Q. A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d< Capacitance of this capacitor is :

Solution:

$\frac{y}{x} = \frac{d}{a} $
$ y = \frac{d}{a} x$
$ dy = \frac{d}{a} \left(dx\right)$
$ \frac{1}{dc } = \frac{y}{KE.adx} + \frac{\left(d-y\right)}{\in_{0} adx } $
$ \frac{1}{dc} = \frac{1}{\in_{0} adx } \left(\frac{y}{k} + d -y\right) $
$ \int dc = \int \frac{\in_{0} adx }{\frac{y}{k} + d - y}$
$ c = \in_{0} a. \frac{a}{d} \int^{d}_{0} \frac{dy}{d+y \left(\frac{1}{k} -1\right)} $
$ = \frac{\in_{0} a^{2}}{\left(\frac{1}{k} - 1\right)d} \left[\ell n \left(d+y \left(\frac{1}{k} -1\right)\right)\right]^{d}_{0} $
$= \frac{k\in_{0} a^{2}}{\left(1-k\right)d} \ell n \left(\frac{d+d\left(\frac{1}{k} -1\right)}{d}\right) $
$ = \frac{k\in_{0} a^{2}}{\left(1-k\right)d} \ell n \left(\frac{1}{k}\right) =\frac{k\in_{0} a^{2}\ell nk}{\left(k-1\right)d} $

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