91Ó°ÊÓ

Given,

$$\begin{gathered} q_1=-5.0nC \\ {q}_2=3.0nC \\ x=3.0×{10}^{-2}m \\ y=4.0×{10}^{-2}\text{m} \end{gathered}$$

The electric field due to the charge is,

$E_1=\frac{k{q}_1}{r_1^2}$

Now, the electric field due to the charge at origin is given by:

$$\begin{gathered} E_1=\frac{k{q}_1}{r_1^2} \\ =\frac{9×{10}^9×5×{{10}^{-}}^9}{{\left(4×{{10}^{-}}^2\right)}^2} \\ =2.8×{10}^4 N/C \end{gathered}$$

Now the direction of the field is in

Therefore,$E_1=\left[-2.8×{10}^4\hat{j}\right]N/C$

Now, the electric field due to the second charge is given by:

$$\begin{gathered} E_2=\frac{k{q}_2}{{r_2}^2} \\ =\frac{k{q}_2}{{\left(\sqrt{\left(x^2+y^2\right)}\right)}^2} \\ ={\frac{9×{10}^9×3×{{10}^{-}}^9}{\sqrt{{\left({\left(0.3\right)}^2+{\left(0.4\right)}^2\right)}^2}}}^2 \\ =1.08×{10}^4 N/C \end{gathered}$$

Therefore, the angle and the electric field and the horizontal is:

$$\begin{gathered} {\mathrm{θ}}_1=arctan\left(\frac{4}{3}\right) \\ =53.{13}^0 \end{gathered}$$

Now, the electric field vector is given by:

$$\begin{gathered} E_2=E_1-\left(cos{\mathrm{θ}}_i+sin\mathrm{θ}\right) \\ =1.08×{10}^4\left(-cos\left(53.13\right)\hat{i}+sin\left(53.13\right)\hat{j}\right) \\ =\left[-6.48×{10}^3\hat{i}+8.64×{10}^3\hat{j}\right]N/m \end{gathered}$$

Therefore, the resultant field at P is$\left[-6.48×{10}^3\hat{i}+8.64×{10}^3\hat{j}\right]N/m$

Now the net electric field is:

$$\begin{gathered} E=E_1+E_2 \\ =-2.8×{10}^4\hat{j}-6.48×{10}^3\hat{i}+8.64×{10}^3\hat{j} \\ =\left[-6.48×{10}^3\hat{i}-1.94×{10}^4\hat{j}\right]N/m \end{gathered}$$