What are the electric fields at points 1, 2, and 3 in FIGURE P22.63? Give your answer in component form.

Given charge $=q=-10nC=-10\times {10}^{-9}C$

Distance of point 1 from the given charge $=r_1=3.0cm=0.03m$

Distance of point 3 from the given charge$=r_3=4.0cm=0.04m$

Distance of point 1 from the given charge $=r_2=\sqrt{3.0^2+4.0^2}=5.0cm=0.05m$

Angle of point 2 from the given charge $={\theta }_2={\sin }^{-1}\left(3/5\right)=36.87°$

Electric field at a point is given by the equation:

$\overrightarrow{E}=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}\hat{r}$

Hence,

for point 1, the magnitude of the electric field will be:

$$\begin{gathered} E_1=\frac{1}{4\pi {ϵ}_0}\frac{\left|q\right|}{{r_1}^2} \\ {E}_1=\frac{9.0\times {10}^9\times 10\times {10}^{-9}}{{\displaystyle 0}{\displaystyle .}{{\displaystyle 03}}^2} \\ {E}_1=1.0\times {10}^{5}N/C \end{gathered}$$

for point 2, the magnitude of the electric field will be:

$$\begin{gathered} E_2=\frac{1}{4\pi {ϵ}_0}\frac{\left|q\right|}{{r_2}^2} \\ {E}_2=\frac{9.0\times {10}^9\times 10\times {10}^{-9}}{{\displaystyle 0.{05}^2}} \\ {E}_2=3.6\times {10}^{4}N/C \end{gathered}$$

for point 3, the magnitude of the electric field will be:

$$\begin{gathered} E_3=\frac{1}{4\pi {ϵ}_0}\frac{\left|q\right|}{{r_3}^2} \\ {E}_3=\frac{9.0\times {10}^9\times 10\times {10}^{-9}}{{\displaystyle 0.{04}^2}} \\ {E}_3=5.6\times {10}^{4}N/C \end{gathered}$$

Since the given electric field is negative in nature, hence the points under the influence of this field will also be negative.

Hence, the component form of these electric fields can be written as:

Component form of electric field at point 1:

$$\begin{gathered} {\overrightarrow{E}}_1=-E_1\cos 90°\hat{i}-E_1\sin 90°\hat{j} \\ {\overrightarrow{E}}_1=(0\hat{i}-1.0\times {10}^5\hat{j}) \\ {\overrightarrow{E}}_1=-1.0\times {10}^5\hat{j}N/C \end{gathered}$$

Component form of electric field at point 2:

$$\begin{gathered} {\overrightarrow{E}}_2=-E_2\cos 36.87°\hat{i}-E_2\sin 36.87°\hat{j} \\ {\overrightarrow{E}}_2=(-3.6\times {10}^4\times 0.8\hat{i}-3.6\times {10}^4\times 0.6\hat{j}) \\ {\overrightarrow{E}}_2=-2.9\times {10}^4\hat{i}-2.2\times {10}^4\hat{j}N/C \end{gathered}$$

Component form of electric field at point 3:

$$\begin{gathered} {\overrightarrow{E}}_3=-E_3\cos 0°\hat{i}-E_3\sin 0°\hat{j} \\ {\overrightarrow{E}}_3=(-5.6\times {10}^4\hat{i}-0\hat{j}) \\ {\overrightarrow{E}}_3=-5.6\times {10}^4\hat{i}N/C \end{gathered}$$

Hence, the electric fields in their component form for the given problem can be calculated as follows:

$$\begin{gathered} {\overrightarrow{E}}_1=-1.0\times {10}^5\hat{j}N/C \\ {\overrightarrow{E}}_2=-2.9\times {10}^4\hat{i}-2.2\times {10}^4\hat{j}N/C \\ {\overrightarrow{E}}_3=-5.6\times {10}^4\hat{i}N/C \end{gathered}$$