Question

A capacitor is made of two square plates each of side ${}^{\prime }{a}^{\prime }$ making a very small angle a between them, as shown in figure. The capacitance will be close to :

• A. $\frac{{ϵ}_0{a}^2}{d}\left(1-\frac{\alpha a}{4d}\right)$
• B. $\frac{{ϵ}_0{a}^2}{d}\left(1+\frac{\alpha a}{d}\right)$
• C. $\frac{{ϵ}_0{a}^2}{d}\left(1-\frac{\alpha a}{2d}\right)$
• D. $\frac{{ϵ}_0{a}^2}{d}\left(1-\frac{3\alpha a}{2d}\right)$

Answer 1

Answer: (C)

Assume small element $dx$ at a distance $x$ from left end
Capacitance for small element $dx$ is
$dC=\frac{{\epsilon }_{0}adx}{d+x\alpha }$
$C=\underset{0}{\overset{a}{\int }}\frac{{\epsilon }_{0}adx}{d+x\alpha }$
$=\frac{{\epsilon }_{0}a}{\alpha }In\left(\frac{1+\alpha a}{d}\right){|}_0^a$$\left(In\left(1+x\right)\approx x-\frac{x^2}{2}\right)$
$=\frac{{\epsilon }_0{a}^2}{d}\left(1-\frac{\alpha a}{2d}\right)$