15. (a) First, the equivalent capacitance of the two 4.00 µF capacitors connected in series is given by 4.00 F/2 = 2.00 F. This combination is then connected in parallel with two other 2.00-F capacitors (one on each side), resulting in an equivalent capacitance C= 3(2.00 F) = 6.00 F. This is now seen to be in series with another combination, which consists of the two 3.0-F capacitors connected in parallel (which are themselves equivalent to C'= 2(3.00 F) = 6.00 F). Thus, the equivalent capacitance of the circuit is ()eq600 F 600 F300 F...CCC.CC..µµ′===′++(b) Let V= 20.0 V be the potential difference supplied by the battery. Then q = CeqV= (3.00 F)(20.0 V) = 6.00 ×10–5C. (c) The potential difference across C1is given by
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