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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$. If the distance $d$ is doubled $(d' = 2d)$,the new capacit

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$\frac{Q^2}{2C}$

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$. If the distance $d$ is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{C}{2}$.Since the capacitor is disconnected from the battery,the charge $Q$ remains constant.The initial energy stored is $U_1 = \frac{Q^2}{2C}$.The final energy stored is $U_2 = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/2)} = \frac{Q^2}{C}$.The work done $W$ is equal to the change in potential energy: $W = U_2 - U_1 = \frac{Q^2}{C} - \frac{Q^2}{2C} = \frac{Q^2}{2C}$.

If the distance between the plates of a capacitor with capacitance $C$ and charge $Q$ is doubled,the work done is:

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