First slide

Capacitors and capacitance

Question

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

Difficult

A

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

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

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

$\frac{\in_{0} a^{2}}{d} (1 - \frac{3 α a}{2 d})$

Solution

$\begin{array}{l} \Rightarrow S e l e c t i n s m a l l s t r i p o f d x, a t a d i s \tan c e x f r o m l e f t . \\ \Rightarrow \tan α = \frac{y}{x} \\ \Rightarrow y = x \tan α = x α (F o r s m a l l a n g l e s \tan α = α) \\ \Rightarrow d i s \tan c e b e t w e e n p l a t e s a t a d i s \tan c e x = d + x α \\ \Rightarrow C a p a c i \tan c e o f t h i s e l e m e n t c a n b e w r i t t e n a s, \\ \Rightarrow d C = \frac{ε_{0} d A}{d + α x} \\ \Rightarrow d C = \frac{ε_{0} a d x}{d + α x} \\ \Rightarrow A l l s u c h e l e m e n t s a r e c o n n e c t e d i n p a r a l l e l, i n t e g r a t i o n w i l l g i v e t o t a l c a p a c i \tan c e \\ \Rightarrow \int_{0}^{a} d C = \int_{0}^{a} \frac{ε_{0} a d x}{d + α x} \\ \Rightarrow C = ε_{0} a \int_{0}^{a} \frac{d x}{d + α x} \\ \Rightarrow C = \frac{ε_{0} a}{α} {[\ln (d + α x)]}_{0}^{a} \\ (∵ \int \frac{1}{a x} d x = \frac{\log a x}{\frac{d}{d x} a x} + C) \\ \Rightarrow C = \frac{ε_{0} a}{α} \ln (1 + \frac{α a}{d}) \\ \Rightarrow C ≃ \frac{ε_{0} a^{2}}{d} (1 - \frac{α a}{2 d}) [\ln (1 + x) = x (1 - \frac{x}{2})] \end{array}$

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