First slide

Capacitors and capacitance

Question

Two square plates of side ‘a’ are arranged as shown in the figure. The minimum separation between plates is ‘d’ and one of the plate is inclined at small angle $α$ with plane parallel to another plate. The capacitance of capacitor is (given $α$ is very small)

Moderate

A

$\frac{\in_{0} a^{2}}{d} (1 - \frac{α a}{2 d})$
B

$\frac{\in_{0} a^{2}}{d} (1 - \frac{α a}{d})$
C

$\frac{\in_{0} a^{2}}{d} (1 + \frac{α a}{2 d})$
D

$\frac{\in_{0} a^{2}}{d} (1 + \frac{α a}{4 d})$

Solution

For small strip of width dx and separation $d + α x$ , differential capacitance can be given as
$d C = \frac{\in_{0} a d x}{d + α x} A l l s u c h s t r i p s a r e i n p a r a l l e l, s o w e w i l l i n t e g r a t e t h e e q u a t i o n C . = \frac{\in_{0} a}{α} {[I n (d + α x)]}_{0}^{a} = \frac{\in_{0} a}{α} I n (1 + \frac{α a}{d}) U s e, I n (1 + x) ≃ x - \frac{x^{2}}{2} (f o r s m a l l v a l u e s o f x) \approx \frac{\in_{0} a}{α} [\frac{α a}{d} - \frac{α^{2} a^{2}}{2 d^{2}}] \approx \frac{\in_{0} a^{2}}{d} (1 - \frac{α a}{2 d})$

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