Obtaining an expression for electric field intensity
Showing behaviour at large distance
Net Electric Field at point $\mathit{P}=\underset{0}{\overset{2\mathit{\pi }\mathit{a}}{\int }}\mathit{d}\mathit{E}\cos \mathit{\theta }$
$\mathit{d}\mathit{E}=$ Electric field due to a small element having charge $\mathit{d}\mathit{q}$
$=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{\mathit{d}\mathit{q}}{{\mathit{r}}^2}$
Let
$=\frac{\mathit{d}\mathit{q}}{\mathit{d}\mathit{l}}$
$\mathit{d}\mathit{q}=\mathit{\lambda }\mathit{d}\mathit{l}$
Hence $\mathit{E}=\underset{0}{\overset{2\mathit{\pi }\mathit{a}}{\int }}\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot \frac{\mathit{\lambda }\mathit{d}\mathit{l}}{{\mathit{r}}^2}\times \frac{\mathit{x}}{\mathit{r}}$ , where $\cos \mathit{\theta }=\frac{\mathit{x}}{\mathit{r}}$
$=\frac{\mathit{\lambda }\mathit{x}}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^3}\left(2\mathit{\pi }\mathit{a}\right)$
$=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{\mathit{Q}\mathit{x}}{{\left({\mathit{x}}^2+{\mathit{a}}^2\right)}^{\frac{3}{2}}},$
where total charge $\mathit{Q}=\mathit{\lambda }\times 2\mathit{\pi }\mathit{a}$
At large distance i.e., $\mathit{x}>>\mathit{a}$
$\mathit{E}\simeq \frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot \frac{\mathit{Q}}{{\mathit{x}}^2}$
This is the electric field due to a point charge at distance $\mathit{x}$ .