Question

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

• A. $\frac{\epsilon_{0}a^{2}}{d}\left(1-\frac{\alpha a}{4d}\right)$
• B. $\frac{\epsilon_{0}a^{2}}{d}\left(1+\frac{\alpha a}{d}\right)$
• C. $\frac{\epsilon_{0}a^{2}}{d}\left(1-\frac{\alpha a}{2d}\right)$
• D. $\frac{\epsilon_{0}a^{2}}{d}\left(1-\frac{3\alpha a}{2d}\right)$

Answer 1

Answer: (C)

$\frac{\epsilon_{0}a^{2}}{d}\left(1-\frac{\alpha a}{2d}\right)$

Answer 2

Assume small element $dx$ at a distance $x$ from left end
Capacitance for small element $dx$ is
$dC=\frac{\varepsilon_{0}a\,dx}{d+x\,\alpha}$
$C=\int\limits^{a}_{{0}}\frac{\varepsilon_{0}a\,dx}{d+x\,\alpha}$
$=\frac{\varepsilon_{0}\,a}{\alpha}In\left(\frac{1+\alpha a}{d}\right)|^a_0$ $\,\left(In\left(1+x\right)\approx x-\frac{x^{2}}{2}\right)$
$=\frac{\varepsilon_{0}a^{2}}{d}\left(1-\frac{\alpha a}{2d}\right)$