91Ó°ÊÓ

Given a positive and a negative charge:

The direction of the electric field due to the positive point charge is always acting away from the point charge and the electric field due to the negative charge acts towards the negative charge.

The electric field is a vector quantity. They cannot be added directly but can be added vectorially.

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{5.0\mathrm{cm}}{5.0\mathrm{cm}}\right)=45° \\ r=\sqrt{(5.0\mathrm{cm}{)}^2+(5.0\mathrm{cm}{)}^2} \\ =7.07\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right) \\ 0.0707\mathrm{m} \end{gathered}$$

Calculate the electric field $E_1$as follows:

$E_1=\frac{k{q}_1}{r_1^2}$

Substitute $9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^{2}fork,3.0\mathrm{nC}for{q}_1\text{,}and0.0707\mathrm{m}for{r}_1$

$$\begin{gathered} E_1=\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(3.0\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{(0.0707\mathrm{m}{)}^2}=5.4×{10}^3\mathrm{N}/\mathrm{C} \\ {\stackrel{→}{\mathrm{E}}}_1=\left({\mathrm{E}}_1\mathrm{³¦´Ç²õθ}\right)\hat{\mathrm{i}}+\left({\mathrm{E}}_1\mathrm{²õ¾\pm ²Ôθ}\right)(-\hat{\mathrm{j}}) \\ \mathrm{Substitute}5.4×{10}^3\mathrm{N}/\mathrm{C}\mathrm{for}{\mathrm{E}}_1\mathrm{and}45°\mathrm{for}\mathrm{θ} \\ {\stackrel{→}{\mathrm{E}}}_1=\left(\left(5.4×{10}^3\mathrm{N}/\mathrm{C}\right)\cos 45°\right)\hat{\mathrm{i}}+\left(\left(5.4×{10}^3\mathrm{N}/\mathrm{C}\right)\sin 45°\right)(-\hat{\mathrm{j}}) \\ =\left(3.82×{10}^3\mathrm{N}/\mathrm{C}\right)(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \\ {\mathrm{E}}_2=\frac{{\mathrm{kq}}_2}{{\mathrm{r}}_2^2} \\ \text{Substitute}9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\mathrm{for}\mathrm{k},3.0\mathrm{nC}\mathrm{for}{\mathrm{q}}_2\text{,}\mathrm{and}0.0707\mathrm{m}\mathrm{for}{\mathrm{r}}_2 \\ {\mathrm{E}}_2=\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(3.0\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{(0.0707\mathrm{m}{)}^2}=5.4×{10}^3\mathrm{N}/\mathrm{C} \\ {\stackrel{→}{\mathrm{E}}}_2=\left({\mathrm{E}}_2\mathrm{³¦´Ç²õθ}\right)(-\hat{\mathrm{i}})+\left({\mathrm{E}}_2\mathrm{²õ¾\pm ²Ôθ}\right)(-\hat{\mathrm{j}}) \\ \text{Substitute}5.4×{10}^3\mathrm{N}/\mathrm{C}\mathrm{for}{\mathrm{E}}_2\mathrm{and}45°\text{for}\mathrm{θ} \\ {\stackrel{→}{\mathrm{E}}}_2=\left(\left(5.4×{10}^3\mathrm{N}/\mathrm{C}\right)\cos 45°\right)(-\hat{\mathrm{i}})+\left(\left(5.4×{10}^3\mathrm{N}/\mathrm{C}\right)\sin 45°\right)(-\hat{\mathrm{j}}) \\ =\left(3.82×{10}^3\mathrm{N}/\mathrm{C}\right)(-\hat{\mathrm{i}}-\hat{\mathrm{j}}) \\ \stackrel{→}{\mathrm{E}}={\mathrm{E}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{E}}_{\mathrm{y}}\hat{\mathrm{j}} \\ \text{Substitute}\left(3.82×{10}^3\mathrm{N}/\mathrm{C}\right)(\hat{\mathrm{i}}-\hat{\mathrm{j}}{)}_{\text{for}}{\stackrel{→}{\mathrm{E}}}_{1\text{and}}\left(3.82×{10}^3\mathrm{N}/\mathrm{C}\right)(-\hat{\mathrm{i}}-\hat{\mathrm{j}}{)}_{\text{for}}{\stackrel{→}{\mathrm{E}}}_{2\text{in}}\stackrel{→}{\mathrm{E}}={\stackrel{→}{\mathrm{E}}}_1+{\stackrel{→}{\mathrm{E}}}_2 \\ \stackrel{→}{\mathrm{E}}=\left(3.82×{10}^3\mathrm{N}/\mathrm{C}\right)(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\left(3.82×{10}^3\mathrm{N}/\mathrm{C}\right)(-\hat{\mathrm{i}}-\hat{\mathrm{j}}) \\ =\left(7.64×{10}^3\mathrm{N}/\mathrm{C}\right)(-\hat{\mathrm{j}}) \end{gathered}$$