The circuit shows two capacitors A and B of c | vedclass

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(D) $1$. Initially,the capacitors are in series. The equivalent capacitance is $C_{eq} = \frac{C \cdot 2C}{C + 2C} = \frac{2C}{3}$.$2$. The charge on

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$c$ alone is correct

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(D) $1$. Initially,the capacitors are in series. The equivalent capacitance is $C_{eq} = \frac{C \cdot 2C}{C + 2C} = \frac{2C}{3}$.$2$. The charge on each capacitor when fully charged is $Q = C_{eq}E = \frac{2CE}{3}$.$3$. When the cell is removed and plates of opposite polarities are connected,the net charge becomes $Q_{net} = Q_B - Q_A = \frac{2CE}{3} - \frac{2CE}{3} = 0$. Since the net charge is zero,the final charge on each capacitor will be zero.$4$. Statement $(a)$ and $(b)$ are incorrect as the final charge is zero.$5$. Initial energy $U_i = \frac{1}{2} C_{eq} E^2 = \frac{1}{2} (\frac{2C}{3}) E^2 = \frac{CE^2}{3}$.$6$. Final energy $U_f = 0$ (since $Q=0$).$7$. Loss of energy $\Delta U = U_i - U_f = \frac{CE^2}{3}$. Thus,statement $(c)$ is correct.

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