Question

Derive an expression for mechanical force per unit area of surface charge density.

Answer 1

First of all need to mark a infinitely small region ds
as we know the total $E=\frac{\sigma }{{ϵ}_{0}k}$
$E=E_1+E_2,$ where $E_1$ is intensity due charges on ds, $E_2$ is intensity due to charges on rest of conductor.
But electric Intensity at $Q=0$.
$\therefore E_1-E_2$ must be zero. $\therefore E_1$ must be equal to $E_2$.
$\therefore E_{net}=2E_2/E_1$
$\therefore \frac{\sigma }{{ϵ}_{0}k}=2E_2\therefore E_2=\frac{\sigma }{2{ϵ}_{0}k}$
$\therefore$ F= charges let the surface charge density be $\sigma$ .ds for region ds.
$\therefore F=\frac{\sigma }{2{ϵ}_{0}k}.\sigma ds.$
$\Rightarrow \frac{F}{ds}=\frac{{\sigma }^2}{2{ϵ}_{0}k}$
$\therefore f=\frac{{\sigma }^2}{2{ϵ}_{0}k}$ which is called mechanical force per unit area.