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. Capacitance of this capacitor is
Answer: (B) $\dfrac{K\varepsilon_0 a^2}{d(K-1)}\ln K$
Divide the capacitor into thin strips of width $dx$ (area $a\,dx$). At distance $x$ from the left edge, the dielectric thickness is $y = \frac{d}{a}x$, and the strip is a dielectric layer in series with an air gap:
$$dC = \frac{\varepsilon_0 a\,dx}{(d - y) + \dfrac yK} = \frac{\varepsilon_0 a\,dx}{d\left[1 - \dfrac xa\left(1 - \dfrac1K\right)\right]}$$
The strips are in parallel, so
$$C = \frac{\varepsilon_0 a}{d}\int_0^a \frac{dx}{1 - \beta x/a},\quad \beta = 1 - \frac1K$$
$$C = \frac{\varepsilon_0 a^2}{d\beta}\ln\frac{1}{1-\beta} = \frac{\varepsilon_0 a^2}{d}\cdot\frac{K}{K-1}\ln K$$
Solution by Sreeraj P, M.Sc Physics