A parallel plate capacitor has plate area ' \(\mathrm{A}\) ' and separation between plates is 'd'. It is charged to a potential difference of \(V_0\) volt. The charging battery is then disconnected and plates are pulled apart three times the initial distance. The work done to increase the distance between the plates is \(\left(\varepsilon_0=\right.\) permittivity of free space \()\)

Let the initial capacitance be \(\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Let the charge on the capacitor be \(\mathrm{Q}_{\text {initial }}=\mathrm{C}_0 \mathrm{~V}_0\) Plate separation is increased by 3 times i.e., \(\mathrm{d}^{\prime}=3 \mathrm{~d}\)
\(C_{\text {final }}=\frac{\varepsilon_0 A}{3 d}=\frac{1}{3}\left(\frac{\varepsilon_0 A}{d}\right)=\frac{C_0}{3}\)
Let \(\mathrm{Q}_{\text {final }}\) be the final charge on the capacitor and \(V_{\text {final }}\) be the final potential on the capacitor.
\(\therefore \quad \mathrm{Q}_{\text {final }}=\mathrm{C}_{\text {fiñal }} \mathrm{V}_{\text {final }}=\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{\text {fial }}\)
As the capacitor is isolated,
\(\mathrm{Q}_{\text {final }}=\mathrm{Q}_{\text {initial, }} \)
\( \mathrm{C}_0 \mathrm{~V}_0 =\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{\text {fial }} \)
\( \therefore \mathrm{V}_{\text {final }}=3 \mathrm{~V}_0 \)
\( \text { Work done } =\text { Final P.E }- \text { Initial P.E } \)
\( =\frac{1}{2} \mathrm{C}_{\text {finu }}\left(\mathrm{V}_{\text {final }}\right)^2-\frac{1}{2} \mathrm{C}_0\left(\mathrm{~V}_0\right)^2 \)
\( =\frac{1}{2} \frac{\mathrm{C}_0}{3} 9 \mathrm{~V}_0^2-\frac{1}{2} \mathrm{C}_0 \mathrm{~V}_0^2 \)
\( =\mathrm{C}_0 \mathrm{~V}_0^2\left(\frac{3}{2}-\frac{1}{2}\right) \)
\( =\mathrm{C}_0 \mathrm{~V}_0^2 \)
\( =\frac{\varepsilon_0 \mathrm{~A} \mathrm{~V}_0^2}{\mathrm{~d}} \ldots\left(\because \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)\)