29.28. Model: The electric potential at the dot is the sum of the potentials due to each charge.
Visualize: Please refer to Figure EX29.28.
Solve: The electric potential at the dot is
V=
1 q1
1 q2
1 q3
+
+
4πε 0 r1 4πε 0 r2 4πε 0 r3
⎡
−5.0 ×10−9 C 5.0 ×10−9 C ⎤
q
= ( 9.0 ×109 Nm 2 /C2 ) ⎢
+
+
⎥ = 3140 V
2
2
0.040 m
0.020 m ⎦⎥
⎣⎢ ( 0.020 m ) + ( 0.040 m )
Solving yields q = 1.00×10–8 C = 10.0 nC.
Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
29.46.
Model:
Energy is conserved.
Visualize:
Solve:
(a) The electric field inside a parallel-plate capacitor is constant with strength
E=
3
ΔV ( 25 ×10 V )
=
= 2.1×106 V/m.
0.012 m
d
(b) Assuming the initial velocity is zero, energy conservation yields
Ui = Kf + U f
1
0 = mevf 2 + ( −e )( Ed )
2
⇒ vf =
Assess:
2 (1.60 ×10−19 C )( 2.1× 106 V/m ) ( 0.012 m )
9.11×10−31 kg
= 9.4 ×107 m/s
This speed is about 31% the speed of light. At that speed, relativity must be taken into account.
30.14. Model: The electric field is the negative of the derivative of the potential function.
Solve: (a) From Equation 30.11, the component of the electric field in the s-direction is Es = − dV ds . For the
given potential,
dV d
V
= (100 x 2 V ) = 200 x
⇒ Ex = −200 x V m
dx dx
m
At x = 0 m, Ex = 0 V/m.
(b) At x = 1 m, Ex = −200 (1) V/m = −200 V/m.
Assess: The potential increases with x, so the electric field must point in the −x-direction.
30.56. Model: Capacitance is a geometric property of two electrodes.
Visualize:
Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q ΔVC . The potential
difference across the capacitor is
ΔVC =
1 Q
1 Q
Q ⎡1 1⎤
−
=
⎢ − ⎥
4πε 0 R1 4πε 0 R2 4πε 0 ⎣ R1 R2 ⎦
−1
⎡1 1⎤
RR
R1R2
⇒ C = 4πε 0 ⎢ − ⎥ = 4πε 0 1 2 = 4πε 0
= 100 × 10−12 F
−
×
R
R
R
R
1.0
10−3 m
⎣ 1
2⎦
2
1
⇒ R1R2 = (100 ×10−12 F )(1.0 ×10−3 m )( 9.0 ×109 N m 2 /C2 ) = 900 × 10−6 m 2
Using R2 = R1 + 1.0 mm,
R1 ( R1 + 1.0 ×10−3 m ) = 900 ×10−6 m ⇒ R12 + (1.0 × 10−3 m ) R1 − 900 × 10−6 m = 0
⇒ R1 =
−1.0 ×10−3 m ±
(1.0 ×10
2
−3
m ) + 3600 × 10−6 m 2
2
= 0.0295 m = 2.95 cm
The outer radius is R2 = R1 + 0.001 m = 0.0305 m = 3.05 cm. So, the diameters are 5.9 cm and 6.1 cm.
30.58.
Visualize:
The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the
figure.
Solve: Because C1 and C2 are in series, their equivalent capacitance Ceq 12 is
1
1
1
1
1
1
= +
=
+
=
⇒ Ceq 12 = 12 μF
Ceq 12 C1 C2 20 μ F 30 μ F 12 μ F
Then, Ceq 12 and C3 are in parallel. So,
Ceq = Ceq 12 + C3 = 12 μF + 25 μF = 37 μF
30.62. Model: Assume the battery is an ideal battery.
Visualize:
The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the
figure.
Solve: Because C2 and C3 are in series,
1
1
1
1
1
10
−1
24
=
+
=
+
= ( μ F ) ⇒ Ceq 23 = 10
μ F = 2.4 μ F
Ceq 23 C2 C3 4 μ F 6 μ F 24
Ceq 23 and C1 are in parallel, so
Ceq = Ceq 23 + C1 = 2.4 μF + 5 μF = 7.4 μF
A potential difference of ΔVC = 9 V across a capacitor of equivalent capacitance 7.4 μF produces a charge
Q = CeqΔVC = (7.4 μF)(9 V) = 66.6 μC
Because Ceq is a parallel combination of C1 and Ceq 23, these capacitors have ΔV1 = ΔVeq 23 = ΔVC = 9 V. Thus the
charges on these two capacitors are
Q1 = (5 μF)(9 V) = 45 μC
Qeq 23 = (2.4 μF)(9 V) = 21.6 μC
Because Qeq 23 is due to a series combination of C2 and C3, Q2 = Q3 = 21.6 μC. This means
ΔV2 =
Q2 21.6 μ C
=
= 5.4 V
C2
4 μF
ΔV3 =
Q3 21.6 μ C
=
= 3.6 V
C3
6 μF
In summary, Q1 = 45 μC, V1 = 9 V; Q2 = 21.6 μC, V2 = 5.4 V; and Q3 = 21.6 μC, V3 = 3.6 V.
30.64. Model: Capacitance is a geometric property.
Visualize: Please refer to Figure P30.64. Shells R1 and R2 are a spherical capacitor C. Shells R2 and R3 are a
spherical capacitor C′. These two capacitors are in series.
Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q ΔVC . The potential
differences across the capacitors C and C′ are
ΔVC =
1 Q
Q
Q ⎡1 1⎤
−
=
⎢ − ⎥
4πε 0 R1 4πε 0 R2 4πε 0 ⎣ R1 R2 ⎦
ΔVC′ =
Q
Q ⎡1
1 Q
1⎤
−
=
⎢ − ⎥
4πε 0 R2 4πε 0 R3 4πε 0 ⎣ R2 R3 ⎦
1
1
⎛1 1 ⎞
⇒ C = ( 4πε 0 ) ⎜ − ⎟
⎝ R1 R2 ⎠
−1
⎛ 1
1 ⎞
C ′ = ( 4πε 0 ) ⎜ − ⎟
R
R
3 ⎠
⎝ 2
−1
Because these two capacitors are in series,
1
1 1
1 ⎛1 1
1
1 ⎞
1 ⎛ R3 − R1 ⎞
= + =
⎜ − + − ⎟=
⎜
⎟
Cnet C C ′ 4πε 0 ⎝ R1 R2 R2 R3 ⎠ 4πε 0 ⎝ R1R3 ⎠
⎛ RR ⎞
⎡ ( 0.010 m )( 0.030 m ) ⎤
1
−12
Cnet = 4πε 0 ⎜ 1 3 ⎟ =
⎥ = 1.67 ×10 F = 1.67 pF
9
2
2 ⎢
⎝ R3 − R1 ⎠ 9.0 ×10 N m /C ⎣ 0.030 m − 0.010 m ⎦
Assess:
Cnet depends on only the inner and outer shells, not on R2.