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(C) The initial capacitance is $C = \frac{\varepsilon_0 A}{d}$.Since the capacitor is disconnected from the battery,the charge $Q$ on the plates remai

Vedclass · Accepted Answer

$\frac{\varepsilon_0 A V^2}{2 d}$

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(C) The initial capacitance is $C = \frac{\varepsilon_0 A}{d}$.Since the capacitor is disconnected from the battery,the charge $Q$ on the plates remains constant.The initial energy stored is $U_i = \frac{Q^2}{2C}$.When the separation is increased to $2d$,the new capacitance becomes $C' = \frac{\varepsilon_0 A}{2d} = \frac{C}{2}$.The final energy stored is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/2)} = \frac{Q^2}{C}$.The work done $W$ is the change in potential energy: $W = U_f - U_i = \frac{Q^2}{C} - \frac{Q^2}{2C} = \frac{Q^2}{2C}$.Substituting $Q = CV$,we get $W = \frac{(CV)^2}{2C} = \frac{1}{2} C V^2$.Substituting $C = \frac{\varepsilon_0 A}{d}$,we get $W = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) V^2 = \frac{\varepsilon_0 A V^2}{2d}$.

$A$ parallel plate capacitor has a plate separation $d$ and plate area $A$. If it is charged to a potential $V$ and then disconnected from the battery,calculate the work done in increasing the separation between the plates to $2d$.

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