65. (a) The potential across C 1 is 10 V, so the charge on it is q 1 = C 1 V 1 = (10.0 µ F)(10.0 V) = 100 µ C. (b) Reducing the right portion of the circuit produces an equivalence equal to 6.00 µ F, with 10.0 V across it. Thus, a charge of 60.0 µ C is on it -- and consequently also on the bottom right capacitor. The bottom right capacitor has, as a result, a potential across it equal to V = q C = 60 µ C 10 µ F = 6.00 V
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