question_answer

A capacitor is made of two square plates each of side ?a? making a very small angle \[\alpha \] between them, as shown in figure. The capacitance will be close to                                        [JEE MAIN Held on 08-01-2020 Evening]

A) \[\frac{{{\varepsilon }_{0}}{{a}^{2}}}{d}\left( 1-\frac{3\alpha a}{2d} \right)\]

B) \[\frac{{{\varepsilon }_{0}}{{a}^{2}}}{d}\left( 1-\frac{\alpha a}{2d} \right)\]

C) \[\frac{{{\varepsilon }_{0}}{{a}^{2}}}{d}\left( 1+\frac{\alpha a}{d} \right)\]

D) \[\frac{{{\varepsilon }_{0}}{{a}^{2}}}{d}\left( 1-\frac{\alpha a}{4d} \right)\]

Correct Answer: B

Solution :

            \[y=\tan \,\alpha x\] \[\therefore \,\,\,\,\,\,\,dC=\frac{adx60}{d+x\tan \alpha }\] \[\therefore \,\,\,\,\,\,{{C}_{eq}}=\int{dc}=a{{\varepsilon }_{0}}\int\limits_{x=0}^{x=a}{\frac{dx}{x\tan \alpha +d}}\] [By Binomial expansion]             \[\Rightarrow \,\,\,\,\,\,\,{{C}_{eq}}=\frac{a{{\varepsilon }_{0}}}{d}\int\limits_{0}^{a}{\left( 1-\frac{x\tan \alpha }{d} \right)}dx=\frac{a{{\varepsilon }_{0}}}{d}\left( x-\frac{{{x}^{2}}\tan \alpha }{d\,2} \right)_{0}^{a}\] \[\Rightarrow \,\,\,\,\,\,\,{{C}_{eq}}=\frac{{{a}^{2}}{{\varepsilon }_{0}}}{d}\left( 1-\frac{a\tan \alpha }{2d} \right)=\frac{{{\varepsilon }_{0}}{{a}^{2}}}{d}\left( 1-\frac{\alpha a}{2d} \right)\]