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The Electric fiels at point charge is

$\stackrel{鈫�}{E}=\frac{1}{4蟺{蔚}_0}\frac{q}{r^2}\widehat{r}$

Electric field lines due to provided energies on a position $P$and associated resolution along the $x,y$axes are depicted in the diagram below.

The electric field at point $A$is,

${\stackrel{鈫�}{E}}_1=E_1(鈭�\widehat{i})$

The electric field magnitude of charge is

$E_1=\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}$

Substituting and solving,

$\begin{array}{l}{\stackrel{鈫�}{E}}_1=\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}(鈭�\widehat{i})\\ {\stackrel{鈫�}{E}}_1=鈭�\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}\widehat{i}\end{array}$

The electric filed at corner $B$is,

${\stackrel{鈫�}{E}}_2=E_{2x}\widehat{i}+E_{2y}\widehat{j}$

The distance $BP$from the diagram is

$BP=\sqrt{B{C}^2+C{P}^2}$

Substituting $L=BC=CP$

$\begin{array}{l}BP=\sqrt{L^2+L^2}\\ BP=\sqrt{2}L\end{array}$

The magnitude at charge $B$is

$E_2=\frac{1}{4蟺{蔚}_0}\frac{4Q}{B{P}^2}$

Substituting $BP=\sqrt{2L}$

$\begin{array}{l}{E}_2=\frac{1}{4蟺{蔚}_0}\frac{4Q}{(\sqrt{2}L{)}^2}\\ E_2=\frac{1}{4蟺{蔚}_0}\frac{4Q}{2L^2}\end{array}$

The horizontal component is

$E_{2x}=E_2\cos 鈦�{45}^{鈭�}$

Substituting,

$\begin{array}{l}{E}_{2x}=\frac{1}{4蟺{蔚}_0}\frac{4Q}{2L^2}\cos 鈦�{45}^{鈭�}\\ E_{2x}=\frac{1}{4蟺{蔚}_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\end{array}$

The vertical component is

$E_{2y}=E_2\sin 鈦�{45}^{鈭�}$

Substituting,

$\begin{array}{l}{E}_{2y}=\frac{1}{4蟺{蔚}_0}\frac{4Q}{2L^2}\sin 鈦�{45}^{鈭�}\\ E_{2y}=\frac{1}{4蟺{蔚}_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\end{array}$

The electric charge at corner is

${\stackrel{鈫�}{E}}_3=E_3(鈭�\widehat{j})$

The magnitude at charge is

$E_3=\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}$

Substituting

${\stackrel{鈫�}{E}}_3=\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}(鈭�\widehat{j})$

The resultant horizontal direction is

$E_x=E_1+E_{2x}$

The net electric field is

$\stackrel{鈫�}{E}=E_x\widehat{i}+E_y\widehat{j}$

solving,

$\begin{array}{l}\stackrel{鈫�}{E}=\left(鈭�\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}+\frac{1}{4蟺{蔚}_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\right)\widehat{i}+\left(鈭�\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}+\frac{1}{4蟺{蔚}_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\right)\widehat{j}\\ E=\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}(鈭�1+\sqrt{2})\widehat{i}+\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}(鈭�1+\sqrt{2})\widehat{j}\\ E=\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}(\sqrt{2}鈭�1)(\widehat{i}+\widehat{j})\end{array}$

The net electric filed is$P=\frac{1}{4蟺{蔚}_0}\frac{Q}{L^2}(\sqrt{2}鈭�1)[\widehat{i}+\widehat{j}].$