$(A)\;\frac{KV^2}{2d^2} \\ (B)\;\frac{KV^2}{2d^2}\in _0 \\ (C)\; \frac{\in _0 (K-1)V^2}{2d^2K} \\ (D)\;\frac{K(K-1) \in _0 V^2 }{2d^2} $

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Resultant force $F= F_0 -F^1$

Where $F_0$= Force acting per unit area on a plate due to other plate.

$F'$= Force acting per unit area from the dielectric.

$ F= \large\frac{\Large\frac{q^2}{2 \in _0 KA}}{A}$

$\quad= \large\frac{\bigg( \Large\frac{\in _0 K AV}{d} \bigg)^2}{2 \in _0 K_A} \times \frac{1}{A}$

=> $ F= \large\frac{\in _0 KV^2}{2 d^2}$

=> $F_0=F \times K$

=>$F'= F_0-\large\frac{F_0}{K}$$=F_0 \bigg ( 1- \frac{1}{K}\bigg)$

=>$F'= KF \bigg(1- \frac{1}{K}\bigg)$

$ \large\frac{K(K-1) \in _0 V^2 }{2d^2} $

Hence D is the correct answer.

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