Question

A parallel plate capacitor is made of two square plates of side $'a'$, separated by $a$ distance $d$ $(d << a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$, as shown in the figure.
Capacitance of this capacitor is :

• A. $\frac{1}{2} \frac{k\in_{0} a^{2}}{d} $
• B. $\frac{k \in_{0} a^{2}}{d} ln\, K $
• C. $\frac{k \in_{0} a^{2}}{d\left(K-1\right)} ln\,K$
• D. $\frac{k\in_{0}a^{2}}{2d \left(K+1\right)} $

Answer 1

Answer: (C)

$\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} $