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 is $2d$. What is the work required to separate the plates?

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

Answer 1

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

Let $C_1$ and $C_2$ be the capacitance of capacitor of separation $d$ and $2d$ respectively.
$⟹$ $C_1=\frac{A{ϵ}_0}{d}$ and $C_2=\frac{A{ϵ}_0}{2d}$
As there is no transfer of charge in this case thus, charge will remain conserved.
$⟹$ $q_1=q_2=q$
We know that $q$ = $CV$ so,
$q=\frac{A{ϵ}_0}{d}\times V$---------(i)
We know that work done is given by change in potential energy
$⟹$$W.D=\Delta U=U_f-U_i$
$⟹$$W.D=\frac{q^2}{2C_2}-\frac{q^2}{2C_1}$
using equ (i) and putting the values of $C_1$ and $C_2$ we get
$W.D$ = $\frac{{ϵ}_{0}A{V}^2}{2d}$