Question 535: Electrostatics & Capacitance

Question

A parallel plate capacitor has plate area '$\mathrm{A}$' and separation between plates is '$d$'. It is charged to a potential difference of $\mathrm{V}_0$ 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)

Accepted Answer

Let the initial capacitance be $\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$
Let the charge on the capacitor be $Q_{	ext {initial }}=C_0 V_0$
Plate separation is increased by 3 times i.e., $\mathrm{d}^{\prime}=3 \mathrm{~d}$
$\mathrm{C}_{	ext {final }}=\frac{\varepsilon_0 \mathrm{~A}}{3 \mathrm{~d}}=\frac{1}{3}\left(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}ight)=\frac{\mathrm{C}_0}{3}$
Let $\mathrm{Q}_{	ext {final }}$ be the final charge on the capacitor and $V_{	ext {final }}$ be the final potential on the capacitor.
$	herefore \quad \mathrm{Q}_{	ext {final }}=\mathrm{C}_{	ext {final }} \mathrm{V}_{	ext {final }}=\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{	ext {final }}
$
As the capacitor is isolated,
$\begin{aligned}
& \mathrm{Q}_{	ext {final }}=\mathrm{Q}_{	ext {initial }} 	ext {, } \
& \mathrm{C}_0 \mathrm{~V}_0=\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{	ext {final }} \
& 	herefore \quad \mathrm{V}_{	ext {final }}=3 \mathrm{~V}_0 \
& 	ext { Work done }=	ext { Final P.E }- 	ext { Initial P.E } \
& =\frac{1}{2} \mathrm{C}_{	ext {final }}\left(\mathrm{V}_{	ext {final }}ight)^2-\frac{1}{2} \mathrm{C}_0\left(\mathrm{~V}_0ight)^2 \
& =\frac{1}{2} \frac{\mathrm{C}_0}{3} 9 \mathrm{~V}_0^2-\frac{1}{2} \mathrm{C}_0 \mathrm{~V}_0^2 \
& =\mathrm{C}_0 \mathrm{~V}_0^2\left(\frac{3}{2}-\frac{1}{2}ight) \
& =\mathrm{C}_0 \mathrm{~V}_0^2 \
& =\frac{\varepsilon_0 \mathrm{AV}_0^2}{\mathrm{~d}} \quad \ldots\left(\because \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}ight) \
\end{aligned}$

Answer

$\frac{3 \varepsilon_0 \mathrm{AV}_0{ }^2}{\mathrm{~d}}$

Answer

$\frac{\varepsilon_0 \mathrm{AV}_0^2}{2 \mathrm{~d}}$

Answer

$\frac{\varepsilon_0 \mathrm{AV}_0{ }^2}{3 \mathrm{~d}}$

Answer

$\frac{\varepsilon_0 \mathrm{AV}_0{ }^2}{\mathrm{~d}}$

Electrostatics & Capacitance

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