Question

8. Three point-like charges are placed at the comers of a rectangle as shown in the figure, a- 260 cm and b- 62.0 cm. Find the magnitude of the electric force exerted on the charge qì. Let q,--1.40pC, q,-+2.60 pC, q,--4.00 μC. iOq 93

Answer 1

A point charge $q_1$ is located at point $\vec{r}_1$. Another point charge $q_2$ is located at point $\vec{r}_2$.

Electric force on $q_1$ due to $q_2$ is  $\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2(\vec{r}_1-\vec{r}_2)}{|\vec{r}_1-\vec{r}_2|^3}$

The origin is at the vertex of rectangle where no charge is present.

Charges and their locations are

$q_1=-1.40\,\mu C$ located at $\vec{r}_1=(a\hat{j})$

$q_2=2.60\,\mu C$ located at $\vec{r}_2=(b\hat{i}+a\hat{j})$

$q_3=-4.00\,\mu C$ located at $\vec{r}_3=(b\hat{i})$

$a=26.0\,cm=0.26\,m$

$b=62.0\,cm=0.62\,m$

Force on charge $q_1$ due to charges $q_2,q_3$ is

$\vec{F}_{1,net}=\vec{F}_{12}+\vec{F}_{13}$

$\vec{F}_{1,net}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2(\vec{r}_1-\vec{r}_2)}{|\vec{r}_1-\vec{r}_2|^3}+\frac{1}{4\pi\epsilon_0}\frac{q_1q_3(\vec{r}_1-\vec{r}_3)}{|\vec{r}_1-\vec{r}_3|^3}$

$\vec{F}_{1,net}=\frac{q_1}{4\pi\epsilon_0}\left ( \frac{q_2(\vec{r}_1-\vec{r}_2)}{|\vec{r}_1-\vec{r}_2|^3}+\frac{q_3(\vec{r}_1-\vec{r}_3)}{|\vec{r}_1-\vec{r}_3|^3} \right )$

$\vec{F}_{1,net}=\frac{q_1}{4\pi\epsilon_0}\left ( \frac{q_2(a\hat{j}-b\hat{i}-a\hat{j})}{|a\hat{j}-b\hat{i}-a\hat{j}|^3}+\frac{q_3(a\hat{j}-b\hat{i})}{|a\hat{j}-b\hat{i}|^3} \right )$

$\vec{F}_{1,net}=\frac{q_1}{4\pi\epsilon_0}\left ( \frac{q_2(-b\hat{i})}{|-b\hat{i}|^3}+\frac{q_3(a\hat{j}-b\hat{i})}{|a\hat{j}-b\hat{i}|^3} \right )$

$\vec{F}_{1,net}=\frac{q_1}{4\pi\epsilon_0}\left (- \frac{q_2\hat{i}}{b^2}+\frac{q_3(a\hat{j}-b\hat{i})}{(a^2+b^2)^{3/2}} \right )$

$\vec{F}_{1,net}=-\frac{1.40*10^{-6}}{4\pi\epsilon_0}\left ( -\frac{2.60*10^{-6}\hat{i}}{0.62^2}-\frac{4.00*10^{-6}(0.26\hat{j}-0.62\hat{i})}{(0.26^2+0.62^2)^{3/2}} \right )$

$\vec{F}_{1,net}=8.99*10^9*{1.40*10^{-6}}\left ( \frac{2.60*10^{-6}\hat{i}}{0.62^2}+\frac{4.00*10^{-6}(0.26\hat{j}-0.62\hat{i})}{(0.26^2+0.62^2)^{3/2}} \right )$

$\vec{F}_{1,net}=(-0.017581\,\hat{i}+0.043062\,\hat{j})\,N$

Magnitude of the net force $|\vec{F}_{1,net}|=((-0.017581)^2+0.043062^2)^{0.5}=0.0465\,N$