question_answer

A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d<<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) \[\frac{1}{2}\,\frac{k\,{{\in }_{0}}{{a}^{2}}}{d}\]

B) \[\frac{k\,{{\in }_{0}}{{a}^{2}}}{d}\,\text{In}\,K\]

C) \[\frac{k\,{{\in }_{0}}{{a}^{2}}}{d\,(K-1)}\,\text{In}\,k\]

D) \[\frac{k\,{{\in }_{0}}{{a}^{2}}}{2d(K+1)}\]

Correct Answer: C

Solution :

\[\frac{y}{x}\,=\,\frac{d}{a}\]

\[dy=\,\frac{\text{d}}{a}\,(\text{d}x)\]

\[\frac{1}{dc}\,=\,\frac{y}{\text{KE}.adx}+\frac{(d-y)}{{{\in }_{0}}\,adx}\]

\[\frac{1}{dc}\,=\,\frac{1}{{{\in }_{0}}}adx\left( \frac{y}{k}+d-y \right)\]

\[\int{dc}\,=\,\int{\frac{{{\varepsilon }_{0}}adx}{\frac{y}{k}+d-y}}\]

\[c={{\varepsilon }_{0}}a.\frac{a}{d}\int_{0}^{d}{\frac{dy}{d+y\left( \frac{1}{k}-1 \right)}}\]

\[=\frac{{{\varepsilon }_{0}}{{a}^{2}}}{\left( \frac{1}{k}\,-\,1 \right)}\left[ \text{In}\left( d+y\left( \frac{1}{k}\,-\,1 \right) \right) \right]_{0}^{\text{d}}\]

\[=\,\frac{k{{\varepsilon }_{0}}{{a}^{2}}}{(1\,-\,k)d}\,\text{In}\left( \frac{d+d\left( \frac{1}{k}\,-\,1 \right)}{d} \right)\]

\[=\,\frac{k{{\varepsilon }_{0}}{{a}^{2}}}{(1-k)d}\,\text{In}\,\left( \frac{1}{k} \right)\,=\,\frac{k{{\varepsilon }_{0}}{{a}^{2}}\,\text{In}\,k}{(k\,-\,1)d}.\]