91Ó°ÊÓ

Given charge $=q=5.0nC=5.0×{10}^{-9}C$

Distance of point 1 from the given charge $=r_1=\sqrt{1.0^2+2.0^2}=2.24cm=0.0224m$

Distance of point 2 from the given charge $=r_2=1.0cm=0.01m$

Distance of point 3 from the given charge $=r_3=\sqrt{1.0^2+2.0^2}=2.24cm=0.0224m$

Angle made by points 1 and 3$={\mathrm{θ}}_1={\mathrm{θ}}_3={\tan }^{-1}\left(2/1\right)=63.43°$

Electric field at a point is given by the equation:

$\stackrel{→}{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{q}{r^2}\hat{r}$

Hence,

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

$$\begin{gathered} E_1=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{\left|q\right|}{{r_1}^2} \\ {E}_1=\frac{9.0×{10}^9×5.0×{10}^{-9}}{{\displaystyle 0.{0224}^2}} \\ {E}_1=9.0×{10}^{4}N/C \end{gathered}$$

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

$$\begin{gathered} E_2=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{\left|q\right|}{{r_2}^2} \\ {E}_2=\frac{9.0×{10}^9×5.0×{10}^{-9}}{{\displaystyle 0.{01}^2}} \\ {E}_2=4.5×{10}^{5}N/C \end{gathered}$$

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

$$\begin{gathered} E_3=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{\left|q\right|}{{r_3}^2} \\ {E}_3=\frac{9.0×{10}^9×5.0×{10}^{-9}}{{\displaystyle 0.{0224}^2}} \\ {E}_3=9.0×{10}^{4}N/C \end{gathered}$$

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

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

Component form of electric field at point 1:

$$\begin{gathered} {\stackrel{→}{E}}_1=E_1\cos 63.43°\hat{i}+E_1\sin 63.43°\hat{j} \\ {\stackrel{→}{E}}_1=(9.0×{10}^4×0.45\hat{i}+9.0×{10}^4×0.9\hat{j}) \\ {\stackrel{→}{E}}_1=4.05×{10}^4\hat{i}+8.1×{10}^4\hat{j}N/C \end{gathered}$$

Component form of electric field at point 2:

$$\begin{gathered} {\stackrel{→}{E}}_2=E_2\cos 0°\hat{i}+E_2\sin 0°\hat{j} \\ {\stackrel{→}{E}}_2=(4.5×{10}^5\hat{i}+0\hat{j}) \\ {\stackrel{→}{E}}_2=4.5×{10}^5\hat{i}N/C \end{gathered}$$

Component form of electric field at point 3:

$$\begin{gathered} {\stackrel{→}{E}}_3=E_3\cos 63.43°\hat{i}-E_3\sin 63.43°\hat{j} \\ {\stackrel{→}{E}}_3=(9.0×{10}^4×0.45\hat{i}-9.0×{10}^4×0.9\hat{j}) \\ {\stackrel{→}{E}}_3=4.05×{10}^4\hat{i}-8.1×{10}^4\hat{j}N/C \end{gathered}$$

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

$$\begin{gathered} {\stackrel{→}{E}}_1=4.05×{10}^4\hat{i}+8.1×{10}^4\hat{j}N/C \\ {\stackrel{→}{E}}_2=4.5×{10}^5\hat{i}N/C \\ {\stackrel{→}{E}}_3=4.05×{10}^4\hat{i}-8.1×{10}^4\hat{j}N/C \end{gathered}$$