Question

A parallel plate capacitor is made of two square plates of side ${}^{\prime }{a}^{\prime }$, 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. $\frac{k{\in }_0{a}^2}{d}lnK$
• C. $\frac{k{\in }_0{a}^2}{d\left(K-1\right)}lnK$
• D. $\frac{k{\in }_0{a}^2}{2d\left(K+1\right)}$

Answer 1

Answer: (C)

$\frac{y}{x}=\frac{d}{a}$
$y=\frac{d}{a}x$
$dy=\frac{d}{a}\left(dx\right)$
$\frac{1}{dc}=\frac{y}{KE.adx}+\frac{\left(d-y\right)}{{\in }_{0}adx}$
$\frac{1}{dc}=\frac{1}{{\in }_{0}adx}\left(\frac{y}{k}+d-y\right)$
$\int dc=\int \frac{{\in }_{0}adx}{\frac{y}{k}+d-y}$
$c={\in }_{0}a.\frac{a}{d}{\int }_0^d\frac{dy}{d+y\left(\frac{1}{k}-1\right)}$
$=\frac{{\in }_0{a}^2}{\left(\frac{1}{k}-1\right)d}{\left[\ell n\left(d+y\left(\frac{1}{k}-1\right)\right)\right]}_0^d$
$=\frac{k{\in }_0{a}^2}{\left(1-k\right)d}\ell n\left(\frac{d+d\left(\frac{1}{k}-1\right)}{d}\right)$
$=\frac{k{\in }_0{a}^2}{\left(1-k\right)d}\ell n\left(\frac{1}{k}\right)=\frac{k{\in }_0{a}^2\ell nk}{\left(k-1\right)d}$