Four positive point charges (+q) are kept at the four corners of a square of a side ‘l’. The net electric field at the mid point of any one side of the square is,
$\left(\mathrm{take}, \frac{1}{4{\mathrm{\pi \epsilon }}_0}=\mathrm{K}\right)$

$F_3=F_4$ and these are in opposite direction, so it will cancel each other out.

${\mathrm{F}}_1=\frac{1}{4\pi {\epsilon }_0}\frac{\mathrm{q}}{{\left(\frac{\sqrt{5}}{2}l\right)}^2}=\mathrm{K}\frac{4\mathrm{q}}{5l^2}$

${\mathrm{F}}_2=\frac{1}{4\pi {\epsilon }_0}\frac{\mathrm{q}}{{\left(\frac{\sqrt{5}}{2}l\right)}^2}=\mathrm{K}\frac{4\mathrm{q}}{5l^2}$

There horizontal components cancel each other, and net force is due to vertical components.

$F_{net}=F_1\sin \theta +F_2\sin \theta =2\frac{4q}{5l^2}\times \frac{1}{\frac{\sqrt{5}}{2}}=\frac{16Kq}{5\sqrt{5}{l}^2}$