Question

$A$ parallel plate capacitor has plates of area $A$ and separation $d$ and is charged to a potential difference $V$. The charging battery is then disconnected, and the plates are pulled apart until their separation becomes $2d$. What is the work required to separate the plates?

​​​​​​​$[$1 Mark$]$

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

Answer 1

The correct option is D $\frac{{ϵ}_{0}A{V}^2}{2d}$

Initially the capacitance $C_1$ of parallel plate capacitor is
$C_1=\frac{{ϵ}_{0}A}{d}$

Now, as the battery is disconnected, the charge on the capacitor will remain constant.

Charge on capacitor, $Q=C_{1}V$

Now, Force between the plates of a capacitor

$F=\frac{Q^2}{2A{ϵ}_0}$

Work required to move apart the plates of capacitor by distance $d$ ($d\to 2d$)

$W=Fd=\frac{Q^2}{2A{ϵ}_0}d$

$=\frac{(C_{1}V{)}^2}{2A{ϵ}_0}d$

$=C_1^2\frac{V^{2}d}{2A{ϵ}_0}$

$=\frac{{ϵ}_0^2{A}^2}{d^2}\frac{V^{2}d}{2A{ϵ}_0}$

$W=\frac{{ϵ}_{0}A{V}^2}{2d}$

$\therefore$ option (d) is correct