Question

Question: A +2.00nC point charge is at the origin, and a second -5.00nC point charge is on the x-axis at x = 0.800 m. (a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: (i) x = 0.200 m; (ii) x = 1.20 m; (iii) x = -0.200 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

Short answer

Answer

Electric field at given point is$575 N/C,-269 N/C,-405 N/C$

Electric force at given point is $-9.21\times {10}^{-19} N,4.31\times {10}^{-17} N,6.49\times {10}^{-17} N$

Step-by-step solution

Data and Formula

Given data;

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

Also, some of the given data’s are

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

Equation;

Electric field due to charge

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

Electric force due to electric field

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

Draw the diagram and find the electric field

(a) Find the electric field

(i) For$x_i=0.200 m$

From equation (1), net electric field

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

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

(ii) For$x_{ii}=1.20 m$

From equation (1), net electric field

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

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

(iii) For$x_{iii}=-0.200 m$

From equation (1), net electric field

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

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

Find electric force

(b)

From Equation (2)

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

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

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

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

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

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