question_answer

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 is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is: [JEE Main 09-Jan-2019 Morning]

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

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

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

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

Correct Answer: B

Solution :

[b]

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

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

\[{{C}_{1}}\,\,=\,\,\frac{{{\varepsilon }_{0}}adx}{d-y}\,\,;\,\,\,\,\,{{C}_{2}}\,=\,\frac{K{{\varepsilon }_{0}}adx}{y}\]

\[{{C}_{eq}}\,\,=\,\,\frac{{{C}_{1}}\,{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\,\,=\,\,\,\,\,\frac{K{{\varepsilon }_{0}}adx}{Kd\,+\,(1\,\,-\,\,K)y}\]

\[\int\limits_{0}^{a}{\frac{K{{\varepsilon }_{0}}adx}{Kd\,+\,(1-K)\,\frac{dx}{a}}}\,\,\,=\,\,\,\frac{K{{\varepsilon }_{0}}{{a}^{2}}\,In\,\,K}{d(K-1)}\]