A parallel plate capacitor of plate area A and plate separation $d$ charged by a battery of voltage V. The battery is then disconnected. The work needed to pull the plates to separation $2d$ is 

Energy stored initially is $U_i=\frac{Q^2}{2C}.$ if d is doubled, C becomes$C / 2$.Hence, energy stored when d is doubled is $U_f=\frac{Q^2}{C}$
$\therefore$ Work needed is
$W=U_f-U_i=\frac{Q^2}{2\left({\displaystyle \frac{C}{2}}\right)}-\frac{Q^2}{2C}=\frac{Q^2}{2C}=\frac{1}{2}C{V}^2$
$(\because Q=CV)$
Now$C=\frac{A{\epsilon }_0}{d}$. Hence
$W=\frac{A{\epsilon }_0{V}^2}{2d}$.