$FIGUREP23.38$shows three charges at the corners of a square. Write the electric field at point $P$in component form.

The Electric fiels at point charge is

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\widehat{r}$

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,

${\overrightarrow{E}}_1=E_1(-\widehat{i})$

The electric field magnitude of charge is

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}$

Substituting and solving,

$\begin{array}{l}{\overrightarrow{E}}_1=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}(-\widehat{i})\\ {\overrightarrow{E}}_1=-\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}\widehat{i}\end{array}$

The electric filed at corner $B$is,

${\overrightarrow{E}}_2=E_{2x}\widehat{i}+E_{2y}\widehat{j}$

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

$E_2=\frac{1}{4\pi {\epsilon }_0}\frac{4Q}{B{P}^2}$

Substituting $BP=\sqrt{2L}$

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

The horizontal component is

$E_{2x}=E_2\cos {45}^{\circ }$

Substituting,

$\begin{array}{l}{E}_{2x}=\frac{1}{4\pi {\epsilon }_0}\frac{4Q}{2L^2}\cos {45}^{\circ }\\ E_{2x}=\frac{1}{4\pi {\epsilon }_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\end{array}$

The vertical component is

$E_{2y}=E_2\sin {45}^{\circ }$

Substituting,

$\begin{array}{l}{E}_{2y}=\frac{1}{4\pi {\epsilon }_0}\frac{4Q}{2L^2}\sin {45}^{\circ }\\ E_{2y}=\frac{1}{4\pi {\epsilon }_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\end{array}$

The electric charge at corner is

${\overrightarrow{E}}_3=E_3(-\widehat{j})$

The magnitude at charge is

$E_3=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}$

Substituting

${\overrightarrow{E}}_3=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}(-\widehat{j})$

The resultant horizontal direction is

$E_x=E_1+E_{2x}$

The net electric field is

$\overrightarrow{E}=E_x\widehat{i}+E_y\widehat{j}$

solving,

$\begin{array}{l}\overrightarrow{E}=\left(-\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}+\frac{1}{4\pi {\epsilon }_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\right)\widehat{i}+\left(-\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}+\frac{1}{4\pi {\epsilon }_0}\frac{4Q}{\left(2\sqrt{2}{L}^2\right)}\right)\widehat{j}\\ E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}(-1+\sqrt{2})\widehat{i}+\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}(-1+\sqrt{2})\widehat{j}\\ E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L^2}(\sqrt{2}-1)(\widehat{i}+\widehat{j})\end{array}$

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