Question: Three negative point charges lie along a line as shown in Fig. Find the magnitude and direction of the electric field this combination of charges produces at point P, which lies 6.00 cm from the -2.00 mC charge measured perpendicular to the line connecting the three charges

Given data;

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

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)

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

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

For the, $q_2=-2.0\mu C$put this value in equation (1)

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

The net electric field at point in direction

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

From the above diagram

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

Were, $\cos \alpha =\frac{0.06}{0.10}$

Put this all values in equation (2)

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

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