Physics - Electrostatics Question with Solution | TestHub

$\mathit{\sigma }=\frac{\mathit{Q}}{\mathit{\pi RL}}$

$\mathit{L}$= Slant height

$\mathit{dQ}=\mathit{\sigma }\times \left(2\mathit{\pi r}\right)\mathit{dx}$

= Charge on the elementary ring

$\mathit{dQ}=\frac{\mathit{Q}}{\mathit{\pi RL}}\times 2\mathit{\pi rdx}=\frac{\mathit{Q}}{\mathit{LR}}\left(2\mathit{rdx}\right)$... (1)

By similarity of triangles :

$\frac{\mathit{R}}{\mathit{L}}=\frac{\mathit{r}}{\mathit{x}}\mathit{}\Rightarrow \mathit{r}=\frac{\mathit{R}}{\mathit{L}}\mathit{x}$... (2)

By (1) and (2) we get

$\mathit{dQ}=\frac{\mathit{Q}}{\mathit{LR}}\times 2\frac{\mathit{R}}{\mathit{L}}\mathit{xdx}=\frac{2\mathit{Q}}{{\mathit{L}}^2}\mathit{xdx}$

$\therefore \;$the electric potential at A due to ring is

$\mathit{dV}=\frac{\mathit{kdQ}}{\mathit{x}}=\frac{\mathit{k}}{\mathit{x}}\frac{\left(2\mathit{Q}\right)}{{\mathit{L}}^2}\mathit{xdx}$

$\therefore \;$Total potential at A is

$\mathit{V}=\int \mathit{dv}=\frac{2\mathit{kQ}}{{\mathit{L}}^2}{\int }_0^{\mathit{L}}\mathit{dx}=\frac{2\mathit{kQ}}{{\mathit{L}}^2}\times \mathit{L}$

V=2kQL

$\therefore \;$Work done to shift a charge q from infinity to point$\mathit{A}$is =$\mathrm{qV}$

$\mathit{W}=\frac{2\mathit{kqQ}}{\mathit{L}}\mathit{}=2\times \frac{1}{4\mathit{\pi }{\mathit{ϵ}}_0}\frac{\mathit{qQ}}{\mathit{L}}=\frac{\mathit{qQ}}{2\mathit{\pi }{\mathit{ϵ}}_0\mathit{L}}$