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) ${k\in_0a^2\over 2d(K+1)}$
Ans: (3) ${k\in_0a^2\over d(K-1)}\ln K$
Sol:
${y\over x}={d\over a}$
$y={d\over a}x$
$\mathrm dy={d\over a}(\mathrm dx)$
${1\over\mathrm dc}={y\over KE.adx}+{(d-y)\over\in_0adx}$
${1\over\mathrm dc}={1\over\in_0adx}\left({y\over K}+d-y\right)$
$\int \mathrm dc=\int{\in_0adx\over{y\over K}+d-y}$
$c=\in_0a.{a\over d}\int\limits_0^d{dy\over d+y\left({1\over k}-1\right)}$
$={\in_0a^2\over\left({1\over k}-1\right)d}\left[\ln\left(d+y\left({1\over k}-1\right)\right)\right]_0^d$
$={k\in_0a^2\over(1-k)d}\ln\left({d+d\left({1\over k}-1\right)\over d}\right)$
$={k\in_0a^2\over(1-k)d}\ln\left({1\over k}\right)$
$={k\in_0a^2\over d(K-1)}\ln K$