Question

A parallel plate capacitor has plate of area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

• A. $\frac{{\epsilon }_{0}A{V}_0^2}{3d}$
• B. $\frac{{\epsilon }_{0}A{V}_0^2}{2d}$
• C. $\frac{{\epsilon }_{0}A{V}_0^2}{4d}$
• D. $\frac{{\epsilon }_{0}A{V}_0^2}{d}$

Answer 1

Answer: (D)

$C=\frac{{\epsilon }_{0}A}{d}$ and charge $Q=C{V}_0$
When spacing is made $(3d)$
$C^{\prime }=\frac{{\epsilon }_{0}A}{3d}=\frac{C}{3}$
As battery is disconnected, charge $Q$ is constant.
$\therefore$ Work done $=$ Final energy $-$ Initial energy
$=\frac{Q^2}{2C^{\prime }}-\frac{Q^2}{2C}$
$=\frac{Q^2\times 3}{2C}-\frac{Q^2}{2C}$
$=\frac{2\times Q^2}{2C}=\frac{{\left(C{V}_0\right)}^2}{C}$
$=C{V}_0^2=\frac{{\epsilon }_{0}A{V}_0^2}{d}$