$(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} $

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$C= \large\frac{A \in _0}{d}$

$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.

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