A capacitor is made of two square plates each of side ‘ a ’ making a very small angle α…

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:

1. $\frac{{\in }_0{a}^2}{d}\left(1-\frac{\alpha a}{2d}\right)$
2. $\frac{{\in }_0{a}^2}{d}\left(1-\frac{\alpha a}{4d}\right)$
3. $\frac{{\in }_0{a}^2}{d}\left(1+\frac{\alpha a}{d}\right)$
4. $\frac{{\in }_0{a}^2}{d}\left(1-\frac{3\alpha a}{2d}\right)$

Solution

$dc=\frac{{\in }_{0}adx}{d+\alpha x}$
$\Rightarrow c=\frac{{\in }_{0}a}{\alpha }{\left[In\left(d+\alpha x\right)\right]}_0^a$
$=\frac{{\in }_{0}a}{\alpha }In\left(1+\frac{\alpha a}{d}\right)\approx \frac{{\in }_0{a}^2}{d}\left(1-\frac{\alpha a}{2b}\right)$

Asked in: JEE Main 2020 (08 Jan Shift 2)

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