cm
23.19. Find the magnitude and direction of the electric field at point P due to the electric dipole.
You may assume that r>>d
(0,d/2)
+q
(r,0)
-q
(0,-d/2)
We begin by writing the exact field.
 

E = E+ + E−
q
q
ˆ
ˆr
=
2 r+ −
4 π ε0 r+
4 π ε0 r− 2 −

d
r+ = (r − 0) iˆ + (0 − ) ˆj
2
d
r+ = r 2 + ( ) 2
2
d

r iˆ − jˆ
r+
2
ˆr+ = =
r+
d
r 2 + ( )2
2

d
r− = (r − 0) ˆi + (0 − (− )) ˆj
2
d
r− = r2 + ( )2
2
d

r iˆ + jˆ
r−
2
ˆr− = =
r−
d
r 2 + ( )2
2
We can substitute in. In the last step, since d/2r is very small we ignore this term. Remember that
the direction of p is in the +j direction in this problem (from - to + charge).
 

E = E+ + E−
q
q
=
rˆ −
rˆ−
2 +
4 π ε0 r+
4 π ε0 r− 2
d
d
r ˆi − ˆj
r ˆi + ˆj
q
q
2
2
=
⋅
−
⋅
d
d
4 π ε0 (r2 + ( )2 ) r2 + ( d )2 4 π ε0 (r 2 + ( ) 2 ) r 2 + ( d ) 2
2
2
2
2
−qd
ˆj
=
d 2 3/ 2
2
4 π ε0 (r + ( ) )
2
−p
ˆj
=
d 2 3/2
3
4 π ε0 r (1+ ( ) )
2r

−p
≈
4 π ε0 r 3
22.22 Figure 23-33 shows two parallel conducting rings with their central axes along a common
line . Ring 1 has uniform charge q1 and radius R; ring 2 has uniform charge q2 and the same
radius R. The rings are separated by a distance 3R. The net electric field at P on the common line
at a distance R from ring 1 is zero. What is the ratio q1 / q2 ?
dq
r
q1
q2
z
R
3R
We begin by computing the field due to the ring on the left. We will assume that the point is as a
position z. This will give us a general expression that we can use for either ring, plugging in the
correct distance as appropriate. We begin by defining a charge per unit length.
q1
2π R
Because of the symmetry, we only need to compute the field along the axis (called z).
λ1 =
dE z = dE cosθ
dE =
Ez =
2π
∫ dE
z
0
dq
4 πε0 r2
Ez =
r = R2 + z 2
dq = λ1 R dϕ
z
cosθ =
R2 + z 2
2π
λ1 R dϕ
z
⋅
2
2
2
+ z ) R + z2
0
∫ 4πε (R
0
2π
λ1R z dϕ
2
+ z 2 ) 3/2
0
0
λ1R z
=
2ε 0 (R 2 + z2 )3/ 2
=
∫ 4πε (R
We can use this expression to compute the field for each loop, add them together and set the sum
equal to zero field.
E1z =
λ1R z
2ε 0 (R 2 + z2 )3/ 2
λ1R 2
2ε 0 (2R 2 ) 3/2
λ 2 R (2R)
E 2z =
2ε 0 (R 2 + (2R) 2 )3/ 2
=
2λ 2 R 2
2ε 0 (5R 2 ) 3/2
0 = E1z + E 2z
=
λ1R 2
2λ 2 R 2
+
2ε 0 (2R 2 ) 3/2 2ε 0 (5R 2 )3/ 2
λ
2 λ2
= 13/ 2 +
(2)
(5) 3/2
2λ
= 3/1 2
(5)
2
= 2⋅ ( ) 3/2
5
2
= 2⋅ ( ) 3/2
5
=
λ1
(2)3/ 2
λ1
λ2
q1
q2
Since the rings have the same circumference, we can see that the ratio of the charge densities equals
the ratio of the charges.
23.24 In Fig 22-41a, two curved plastic rods, one of charge +q and one of charge -q, for a circle of
radius R in the XY plane.. The x axis passes through their connection points and the charge is
distributed uniformly on both rods. What are the magnitude and direction of the electric field E
produces at P, the center of the circle.
Because of the symmetry in the problem, we can see that the net field will point downward. We
also can see that the contribution from the bottom half circle is equal to the contribution from the
top half circle. Because of this, we only need to compute the downward component due to the top
half circle and multiply by 2.
dq
R
θ
dE
θ
We begin by defining a charge per unit length
q
πR
We now find the component of interest and integrate...
λ=
Ey =
π /2
∫
dE y
−π /2
dE y = −dE cos θ
Ey =
π /2
∫
−π /2
E y tot
The last factor of 2 is because there are two half rings.
λ R dθ
⋅ cosθ
4πε0 R 2
λ π /2
∫ cosθ dθ
4πε0 R −π /2
λ
=
⋅2
4 πε0 R
λ
=
2πε0 R
λ
= 2E y =
πε 0 R
=−
dq
dE =
4 πε0 r2
r =R
dq = λ R dθ
−
22.27 In Fig. 22-43, a nonconducting rod of length L=8.15 cm has charge -q = 4.23fC uniformly
distributed along its length. (a) What is the linear charge density of the rod What are the (b)
magnitude and (c) direction of the electric field at point P, a distance a=12cm from the end of the
rod? What is the electric field magnitude produced at a distance a=50 m by (d) the rod and (e) a
particle of charge -q =-4.23 fC that replaced the rod.
a
-L
0
dE
(a) The charge per unit length is
q −4.23 × 10 −15C
λ= =
= −5.19 × 10 −14 C / m
L
0.0815m
First we write dq for a little length of charge
dq = λ dx

