Question

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 :

• A. \(\frac{K\epsilon_0a^2}{2d(K+1)}\)
• B. \(\frac{K\epsilon_0a^2}{d(K-1)}\ln K\)
• C. \(\frac{K\epsilon_0a^2}{d}\ln K\)
• D. \(\frac12\frac{K\epsilon_0a^2}{d}\)

Hint:

Slice into thin vertical strips of width \(dx\); each is a dielectric slab of thickness \(t(x)\) in series with an air gap, all strips being in parallel. Integrate the strip capacitances.

Answer 1

The Correct Option is B

The dielectric wedge has thickness \(t=\frac{d}{a}x\) at distance \(x\) from the thin end (0 at \(x=0\), \(d\) at \(x=a\)).
A strip of width \(dx\) (area \(a\,dx\)) has air gap \(d-t\) in series with dielectric of thickness \(t\): \(dC=\frac{\epsilon_0a\,dx}{(d-t)+\frac{t}{K}}=\frac{\epsilon_0a\,dx}{d-\frac{d}{a}x\left(1-\frac1K\right)}\).
Let \(u=\frac xa\): \(C=\frac{\epsilon_0a^2}{d}\int_0^1\frac{du}{1-\left(1-\frac1K\right)u}=\frac{\epsilon_0a^2}{d}\cdot\frac{-\ln\left(\frac1K\right)}{1-\frac1K}\).
\(C=\frac{\epsilon_0a^2}{d}\cdot\frac{K\ln K}{K-1}=\frac{K\epsilon_0a^2}{d(K-1)}\ln K\).
(Check: for \(K\to1\), \(C\to\frac{\epsilon_0a^2}{d}\).)
Answer: (2) \(\frac{K\epsilon_0a^2}{d(K-1)}\ln K\)