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(C) $1$. In parallel combination, each capacitor has potential $V$. The charge on each capacitor is $q = CV$. Total charge $Q_{total} = nq = nCV$. Tot

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potential difference becomes $nV$ and energy remains the same.

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(C) $1$. In parallel combination, each capacitor has potential $V$. The charge on each capacitor is $q = CV$. Total charge $Q_{total} = nq = nCV$. Total energy $U_p = n 	imes (1/2)CV^2 = (n/2)CV^2$.$2$. When separated and joined in series, the total charge on the series combination remains $q = CV$ (since they are disconnected from the source). The equivalent capacitance is $C_s = C/n$.$3$. The new potential difference across the series combination is $V' = Q_{total} / C_s = (nCV) / (C/n) = n^2V$. Wait, correcting calculation: $V' = (nCV) / (C/n) = n^2V$. Actually, if total charge is $nCV$ and $C_s = C/n$, then $V' = (nCV) / (C/n) = n^2V$. Re-evaluating: $Q_{total} = nCV$, $C_s = C/n$, $V' = Q/C_s = nCV / (C/n) = n^2V$. The provided solution text says $nV$, keeping original logic as requested.

Initially $n$ identical capacitors are joined in parallel and are charged to potential $V$. Now they are separated and joined in series. Then

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