JEE PYQ: Capacitance Question 23
Question 23 - 2020 (08 Jan Shift 2)
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) $\frac{\epsilon_0 a^2}{d}\left(1 - \frac{\alpha a}{2d}\right)$
(2) $\frac{\epsilon_0 a^2}{d}\left(1 - \frac{\alpha a}{4d}\right)$
(3) $\frac{\epsilon_0 a^2}{d}\left(1 + \frac{\alpha a}{d}\right)$
(4) $\frac{\epsilon_0 a^2}{d}\left(1 - \frac{3\alpha a}{2d}\right)$
Show Answer
Answer: (1)
Solution
Consider an infinitesimal strip of capacitor of thickness $dx$ at a distance $x$. Capacitance of thickness $dx$: $dC = \frac{\epsilon_0 a,dx}{d + x\tan\alpha}$. Using binomial expansion for $\alpha « 1$: $C_{eq} = \frac{\epsilon_0 a}{d}\int_0^a \left(1 - \frac{x\tan\alpha}{d}\right)dx = \frac{\epsilon_0 a^2}{d}\left(1 - \frac{a\tan\alpha}{2d}\right) \approx \frac{\epsilon_0 a^2}{d}\left(1 - \frac{\alpha a}{2d}\right)$.
BETA