We then write the r, r, rˆ for the charge dq.
r = (a − x) 2

r = (a − x) ˆi

r
rˆ = = iˆ
r
We can then compute the x and y components of the Electric Field. You may need

E=∫
dq
rˆ =
4πε 0 r2
0
λ dx
∫ 4πε (a − x)
−L
2
ˆi
0
λ ˆi
dx
∫
4πε0 − L (a − x)2
λ ˆi
1
1
⋅( −
)
4πε0 a a + L
λ ˆi
a+L− a
⋅(
)
4πε0 a(a + L)
λ ˆi
L
⋅(
)
4πε0 a(a + L)
λL
1
⋅(
) ˆi
4πε0 a(a + L)
0
=
=
=
=
=
5.19 × 10−14 C / m ⋅ 0.0815m
1
⋅(
)
4πε0
0.12m (0.12m + 0.0815m)
= 0.00157N / C
=
The field points to the left
At a distance of 50 m, both the rod and the point produce a field E = 1.52 × 10−8 N / C . The rod
looks quite “pointlike” from 50 m.
22-29. A thin nonconducting rod of finite length L has a charge q spread uniformly along it. Find
the electric field at a point above its center.
(0,d)
-L / 2
First we write dq for a little length of charge
L/2
dq = λ dx

We then write the r, r, rˆ for the charge dq.
r = (0 − x)2 + (d − 0)2

r = (0 − x) iˆ + (d − 0) ˆj

ˆ
ˆ
ˆr = r = −x i + d j
r
x 2 + d2
We can then compute the x and y components of the Electric Field. You may need
∫ (a

E=
dq
∫ 4 πε r
2
rˆ =
0
2
dx
x
2 3/ 2 = 2
+x )
a a2 + x 2
λ dx
− x ˆi + d ˆj
2
2 ⋅
+ d ) x2 + d 2
0
∫ 4πε (x
Ex = 0
⎤ L /2
dx
λd
x
Ey = ∫
∫ (x 2 + d 2) 3/2 = 4 πε ⋅ d 2 d 2 + x 2 ⎥⎦
0
− L /2
⎡
⎤
λ
L /2
−L / 2
=
⋅⎢ 2
−
⎥
2
4πε0 d ⎣ d + L / 4
d 2 + L2 / 4 ⎦
λ
L
=
⋅
2
4πε0 d
d + L2 / 4
λ
L
=
⋅
2πε0 d
4d 2 + L2
λ d dx
λd
2
2 3/2 =
4 πε0 (x + d )
4 πε0
Now we can compute with numbers
λ=
7.81 × 10 −12C
0.145m
d = 0.06m
Ey ==
= 5.386 × 10 −11C / m
5.386 × 10 −11C / m
0.145m
⋅
= 12.43N / C
2πε 0 ⋅ 0.06m
4(0.06m)2 + (0.145m)2
22.30 This problem is worked out in detail in section 23-7 of the book. We also worked this
problem in detail in class. Please review the derivation there.
σ = 5.3× 10−6 C / m2
R = 2.5 ×10−2 m
z = 12 ×10−2 m
σ
z
Ez =
⋅ (1− 2
)
2ε 0
z + R2
= 6.29 ×10 3 N / C
22.42 In Millikan’s experiment, an oil drop of radius r = 1.64 µ m and density ρ = 0.851g / cm 3
is suspended in chamber C (Fig. 22-14) when a downward electric field E = 1.92 × 105 N / C is
applied. Find the charge on the drop in terms of e
FE
mg
If the droplet is suspended, we know that the net force is zero. This means that the weight is
balanced by an upward electric force. We write this balance
FE = mg
qE = mg
q=
mg
E
We can write the mass in terms of the volume of the drop.
4
g 4
m = ρ ⋅ π r 3 = 0.851 3 ⋅ π (1.64 × 10 −4 cm) 3
3
cm 3
−11
= 1.57 × 10 g
= 1.57 × 10
−14
kg
mg 1.57 × 10 −14 kg ⋅9.8m / s 2
=
= 8.03 × 10 −19 C
E
1.92 × 10 5 N / C
= 5e
q=
22.50 An electric dipole consists of charges +2e and -2e separated by 0.78 nm. It is in an electric
field of strength 3.4 × 106 N / C . Calculate the magnitude of the torque on the dipole when the
dipole moment is (a) parallel to , (b) perpendicular to, and (c) antiparallel to the electric field
parallel
τ = p E sin0 = p E
perp
τ = p E sin90 = 0
antiparallel
τ = p E sin180 = − p E