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:
Answer: (A) $\dfrac{\varepsilon_0 a^2}{d}\left(1 - \dfrac{\alpha a}{2d}\right)$
At distance $x$ from the narrow edge the gap is $d + \alpha x$ (small angle). A strip of width $dx$ is a parallel-plate capacitor of area $a\,dx$, and all strips are in parallel:
$$C = \int_0^a \frac{\varepsilon_0 a\,dx}{d+\alpha x} = \frac{\varepsilon_0 a}{\alpha}\ln\left(1 + \frac{\alpha a}{d}\right)$$
Using $\ln(1+y) \approx y - \dfrac{y^2}{2}$ with $y = \dfrac{\alpha a}{d}$:
$$C \approx \frac{\varepsilon_0 a}{\alpha}\left(\frac{\alpha a}{d} - \frac{\alpha^2a^2}{2d^2}\right) = \frac{\varepsilon_0 a^2}{d}\left(1 - \frac{\alpha a}{2d}\right)$$
Solution by Sreeraj P, M.Sc Physics