Question

Repeat Exercise 21.17 for ${\mathit{q}}_{\mathbf{3}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{C}$.

Short answer

Charge $q_3$is located at $+0.144m$

Step-by-step solution

Step 1:

$$\begin{gathered} q_1=3.0\mu C=3.0\times {10}^{-6}Cat{x}_1=0m \\ {q}_2=-5\mu C=-5.0\times {10}^{-6}Cat{x}_2=0.2m \\ {q}_3=+8\mu C=8.0\times {10}^{-6}C \\ \sum F_1=-7N \\ andK=8.989\times {10}^{9}N{m}^2/C^2 \end{gathered}$$

Step 2:

Here the third charge is positioned at the right of the first charge, as the net force exerted by the first is towards negative x-direction. As the electrical force between the first and second charge is attractive, therefore the force between the first and third charge is repulsive.

By using the Newton’s Second Law;

$\sum {\mathrm{F}}_1={\mathrm{F}}_{2\mathrm{on}1}-{\mathrm{F}}_{3\mathrm{on}1}$

Applying coulomb 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,

$$\begin{gathered} r_1=x_2-x_1,r_2=x_3-x_1 \\ if{x}_1=0 \\ {r}_1=x_{2}and{r}_2=x_3 \\ \sum F_1=K\frac{q_1{q}_2}{x_2^2}-K\frac{q_1{q}_3}{x_3^2} \end{gathered}$$

Step 3:

Solving for $x_3$

localid="1668414341308" $$\begin{gathered} \sum {\mathrm{F}}_1-\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{x}}_2^2}=-\mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{x}}_3^2} \\ {\mathrm{x}}_3^2\left[\sum {\mathrm{F}}_1-\frac{{\mathrm{Kq}}_1{\mathrm{q}}_2}{{\mathrm{x}}_2^2}\right]=-{\mathrm{Kq}}_1{\mathrm{q}}_3 \\ {\mathrm{x}}_3=\pm \sqrt{\frac{-{\mathrm{Kq}}_1{\mathrm{q}}_3}{\sum {\mathrm{F}}_1-\frac{{\mathrm{Kq}}_1{\mathrm{q}}_2}{{\mathrm{x}}_2^2}}} \end{gathered}$$

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 by the first charge is towards the left negative x-direction, the force exerted by the third charge is a repulsive force, and it is towards the right of the first charge.

So due to the positive root $x_3=+0.144m$

Hence, Charge $q_3$is located at localid="1668414346613" $+0.144m$.