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Question

(C) The initial capacitance of the parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$. The initial energy stored is $U_i = \frac{q^2}{2C}$.When

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

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(C) The initial capacitance of the parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$. The initial energy stored is $U_i = \frac{q^2}{2C}$.When the distance between the plates is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}$.Since the capacitor is isolated,the charge $q$ remains constant. The new energy stored is $U_f = \frac{q^2}{2C'} = \frac{q^2}{2(C/2)} = \frac{q^2}{C}$.The work done by the external force is equal to the change in potential energy: $W = U_f - U_i$.$W = \frac{q^2}{C} - \frac{q^2}{2C} = \frac{q^2}{2C}$.

$A$ parallel plate capacitor carries a charge $q$. The distance between the plates is doubled by the application of a force. The work done by the force is

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