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(D) Initial capacitance is $C = \frac{\epsilon_0 A}{d}$. Initial charge is $Q_i = CV$.When the distance is reduced to half $(d' = d/2)$,the new capaci

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$CV^2$

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(D) Initial capacitance is $C = \frac{\epsilon_0 A}{d}$. Initial charge is $Q_i = CV$.When the distance is reduced to half $(d' = d/2)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{d/2} = 2C$.Since the battery remains connected,the potential across the capacitor remains $V$.The new charge on the capacitor is $Q_f = C'V = 2CV$.The additional charge supplied by the battery is $\Delta Q = Q_f - Q_i = 2CV - CV = CV$.The energy supplied by the battery is $W = \Delta Q 	imes V = (CV) 	imes V = CV^2$.

$A$ capacitor of capacity $C$ is connected to a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once,while the battery remains connected. Then,the energy supplied by the battery to charge the capacitor to potential $V$ again is:

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