Take a long thin wire XY (as shown in the following figure) of uniform linear charge density $\mathrm{\lambda }$

Consider a point A at a perpendicular distance I from the mid-point
O of the wire, as shown in the following figure.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ X
Let q be the charge on this piece.

$\mathrm{Electric}<mspace\; width="1em"></mspace>\mathrm{field}<mspace\; width="1em"></mspace>\mathrm{due}<mspace\; width="1em"></mspace>\mathrm{to}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{piece},<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\therefore \mathrm{q}=\mathrm{\lambda dx}$
$\mathrm{dE}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\cdot \frac{\mathrm{\lambda dx}}{(\mathrm{AZ}{)}^2}$
$\mathrm{However},\hspace{0.17em}\mathrm{AZ}=\sqrt{{\mathrm{l}}^2+{\mathrm{x}}^2}$
$\therefore <mspace\; width="1em"></mspace>\mathrm{dE}<mspace\; width="1em"></mspace>=\frac{1}{4{\mathrm{\pi ϵ}}_0}\cdot \frac{\mathrm{\lambda dx}}{({\mathrm{l}}^2+{\mathrm{x}}^2)}$

The electric field is resolved into two rectangular components. dE cos $\mathrm{\theta }$ is the perpendicular component and dE sin G is the parallel component. When the whole wire is considered, the component dE sin G is cancelled. Only the perpendicular component dEcos6 affects point A.

Hence, effective electric field at point A due to the element dx is dEj.

$\mathrm{In}<mspace\; width="1em"></mspace>∆\mathrm{AZO},<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\begin{array}{l}\therefore {\mathrm{dE}}_1=\frac{1}{4{\mathrm{\pi ϵ}}_0}\cdot \frac{\mathrm{\lambda dx}\cdot \cos \mathrm{\theta }}{({\mathrm{l}}^2+{\mathrm{x}}^2)}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)\\ \tan \mathrm{\theta }=\frac{\mathrm{x}}{\mathrm{l}}\Rightarrow \mathrm{x}=\mathrm{l}.\tan \mathrm{\theta }\dots \dots \dots \dots \dots \dots \dots \dots .\left(2\right)\end{array}$

On differentiating equation (2), we obtain

$\frac{dx}{d\theta }={\mathrm{lsec}}^2\theta \Rightarrow dx={\mathrm{lsec}}^2\theta d\theta \dots \dots \dots \dots \dots \dots \text{\hspace{0.17em}(3)\hspace{0.17em}}$

From equation (2), we have

${\mathrm{x}}^2+{\mathrm{l}}^2={\mathrm{l}}^2{\tan }^2\mathrm{\theta }+{\mathrm{l}}^2={\mathrm{l}}^2({\tan }^2\mathrm{\theta }+1)={\mathrm{l}}^2{\sec }^2\mathrm{\theta }$

Putting equations (3) and (4) in equation (1), we obtain

$\begin{array}{l}d{E}_1=\frac{1}{4\pi {ϵ}_0}\cdot \frac{\lambda {\sec }^2\theta d\theta \cdot \cos \theta }{l^2{\sec }^2\theta }\\ =\frac{1}{4\pi {ϵ}_0}\cdot \frac{\lambda \cos \theta d\theta }{l}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(5\right)\end{array}$

The wire is so long that 6 tends from $-\frac{\pi }{2}\text{\hspace{0.17em}to\hspace{0.17em}}\frac{\pi }{2}$

By integrating equation (5), we obtain the value of field E as,

$\begin{array}{c}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}d{E}_1={\int }_{-\pi }^{\frac{\pi }{2}}\frac{1}{4\pi {ϵ}_0}\frac{\lambda }{l}\cos \theta d\theta \\ \Rightarrow E_1=\frac{1}{4\pi {ϵ}_0}\frac{\lambda }{l}[\sin \theta {]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\Rightarrow E_1=\frac{1}{4\pi {ϵ}_0}\frac{\lambda }{l}\times 2\Rightarrow E_1=\frac{\lambda }{2\pi {ϵ}_{0}l}\end{array}$

Therefore, the electric field due to long wire is $\frac{\lambda }{2\pi {ϵ}_{0}l}$.