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.
The 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. $\frac{k{\in }_0{a}^2}{d(K-1)}lnK$
• D. $\frac{k{\in }_0{a}^2}{2d(K+1)}$

Answer 1

The correct option is C $\frac{k{\in }_0{a}^2}{d(K-1)}lnK$

Lets consider a strip of thickness $dx$ at a distance of x from the left end a shown in the figure.

$\frac{y}{x}=\frac{d}{a}⟹\left(\frac{d}{a}\right)x$

$C_1=\frac{{ϵ}_{o}adx}{d-y}$ and $C_2=\frac{k{ϵ}_{o}adx}{y}$

$C_{eq}=\frac{C_1.C_2}{C_1+C_2}=\frac{k{ϵ}_{o}adx}{kd+(1-k)y}$

On integrating it from 0 to a, we will get $\frac{k{\in }_0{a}^2}{d(K-1)}lnK$