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(C) Let the capacitance of both capacitors be $C_1 = C_2 = C$. Initially,the energy stored in $C_1$ is $E = \frac{1}{2} CV^2$,where $V$ is the initial

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$E/4$

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(C) Let the capacitance of both capacitors be $C_1 = C_2 = C$. Initially,the energy stored in $C_1$ is $E = \frac{1}{2} CV^2$,where $V$ is the initial potential difference across $C_1$. When the uncharged capacitor $C_2$ is connected in parallel to $C_1$,the charge $Q = CV$ redistributes between the two capacitors until they reach a common potential $V'$. The common potential is given by $V' = \frac{Q_{total}}{C_{total}} = \frac{CV}{C+C} = \frac{V}{2}$. The energy stored in capacitor $C_2$ is $E_{C_2} = \frac{1}{2} C_2 (V')^2$. Substituting the values,$E_{C_2} = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} C \frac{V^2}{4} = \frac{1}{4} \left(\frac{1}{2} CV^2\right)$. Since $E = \frac{1}{2} CV^2$,we get $E_{C_2} = \frac{E}{4}$.

Energy $E$ is stored in a parallel plate capacitor $C_1$. An identical uncharged capacitor $C_2$ is connected to it,kept in contact with it for a while,and then disconnected. The energy stored in $C_2$ is

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