Question

Three-point charges are arranged along the x-axis. Charge${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{C}$is at the origin, and charge${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{C}$ is at x = 0.200 m. Charge_{${\mathit{q}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{C}$}. Where is${\mathit{q}}_{\mathbf{3}}$ located if the net force on_{${\mathit{q}}_{\mathbf{1}}$}is 7.00 N in the -x-direction?

Short answer

Charge$q_3$is located at$-0.144m$..

Step-by-step solution

Step 1:

Given data:

$$\begin{gathered} {\mathrm{q}}_1=3.0\mathrm{\mu C}=3.0\times {10}^{-6}\mathrm{C}\mathrm{at}{\mathrm{x}}_1=0\mathrm{m} \\ {\mathrm{q}}_2=-5\mathrm{\mu C}=-5.0\times {10}^{-6}\mathrm{C}\mathrm{at}{\mathrm{x}}_2=0.2\mathrm{m} \\ {\mathrm{q}}_3=-8\mathrm{\mu C}=-8.0\times {10}^{-6}\mathrm{C} \end{gathered}$$

Net force is

$\sum _{}{F}_1=-7N$

and$K=8.989\times {10}^{9}Nm/C^2$

Step 2:

As the third charge is positioned to the left of the first charge, thus the charge is directed towards x-direction. And electrical force between the first and second charge is an attractive force and also between the first and third charge.

So, to find net force on first charge by using Newton's Second Law.

$\sum {\mathrm{F}}_1=\left|\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}_1^2}\right|-\left|\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{r}}_2^2}\right|$

As $r_1$is the distance between two charges,

role="math" localid="1668158352531" $$\begin{gathered} {\mathrm{r}}_1={\mathrm{x}}_2-{\mathrm{x}}_1,{\mathrm{r}}_2={\mathrm{x}}_3-{\mathrm{x}}_1 \\ \mathrm{if}{\mathrm{x}}_1=0 \\ {\mathrm{r}}_1={\mathrm{x}}_2\mathrm{and}{\mathrm{r}}_2={\mathrm{x}}_3 \\ \sum {\mathrm{F}}_1=\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{x}}_2^2}-\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{x}}_3^2} \end{gathered}$$

Step 3:

Solving for $x_3$

$$\begin{gathered} \sum {\mathrm{F}}_1-\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{x}}_2^2}=-\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{x}}_3^2} \\ {\mathrm{x}}_3^2\left[\sum {\mathrm{F}}_1-\frac{{\mathrm{Kq}}_1{\mathrm{q}}_2}{{\mathrm{x}}_2^2}\right]=-{\mathrm{Kq}}_1{\mathrm{q}}_3 \\ {\mathrm{x}}_3=\pm \sqrt{\frac{-{\mathrm{Kq}}_1{\mathrm{q}}_3}{\sum {\mathrm{F}}_1-\frac{{\mathrm{Kq}}_1{\mathrm{q}}_2}{{\mathrm{x}}_2^2}}} \end{gathered}$$

By putting all the values,

$$\begin{gathered} {\mathrm{x}}_3=\pm \sqrt{\frac{+8.988\times {10}^9\times 3\times {10}^{-6}\times 8\times {10}^{-6}}{-7.0-\left(\frac{8.988\times {10}^9\times 3\times {10}^{-6}\times 5\times {10}^{-6}}{(0.2{)}^2}\right)}} \\ {\mathrm{x}}_3=\pm 0.144\mathrm{m} \end{gathered}$$

As the net force exerted on the first charge is to the left, negative x-direction,

Thus, using the negative root,

$x_3=-0.144m$

Hence, Charge $q_3$is located at $x_3=-0.144m$.