Question

A variable air gap type capacitor consists of two parallel plates : a fixed plate and a moving Plate at a distance x. If a potential voltage is applied across the two plates, then force of attraction between the plates is related to x as

• A. $F\propto x^2$
• B. $F\propto \frac{1}{x^2}$
• C. $F\propto \frac{1}{x}$
• D. $F\propto x$

Answer 1

The correct option is B $F\propto \frac{1}{x^2}$

If voltage = V then electric field E between the plates,

$E=\frac{V}{x}$

Therefore force of attraction,

$F=qE=CV.E=\frac{ϵAV}{x}\times \frac{V}{x}$

$F=\frac{ϵA{V}^2}{x^2}$

$\text{Here}ϵ,\text{A and V are constant.}$

So,
$F\propto \frac{1}{x^2}$