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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Since the battery is removed,the charge $q$ on the plates r

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$2 U$

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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Since the battery is removed,the charge $q$ on the plates remains constant.The energy stored in a capacitor is given by $U = \frac{q^2}{2C} = \frac{q^2 d}{2 \epsilon_0 A}$.When the separation distance $d$ is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}$.The new energy stored is $U' = \frac{q^2}{2C'} = \frac{q^2}{2(C/2)} = 2 	imes \frac{q^2}{2C} = 2U$.Therefore,the energy stored becomes $2U$.

$A$ capacitor is charged by a battery and the energy stored in it is $U$. The battery is now removed and the separation distance between the plates is doubled. The energy stored now is:

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