91Ó°ÊÓ

Given data;

$$\begin{gathered} q_1=q_3=-5.0 \mathrm{μ}C \\ {q}_2=-2.0 \mathrm{μ}C \\ R=0.10 m \\ r=0.06 m \\ K=9×{10}^9 N·m^2/C^2 \end{gathered}$$

Equation;

Electric field due to charge

$\mathit{E}\mathbf{=}\frac{\mathbf{K}\left|q\right|}{{\mathbf{r}}^{\mathbf{2}}}$ $E=\frac{K\left|q\right|}{r^2}$ .......... (1)

$q_1=q_3=-5.0\mathrm{μ}C$And distance between them is also same. Put is value in equation

$$\begin{gathered} E_1=E_3=\frac{9×{10}^9 N·m^2/C^2×\left|-5×{10}^{-6} C\right|}{{(0.100 m)}^2} \\ =4.50×{10}^9 N/C \end{gathered}$$

For the, $q_2=-2.0\mathrm{μ}C$put this value in equation (1)

$$\begin{gathered} E_2=\frac{9×{10}^9 N·m^2/C^2×\left|2×{10}^{-6} C\right|}{{(0.060 m)}^2} \\ =5.00×{10}^6 N/C \end{gathered}$$

The net electric field at point in direction

$E_{net}=E_{1x}+E_{2x}+E_{3x}$ .......... (2)

From the above diagram

$$\begin{gathered} E_{1x}=E_1\cos \mathrm{Î\pm } \\ {E}_{2x}=E_2 \\ {E}_{3x}=E_3\cos \end{gathered}$$

Were, $\cos \mathrm{Î\pm }=\frac{0.06}{0.10}$

Put this all values in equation (2)

$$\begin{gathered} E_{net}=4.50×{10}^9 N/C×\frac{0.06}{0.10}+5.0×{10}^{5}N/C+4.50×{10}^6 N/C×\frac{0.06}{0.10} \\ =10.40×{10}^6 N/C \end{gathered}$$

Hence, the electric field at point P is$E_{net}=10.40×{10}^{6}N/C$ and the direction of the electric field-x is axis .