91Ó°ÊÓ

Given data;

$$\begin{gathered} q_1=+2.00nC=+2.0×{10}^{-9}C \\ {x}_1=0m \\ {q}_2=-5.0nC=-5.0×{10}^{-9}C \\ {x}_i=0.200m \end{gathered}$$

Also, some of the given data’s are

$$\begin{gathered} x_{ii}=1.20m \\ {x}_{iii}=-0.200m \\ K=9.0×{10}^{9}N{m}^2/C^2 \\ {e}^{-}=-1.602×{10}^{-19}C \end{gathered}$$

Equation;

Electric field due to charge

$E=\frac{Kq}{r^2}$ .......... (1)

Electric force due to electric field

$F=Eq$ .......... (2)

(a) Find the electric field

(i) For$x_i=0.200 m$

From equation (1), net electric field

$$\begin{gathered} \stackrel{→}{E_i}=\left(\frac{k\left|q_1\right|}{r_1^2}+\frac{k\left|q_2\right|}{r_2^2}\right)\hat{i} \\ =\frac{9.0×{10}^9×\left|+2.0×{10}^{-9}\right|}{{0.200}^2}+\frac{9.0×{10}^9×\left|-5.0×{10}^{-9}\right|}{{0.6}^2} \\ =557N/C\hat{i} \end{gathered}$$

Hence, the electric field at given point is 557 N/C

(ii) For$x_{ii}=1.20 m$

From equation (1), net electric field

$$\begin{gathered} \stackrel{→}{E_i}=\left(\frac{k\left|q_1\right|}{r_1^2}+\frac{k\left|q_2\right|}{r_2^2}\right)\hat{i} \\ =\frac{9.0×{10}^9×\left|+2.0×{10}^{-9}\right|}{{1.20}^2}+\frac{9.0×{10}^9×\left|-5.0×{10}^{-9}\right|}{{0.4}^2} \\ =-269N/C\hat{i} \end{gathered}$$

Hence, the electric field at given point is -269N/C

(iii) For$x_{iii}=-0.200 m$

From equation (1), net electric field

$$\begin{gathered} \stackrel{→}{E_i}=\left(\frac{k\left|q_1\right|}{r_1^2}+\frac{k\left|q_2\right|}{r_2^2}\right)\hat{i} \\ =\frac{9.0×{10}^9×\left|+2.0×{10}^{-9}\right|}{{1.20}^2}+\frac{9.0×{10}^9×\left|-5.0×{10}^{-9}\right|}{{1.0}^2} \\ =-405N/C\hat{i} \end{gathered}$$

Hence, the electric field at given point is -405N/C

(b)

From Equation (2)

$$\begin{gathered} \stackrel{→}{F_i}=\stackrel{→}{E_i}e \\ =575×(-1.602×{10}^{-19}) \\ =-9.21×{10}^{-17}N\hat{I} \end{gathered}$$

Hence, Electric force at given point is$-9.21×{10}^{-17}N\hat{I}$

$$\begin{gathered} \stackrel{→}{F_i}=\stackrel{→}{E_i}e \\ =-269×(-1.602×{10}^{-19}) \\ =4.31×{10}^{-17}N\hat{I} \end{gathered}$$

Hence, Electric force at given point is$4.31×{10}^{-17}N\hat{I}$

$$\begin{gathered} \stackrel{→}{F_i}=\stackrel{→}{E_i}e \\ =405×(-1.602×{10}^{-19}) \\ =-6.49×{10}^{-17}N\hat{I} \end{gathered}$$

Hence, Electric force at given point is$-6.49×{10}^{-17}N\hat{I}$