A parallel plate capacitor is made of two square plates of side a , separated by a distance d   d…

A parallel plate capacitor is made of two square plates of side $a,$ separated by a distance $d \left(d\ll a\right).$ The lower triangular portion filled with a dielectric of dielectric constant $K,$ as shown in the figure. Capacitance of this capacitor is:

1. $\frac{K{\epsilon }_0{a}^2}{2d\left(K+1\right)}$
2. $\frac{K{\epsilon }_0{a}^2}{d\left(K-1\right)}\ln K$
3. $\frac{K{\epsilon }_0{a}^2}{d}\ln K$
4. $\frac{1}{2}\frac{K{\epsilon }_0{a}^2}{d}$

Solution

Consider a small element at a distance x. It is a two capacitors in series.

$dc=\frac{d{c}_1\times d{c}_2}{d{c}_1+d{c}_2}$

$dc=\frac{{\epsilon }_0.adx}{\frac{x\mathrm{t}\mathrm{a}\mathrm{n}\theta }{k}+\frac{d-x\mathrm{t}\mathrm{a}\mathrm{n}\theta }{1}}$

$\therefore dc=\frac{{\epsilon }_{0}K.adx}{-x\tan \theta \left(K-1\right)+d.k}$

$\therefore C={\int }_0^a\frac{{\epsilon }_{0}K.adx}{-x\tan \theta \left(K-1\right)+d.k}$

$=\frac{{\epsilon }_{0}k.a}{\tan \theta (K-1)}ln\frac{-a\tan \theta \left(K-1\right) +d.k}{d.K}$

$=\frac{{\epsilon }_{0}K{a}^2}{d\left(K-1\right)}lnK$

Asked in: JEE Main 2019 (09 Jan Shift 1)

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