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(B) $1$. In the parallel combination,the total capacitance is $C_p = nC$. The energy stored is $U_p = \frac{1}{2} (nC) V^2$.$2$. When the capacitors a

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Energy remains the same and potential difference is $nV$.

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(B) $1$. In the parallel combination,the total capacitance is $C_p = nC$. The energy stored is $U_p = \frac{1}{2} (nC) V^2$.$2$. When the capacitors are disconnected,each capacitor retains a charge $q = CV$.$3$. When these $n$ capacitors are connected in series,the equivalent capacitance is $C_s = \frac{C}{n}$.$4$. The total charge in the series combination is $q = CV$. The total potential difference $V'$ across the series combination is $V' = \frac{q}{C_s} = \frac{CV}{C/n} = nV$.$5$. The energy stored in the series combination is $U_s = \frac{q^2}{2C_s} = \frac{(CV)^2}{2(C/n)} = \frac{nC^2V^2}{2C} = \frac{1}{2} nCV^2$.$6$. Comparing $U_p$ and $U_s$,we see that $U_p = U_s$. Thus,the energy remains the same and the potential difference becomes $nV$.

$n$ identical capacitors are connected in parallel and charged to a potential $V$. Now,they are disconnected and reconnected in series. What will be the total energy and potential difference of the combination?

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