Physics 153
Key for Chapter 25 Homework
6. Givens: Spherical capacitor with radii of 38.0 mm and 40.0 mm. Find capacitance.
(a) From Eqn. 25-17,
C = 4πε0ab/(b - a) =
(40.0 mm)(38.0 mm)
9
2
= 0.760/[8.99x109 C2 /N*m]
2
(8.99 x 10 N*m /C )(40.0 - 38.0)mm
= (84.5 x 10-12)J/C2 = 84.5 pF.
(b) C = ε0A/d = 84.5 pF -> A = Cd/ε0
= (84.5 x 10-12 F)(2.00 x 10-3 m)/(8.854 x 10-12 C2/N*m2) = 0.0191 m2 = 191 cm2.
17. Givens: Figure 25-33, V = 20.0 V, C1 = C6 = 3.00 µF, C3 = C5 = 2.00C2 = 2.00C4 = 4.00 µF.
(a) C3 and C5 are in series and are equivalent to C35 = [1/4 + 1/4]-1µF = 2.00 µF.
C35, C2 and C4 are in parallel and their equivalent capacitance is C2345 = C2 + C35 + C4
= (2.00 + 2.00 + 2.00)µF = 6.00 µF.
This is now in series with the lower parallel combination of C1 and C6 . These add to give C16
= 6.00 µF.
C16 and C2345 are and series and add reciprocally to give Ceq = (1/6 + 1/6)-1 µF = 3.00 µF.
(b) By our definition for capacitance, Ceq = q/V 4 q = CeqV = 60.0 µC.
(c) Capacitors in series have the same charges on them, so C16 = q/V16 and C2345 = q/V2345.
Since C16 = C2345 = 3.00 µF, the potential drop across the bottom and top sets of capacitors is
the same and must add up to 20.0 V. The potential drop across C1 and C6 (and also across C2
and C4, and the series combination of C3 and C5) is 10.0 V.
(d) The charge carried by C1 is q1 = C1V1 = (3.00 mF)(10.0 V) = 30.0 µC.
(e) The potential difference across C2 is given by V - V1 = (20.0 V - 10.0 V) = 10.0 V.
(f) The charge carried by C2 is q2 = C2V2 = (2.00 µF)(10.0 V) = 20.0 µC.
(g) Since the potential difference is divided equally between C3 and the other C5 (since both
have the same capacitance) capacitor connected in series with it, the potential difference
across it is half the total of V2, or V3 = V2/2 = 5.00 V.
(h) q3 = C3V3 = (4.00 µF)(5.00 V) = 20.0 µC.
26. Givens: Figure 25-42, V0 = 12.0 V, C1 = 4.00 µF, C2 = 6.00 µF and C3 = 3.00 µF.
The assumptions are that as the switch is thrown to the right, the total charge on C1 redistributes
itself on the capacitors such that the total charge is the same as before, and that the series
capacitors have the same charge on each. That is, q = q1 + q2, and q3 = q2. The equivalent
capacitance of the two series capacitors is C23 = [1/C2 + 1/C3]-1 µF = 1µF/[1/6 + 1/3] = 2.00 µF.
As charge has been transferred from C1 to C2 and C3, the potential drops to a new lower value.
Since the potential difference across both the left and right branch is the same, we get
q1 /C1 = q2/C23 (i.e., C = q/V 4 V = q/C). Thus, q1 + q2 = C1V0 = (4.00 µF)(12.0 V) = 48.0 µC.
Since C23 has half the capacitance in the equation above, its charge must be half as much.
That is, q2 = q1/2 and 1.5 q1 = 48.0 µC.
(a) q1 = 2(48.0)µC/3 = 32.0 µC.
(b) q2 = q1/2 = 16.0 µC.
(c) q3 = q2 = 16.0 µC, since they are in series.
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45. Givens: k = 2.8 and the maximum field within the dielectric is E = V/d = 18 MV/m, the
desired capacitance is C = 7.0 x 10-2 mF, and the maximum value for V is 4.0 kV.
From this, d = V/E, and C = κε0AE/V and
A = CV/κε0 E = (7.0 x 10-8 F)(4.0 x 103 V)/[(2.8 x 8.854 x 10-12 C2/N*m2)(18 x 106 V/m)]
= 0.627 m2 @0.63 m2 .
If the area were any less the value of d would have to decrease to maintain the same capacitance,
E would increase, and breakdown would occur.
50. Givens: Figure 25-50 and its data, with A = 10.5 cm2 and 2d = 7.12 mm.
The capacitance of the left half of the capacitor is
C1 = κ1ε0(A/2)/2d = κ1ε0 A/4d.
The right half can be thought of as two capacitors in series.
C2 ’ = k2ε0(A/2)/d = k2 ε0A/2d and C3’ = k3ε0(A/2)/d = k3ε0A/2d.
The series combination is equivalent to C2 = [C2’ + C3 ’]-1 = C2’C3 ’/(C2’ + C3 ’)
= (k2ε0A/2d)(k3ε0 A/2d)/[k2ε0A/2d + k3ε0A/2d], which reduces to
C2 ’ = (ε0 A/2d)[k2k3 /(k2 + k3 )].
The parallel combination of C1 and C2’ is simply C = C1 + C2 ’ = (ε0 A/4d)[k1 + 2k2k3/(k2 + k3 )].
Since k1 = 21.0, k2 = 42.0, and k3 = 58.0,
[k1 + 2k2k3 /(k2 + k3 )] = [21.0 + 2(42.0 x 58.0)/100] = 69.72,
and C = 69.72(ε0A/4d) = 45.5 pF.
54. Givens: A = 100 cm2, Q = 8.9 x 10-7 C, E = 1.4 x 106 V/m.
(a) σ = q/A = (8.9 x 10-7 C)/(100 x 10-4 m2) = 8.9 x 10-5 C/m2 .
E = σ/kε0 = 1.4 x 106 V/m -> k = σ/ε0 E = (8.9 x 10-5 C/m2 )/[(8.85 x 10-12 C2/N*m2)(1.4 x 106
V/m)].
k = 7.18 ~= 7.2.
(b) q - q’ = q/k -> q’ = q(1 - 1/k) = 7.66 x 10-7 C ~= 0.77 mC.