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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 :

(1) ${1\over2}{k\in_0a^2\over d}$

(2) ${k\in_0a^2\over d}\ln K$

(3) ${k\in_0a^2\over d(K-1)}\ln K$

(4) ${k\in_0a^2\over 2d(K+1)}$
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Ans: (3) ${k\in_0a^2\over d(K-1)}\ln K$

Sol: 

${y\over x}={d\over a}$

$y={d\over a}x$

$\mathrm dy={d\over a}(\mathrm dx)$

${1\over\mathrm dc}={y\over KE.adx}+{(d-y)\over\in_0adx}$

${1\over\mathrm dc}={1\over\in_0adx}\left({y\over K}+d-y\right)$

$\int \mathrm dc=\int{\in_0adx\over{y\over K}+d-y}$

$c=\in_0a.{a\over d}\int\limits_0^d{dy\over d+y\left({1\over k}-1\right)}$

$={\in_0a^2\over\left({1\over k}-1\right)d}\left[\ln\left(d+y\left({1\over k}-1\right)\right)\right]_0^d$

$={k\in_0a^2\over(1-k)d}\ln\left({d+d\left({1\over k}-1\right)\over d}\right)$

$={k\in_0a^2\over(1-k)d}\ln\left({1\over k}\right)$

$={k\in_0a^2\over d(K-1)}\ln K$

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