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(B) Initially,the capacitor of capacitance $C$ is charged to a potential $V$ by a battery. The energy stored is $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}

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$\frac{U}{4}$

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(B) Initially,the capacitor of capacitance $C$ is charged to a potential $V$ by a battery. The energy stored is $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$,where $Q$ is the initial charge.When the battery is disconnected and an identical capacitor $C$ is connected in parallel,the total charge $Q$ is redistributed between the two capacitors.Since the capacitors are in parallel,they will share the charge equally,so each capacitor now has a charge $Q' = \frac{Q}{2}$.The new potential difference across each capacitor is $V' = \frac{Q'}{C} = \frac{Q}{2C} = \frac{V}{2}$.The energy stored in each capacitor is $U' = \frac{1}{2} C(V')^2 = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} C \frac{V^2}{4} = \frac{1}{4} \left(\frac{1}{2} CV^2\right) = \frac{U}{4}$.

$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another capacitor of the same capacity is connected in parallel to the first capacitor. Then the energy stored in each capacitor is

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