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(D) For the series combination of $n_1$ capacitors,each of capacitance $C_1$:Equivalent capacitance $(C_{eq})_1 = \frac{C_1}{n_1}$.Potential differenc

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$\frac{9 C_1}{n_1 n_2}$

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(D) For the series combination of $n_1$ capacitors,each of capacitance $C_1$:Equivalent capacitance $(C_{eq})_1 = \frac{C_1}{n_1}$.Potential difference $V_1 = 6 \ V$.Energy stored $U_1 = \frac{1}{2} (C_{eq})_1 V_1^2 = \frac{1}{2} \left( \frac{C_1}{n_1} \right) (6)^2 = \frac{18 C_1}{n_1}$.For the parallel combination of $n_2$ capacitors,each of capacitance $C_2$:Equivalent capacitance $(C_{eq})_2 = n_2 C_2$.Potential difference $V_2 = 2 \ V$.Energy stored $U_2 = \frac{1}{2} (C_{eq})_2 V_2^2 = \frac{1}{2} (n_2 C_2) (2)^2 = 2 n_2 C_2$.Given that the total energy is the same,$U_1 = U_2$:$\frac{18 C_1}{n_1} = 2 n_2 C_2$.Solving for $C_2$:$C_2 = \frac{18 C_1}{2 n_1 n_2} = \frac{9 C_1}{n_1 n_2}$.

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