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(C) $1$. Initially, $n$ capacitors of capacitance $C$ are connected in parallel. The potential difference across each is $V$. The total charge on each

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potential difference becomes $nV$ and energy remains the same

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(C) $1$. Initially, $n$ capacitors of capacitance $C$ are connected in parallel. The potential difference across each is $V$. The total charge on each capacitor is $q = CV$. The total energy stored in the parallel combination is $U_p = n 	imes (\frac{1}{2}CV^2) = \frac{n}{2}CV^2$.$2$. When they are separated and connected in series, the charge $q = CV$ remains on each capacitor. $3$. The total potential difference across the series combination is $V_{total} = V_1 + V_2 + ... + V_n = nV$.$4$. The total energy stored in the series combination is $U_s = n 	imes (\frac{q^2}{2C}) = n 	imes (\frac{(CV)^2}{2C}) = n 	imes (\frac{1}{2}CV^2) = \frac{n}{2}CV^2$.$5$. Thus, the potential difference becomes $nV$ and the total energy remains the same.

$List-I$	$List-II$
$(P)$ The capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected, is	$(1)$ $3 C_0$
$(Q)$ The capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$, is	$(2)$ $C_0 / 2$
$(R)$ The capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$, is	$(3)$ $C_0 / 3$
$(S)$ The capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$, is	$(4)$ $2 C_0 / 3$
	$(5)$ $2 C_0$

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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