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Question

(D) Initial capacitance is $C = \frac{\epsilon_0 A}{d}$.When the distance is halved $(d' = d/2)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}

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$CV^2$

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(D) Initial capacitance is $C = \frac{\epsilon_0 A}{d}$.When the distance is halved $(d' = d/2)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{d/2} = 2C$.The initial charge on the capacitor is $Q_i = CV$.After the distance is halved,the capacitor is connected to the same battery of $V$ volts.The new charge on the capacitor becomes $Q_f = C'V = (2C)V = 2CV$.The additional charge supplied by the battery is $\Delta Q = Q_f - Q_i = 2CV - CV = CV$.The energy supplied by the battery is $W = \Delta Q 	imes V = (CV) 	imes V = CV^2$.

$A$ capacitor of capacitance $C$ is connected to a battery of $V$ volts. Now,the distance between the plates of the capacitor is halved while keeping the charge constant,and it is again charged to $V$ volts. What is the energy supplied by the battery?

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