56. Initially, the total equivalent capacitance is C12 = [(C1)−1+ (C2)−1]−1= 3.0 µF, and the charge on the positive plate of each one is (3.0 µF)(10 V) = 30 µC. Next, the capacitor (call is C1) is squeezed as described in the problem, with the effect that the new value of C1is 12 µF (see Eq. 25-9). The new total equivalent capacitance then becomes C12 = [(C1)−1+ (C2)−1]−1= 4.0 µF, and the new charge on the positive plate of each one is (4.0
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