A parallel plate capacitor of capacitance C i | vedclass

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(C) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = (2C)(2V) = 4CV$.Since the capacitors are connected with opposite polarities,the net ch

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$\frac{3CV^2}{2}$

Vedclass · Answer

(C) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = (2C)(2V) = 4CV$.Since the capacitors are connected with opposite polarities,the net charge on the system is $Q_{net} = |Q_2 - Q_1| = |4CV - CV| = 3CV$.The equivalent capacitance of the parallel combination is $C_{eq} = C + 2C = 3C$.The common potential $V_c$ after connection is given by $V_c = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$.The final energy stored in the configuration is $U = \frac{1}{2} C_{eq} V_c^2$.Substituting the values,$U = \frac{1}{2} (3C) (V)^2 = \frac{3CV^2}{2}$.

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