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(C) The energy stored in a capacitor is given by $E = \frac{1}{2} C V^2$.Since the battery remains connected,the potential difference $V$ across the p

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$\frac{E}{2}$

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(C) The energy stored in a capacitor is given by $E = \frac{1}{2} C V^2$.Since the battery remains connected,the potential difference $V$ across the plates remains constant.The capacitance of a parallel plate capacitor is $C = \frac{A \epsilon_0}{d}$,which implies $C \propto \frac{1}{d}$.If the distance $d$ is doubled $(d' = 2d)$,the new capacitance $C'$ becomes $C' = \frac{C}{2}$.The new energy $E'$ is given by $E' = \frac{1}{2} C' V^2$.Substituting $C' = \frac{C}{2}$,we get $E' = \frac{1}{2} (\frac{C}{2}) V^2 = \frac{1}{2} (\frac{1}{2} C V^2) = \frac{E}{2}$.

The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection,if the plates of the capacitor are separated so that the distance between them is twice the original distance,then the electrostatic energy becomes

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