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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
$\frac{{K{\varepsilon _0}{a^2}}}{{d\left( {K - 1} \right)}}\,\ln \,K$
$\frac{{K{\varepsilon _0}{a^2}}}{{2d\left( {K + 1} \right)}}$
$\frac{{K{\varepsilon _0}{a^2}}}{d}\,\ln \,K$
$\frac{1}{2}\frac{{K{\varepsilon _0}{a^2}}}{d}$
Solution
Let's consider a strip of thickness $dx$ at a distance of $x$ from the left end as shown in the figure.
$\frac{y}{x}=\frac{d}{a}$
$\Rightarrow \quad y=\left(\frac{d}{a}\right) x$
$C_{1}=\frac{\varepsilon_{0} a d x}{(d-y)} \quad ; \quad C_{2}=\frac{k \varepsilon_{0} a d x}{y}$
$C_{e q}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{k \varepsilon_{0} a d x}{k d+(1-k) y}$
Now integrating it from $0$ to $a$
$\int_0^a {\frac{{{\text{k}}{\varepsilon _0}{\text{adx}}}}{{{\text{kd}} + (1 – {\text{k}})\frac{{\text{d}}}{{\text{a}}}{\text{x}}}}} = \frac{{{\text{k}}{\varepsilon _0}{{\text{a}}^2}{\text{lnk}}}}{{{\text{d}}({\text{k}} – 1)}}$
