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12. (a) The potential difference across C1 is V1 = 10.0 V. Thus, q1 = C1V1 = (10.0 µF)(10.0 V) = 1.00 × 10–4 C. (b) Let C = 10.0 µF. We first consider the three-capacitor combination consisting of C2 and its two closest neighbors, each of capacitance C. The equivalent capacitance of this combination is CC Ceq = C + 2 = 1.50 C. C + C2 Also, the voltage drop across this combination is
V= CV1 CV1 = = 0.40V1 . C + Ceq C + 1.50 C

Since this voltage difference is divided equally between C2 and the one connected in series with it, the voltage difference across C2 satisfies V2 = V/2 = V1/5. Thus ⎛ 10.0V ⎞ −5 q2 = C2V2 = (10.0 µ F ) ⎜ ⎟ = 2.00 ×10 C. ⎝ 5 ⎠

19. (a) After the switches are closed, the potential differences across the capacitors are the same and the two capacitors are in parallel. The potential difference from a to b is given by Vab = Q/Ceq, where Q is the net charge on the combination and Ceq is the equivalent capacitance. The equivalent capacitance is Ceq = C1 + C2 = 4.0 × 10–6 F. The total charge on the combination is the net charge on either pair of connected plates. The charge on capacitor 1 is q1 = C1V = (1.0 × 10−6 F ) (100 V ) = 1.0 × 10−4 C and the charge on capacitor 2 is q2 = C2V = 3.0 × 10−6 F 100 V = 3.0 × 10−4 C , so the net charge on the combination is 3.0 × 10–4 C – 1.0 × 10–4 C = 2.0 × 10–4 C. The potential difference is 2.0 × 10−4 C Vab = = 50 V. 4.0 × 10−6 F (b) The charge on capacitor 1 is now q1 = C1Vab = (1.0 × 10–6 F)(50 V) = 5.0 × 10–5 C. (c) The charge on capacitor 2 is now q2 = C2Vab = (3.0 × 10–6 F)(50 V) = 1.5 × 10–4 C.

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26. The charges on capacitors 2 and 3 are the same, so these capacitors may be replaced by an equivalent capacitance determined from
1 1 1 C2 + C3 = + = . Ceq C2 C3 C2 C3

Thus, Ceq = C2C3/(C2 + C3). The charge on the equivalent capacitor is the same as the charge on either of the two capacitors in the combination and the potential difference across the equivalent capacitor is given by