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A parallel plate capacitor is immersed in liquid of dielectric constant K. Find force per unit area acting on plate from the liquid from dielectric in contact with the plate.


$(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} $

1 Answer

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.
answered Jan 13, 2014 by meena.p

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