The force of attraction between the plates of parallel plate capacitor having area A and charge Q is given by

$(A)\;\frac{Q^2}{2A \in _0} \\ (B)\;\frac{Q^2}{4A \in _0} \\ (C)\;\frac{Q^2}{A \in _0} \\ (D)\;\frac{Q}{A \in _0}$

$C= \large\frac{A \in _0}{d}$
$U= \large\frac{1}{2}$$CV^2$
$\qquad= \large\frac{A \in _0V^2}{2d}$
$\large\frac{dU}{dd}=\frac{\in_0 AV^2}{2} \frac{d}{dd} \bigg(\frac{1}{d}\bigg)$
$\qquad= -\large\frac{-\in _0 AV^2}{2d^2}$
Force of attraction between the plates $= \large\frac{-dU}{dd}$
$\qquad=\large\frac{\in _0 AV^2}{2d^2}$
$Q= CV$
=> $F=\ \large\frac{Q^2}{2A \in _0}$
Hence A is the correct answer.

edited Aug 11, 2014