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. $\frac{K{ϵ}_0{a}^2}{d(K-1)}\ln K$
• B. $\frac{K{ϵ}_0{a}^2}{d}\ln K$
• C. $\frac{K{ϵ}_0{a}^2}{2d(K+1)}$
• D. $\frac{1}{2}\frac{K{ϵ}_0{a}^2}{d}$

Answer 1

The correct option is A $\frac{K{ϵ}_0{a}^2}{d(K-1)}\ln K$


Consider a small element $dx$ at a distance $x$ and has height $y$ in the dielectric medium.

From the figure, $\frac{y}{x}=\frac{d}{a}\Rightarrow y=\frac{d}{a}x$

$\Rightarrow dy=\frac{d}{a}\left(dx\right)$

The capacitance $\left(dC\right)$ of the small element $dx$ is given by,

$\frac{1}{dC}=\frac{y}{K{\epsilon }_{0}adx}+\frac{(d-y)}{{\epsilon }_{0}adx}$

$\frac{1}{dC}=\frac{1}{{\epsilon }_{0}adx}(\frac{y}{K}+d-y)$

Therefore, the total capacitance is given by,

$C=\int dC=\int \frac{{\epsilon }_{0}adx}{\frac{y}{K}+d-y}$

$C={\epsilon }_{0}a\frac{a}{d}{\int }_0^d\frac{dy}{d+y(\frac{1}{K}-1)}$

$C=\frac{{\epsilon }_0{a}^2}{(\frac{1}{K}-1)d}{\left[\ln (d+y(\frac{1}{K}-1))\right]}_0^d$

$C=\frac{K{ϵ}_0{a}^2}{(1-K)d}\ln \left(\frac{d+d(\frac{1}{K}-1)}{d}\right)$

$C=\frac{K{ϵ}_0{a}^2}{(1-K)d}\ln \left(\frac{1}{K}\right)$

$C=\frac{K{ϵ}_0{a}^2\ln K}{(K-1)d}$

Hence, option $\left(B\right)$ is correct.