A parallel plate capacitor is made of two square plates of side $a$, separated by a distance $d$ ($d \ll a$). The lower triangular portion is filled with a dielectric of dielectric constant $K$, as shown in the figure. The capacitance of this capacitor is
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When dielectric varies gradually, divide the capacitor into infinitesimal parallel strips and integrate the capacitance.
Step 1: Since $d \ll a$, fringe effects are neglected.
The capacitor can be treated as a parallel combination of infinitesimal capacitors along the horizontal direction.
Step 2: Consider a thin vertical strip of width $dx$ at a distance $x$ from the left end.
The height of dielectric in this strip increases linearly from $0$ to $d$:
\[
\text{Effective separation at }x:\quad d(x)=\frac{d}{a}x
\]
Step 3: The infinitesimal capacitance $dC$ of this strip is:
\[
dC=\frac{\varepsilon(x)\,a\,dx}{d(x)}
\]
where the effective permittivity varies continuously from $\varepsilon_0$ to $K\varepsilon_0$.
Step 4: The equivalent capacitance is obtained by integrating:
\[
C=\int_0^a \frac{\varepsilon_0 a\,dx}{d}\,
\frac{K}{1+(K-1)\frac{x}{a}}
\]
Step 5: Simplify and integrate:
\[
C=\frac{K\varepsilon_0 a^2}{d}\int_0^1
\frac{du}{1+(K-1)u}
\]
\[
C=\frac{K\varepsilon_0 a^2}{d}\left[\frac{\ln(1+(K-1)u)}{K-1}\right]_0^1
\]
Step 6: Evaluating the limits:
\[
C=\frac{K\varepsilon_0 a^2}{d(K-1)}\ln K
\]
