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