Question 110 of 180

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

A$\dfrac{K{\epsilon }_{0}{a}^{2}}{2d\left(K+1\right)}$B$\dfrac{K{\epsilon }_{0}{a}^{2}}{d\left(K-1\right)}\ln ⁡K$C$\dfrac{K{\epsilon }_{0}{a}^{2}}{d}\ln ⁡K$D$\dfrac{1}{2}\dfrac{K{\epsilon }_{0}{a}^{2}}{d}$

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Answer: (B) $\dfrac{K{\epsilon }_{0}{a}^{2}}{d\left(K-1\right)}\ln ⁡K$

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

$dc=\dfrac{d{c}_{1}\times d{c}_{2}}{d{c}_{1}+d{c}_{2}}$

$dc=\dfrac{{\epsilon }_{0}.adx}{\dfrac{xtan\theta }{k}+\dfrac{d-xtan\theta }{1}}$

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

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

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

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