30.56. Model: Assume the battery is ideal.
Visualize:
The pictorial representation shows the capacitor plates connected to a battery, and the capacitor plates moved apart
with insulating handles while they are connected to the battery.
Solve: (a) The initial capacitance of the plates is
Ci =
ε0 A
di
=
( 8.85 × 10
−12
)
C2 /N m 2 ( 0.02 m )
1.0 × 10 −3 m
2
= 3.54 × 10 −12 F
Consequently, an initial voltage ∆Vi = 9.0 V charges the plates to
(
)
Q = ±Ci ∆Vi = ± 3.54 × 10 −12 F ( 9.0 V ) = ±31.9 × 10 −12 C = ±31.9 pC
(b) The new capacitance is Cf = ε 0 A 2di = 12 Ci . The potential difference across the plates is determined by the
battery and is unchanged: ∆Vf = ∆Vi = 9.0. Thus, the new charge on the plates is
(
)
Q = ±Ci ∆Vf = ± 12 3.54 × 10 −12 F ( 9.0 V ) = 16.0 × 10 −12 C = 16.0 pC
30.59. Model: Two capacitors in series combine to give less capacitance.
Solve: Since we have a 75 µF capacitor and we want a 50 µF capacitance, we must connect the second capacitor
in series with the 75 µF capacitor. The capacitance of the second capacitor is calculated as follows:
1
1
1
+ =
⇒ C = 150 µF
75 µ F C 50 µ F
30.61. 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
Ceq 12
=
1
1
1
1
1
+
=
+
=
⇒ Ceq 12 = 12 µF
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
30.66. Model: Capacitors in parallel add to a greater capacitance compared to individual capacitances. On the
other hand, capacitors in series add to a smaller capacitance compared to individual capacitances.
Visualize:
Solve: (a) Three capacitors in series:
−1
1
1
1
1
3
=
+
+
= ( µ F ) ⇒ Ceq = 4 µF
Ceq 12 µ F 12 µ F 12 µ F 12
(b) Two capacitors in parallel and the third in series with this parallel combination:
Ceq 12 = 12 µF + 12 µF = 24 µF
⇒
1
1
1
1
1
1
=
+
=
+
=
⇒ Ceq = 8 µF
Ceq Ceq 12 12 µ F 24 µ F 12 µ F 8 µ F
(c) Two capacitors in series and the third in parallel with this series combination:
−1
1
1
1
1
=
+
= ( µ F ) ⇒ Ceq 12 = 6 µF
Ceq 12 12 µ F 12 µ F 6
⇒ Ceq = Ceq 12 + 12 µF = 6 µF + 12 µF = 18 µF
(d) Three capacitors in parallel:
Ceq = 12 µF + 12 µF +12 µF = 36 µF