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ch25_p61 - 8 3 µ F which clearly has the battery voltage...

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61. (a) In the top right portion of the circuit is a pair of 4.00 µ F which we reduce to a single 8.00 µ F capacitor (which is then in series with the bottom capacitor that the problem is asking about). The further reduction with the bottom 4.00 µ F capacitor results in an equivalence of
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Unformatted text preview: 8 3 µ F, which clearly has the battery voltage across it -- and therefore has charge ( 8 3 µ F)(9.00 V) = 24.0 µ C. This is seen to be the same as the charge on the bottom capacitor. (b) The voltage across the bottom capacitor is V = q C = 24.0 C 4.00 F µ = 6.00 V ....
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