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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 $)$
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The correct answer is:
$\frac{\varepsilon_0 \mathrm{AV}_0{ }^2}{\mathrm{~d}}$
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,
$\begin{aligned}
& \mathrm{Q}_{\text {final }}=\mathrm{Q}_{\text {initial, }} \\
& \begin{aligned}
\mathrm{C}_0 \mathrm{~V}_0 & =\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{\text {fial }} \\
\therefore \quad \mathrm{V}_{\text {final }}=3 \mathrm{~V}_0 & \\
\text { Work done } & =\text { Final P.E }- \text { Initial P.E } \\
& =\frac{1}{2} \mathrm{C}_{\text {finu }}\left(\mathrm{V}_{\text {final }}\right)^2-\frac{1}{2} \mathrm{C}_0\left(\mathrm{~V}_0\right)^2 \\
& =\frac{1}{2} \frac{\mathrm{C}_0}{3} 9 \mathrm{~V}_0^2-\frac{1}{2} \mathrm{C}_0 \mathrm{~V}_0^2 \\
& =\mathrm{C}_0 \mathrm{~V}_0^2\left(\frac{3}{2}-\frac{1}{2}\right) \\
& =\mathrm{C}_0 \mathrm{~V}_0^2 \\
& =\frac{\varepsilon_0 \mathrm{~A} \mathrm{~V}_0^2}{\mathrm{~d}} \quad \ldots\left(\because \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)
\end{aligned}
\end{aligned}$
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,
$\begin{aligned}
& \mathrm{Q}_{\text {final }}=\mathrm{Q}_{\text {initial, }} \\
& \begin{aligned}
\mathrm{C}_0 \mathrm{~V}_0 & =\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{\text {fial }} \\
\therefore \quad \mathrm{V}_{\text {final }}=3 \mathrm{~V}_0 & \\
\text { Work done } & =\text { Final P.E }- \text { Initial P.E } \\
& =\frac{1}{2} \mathrm{C}_{\text {finu }}\left(\mathrm{V}_{\text {final }}\right)^2-\frac{1}{2} \mathrm{C}_0\left(\mathrm{~V}_0\right)^2 \\
& =\frac{1}{2} \frac{\mathrm{C}_0}{3} 9 \mathrm{~V}_0^2-\frac{1}{2} \mathrm{C}_0 \mathrm{~V}_0^2 \\
& =\mathrm{C}_0 \mathrm{~V}_0^2\left(\frac{3}{2}-\frac{1}{2}\right) \\
& =\mathrm{C}_0 \mathrm{~V}_0^2 \\
& =\frac{\varepsilon_0 \mathrm{~A} \mathrm{~V}_0^2}{\mathrm{~d}} \quad \ldots\left(\because \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)
\end{aligned}
\end{aligned}$
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