Question

The plate separation in a parallel plate condenser is d and plate area is A. If it is charged to V volt and battery is disconnected then the work done in increasing the plate separation to $2d$ will be

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

Answer 1

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

$Q=CV=\left(\frac{{\epsilon }_{0}A}{d}\right)V$

After changing, Q will remain the same.

Force between the plates $F=\frac{Q^2}{2A{ϵ}_o}=\frac{{ϵ}_{o}A{V}^2}{2d^2}$

Work done $W=F(2d-d)=Fd$

$⟹W=\frac{{ϵ}_{o}A{V}^2}{2d}$