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# Three point charges are arranged on a line. Charge $q _ { 3 } = + 5.00 n C$ and is at the origin. Charge $q _ { 2 } = - 3.00 n C$ and is at x = +4.00 cm. Charge $q_1$ is at x = +2.00 cm. What is $q_1$ (magnitude and sign) if the net force on $q_3$ is zero?

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#### Given

We are given three charges separated with a distance as shown below. Where $q_{2}$ = - 3.0 nC and $q_{3}$ = +5.0 nC. The charges $q_{1}$ and $q_{2}$ exerted a force on charge $q_{3}$ where the net force that acts on $q_{3}$ is zero.

#### Solution

In this problem $q_{3}$ is affected by two forces $F_{1 \,\text{on}\, 3}$ and $F_{2 \,\text{on}\, 3}$ and the net force is

$F_{\text{net}} = F_{1 \,\text{on}\, 3} + F_{2 \,\text{on}\, 3}$

Also, we are given the charge $q_{1}$ in the positive direction of $x$ which means there is a repulsive force between $q_{3}$ and $q_{1}$ and we can predict that the charge $q_{1}$ is $\textbf{positive}$. Also, as the net force is zero, this means

$F_{1 \,\text{on}\, 3} = F_{2 \,\text{on}\, 3}$

Now we can apply Coulomb's law to find the magnitude of $q_{1}$ by the next steps

$\begin{gather*} F_{1 \,\text{on}\, 3} = F_{2 \,\text{on}\, 3} \\ \cancel k \dfrac{|q_{1} \cancel q_{3}|}{r_{1}^{2}} = \cancel k \dfrac{|q_{2}\cancel q_{3}|}{r_{2}^{2}}\\ \dfrac{|q_{1}|}{r_{1}^{2}} = \dfrac{|q_{2}|}{r_{2}^{2}} \tag{Solve for q_{1}}\\ |q_{1}| = |q_{2}| \left(\dfrac{r_{1}^{2}}{r_{2}^{2}}\right) \tag{1} \end{gather*}$

Now we can plug our values for $q_{2}, r_{1}$ and $r_{2}$ into equation (1) to get $q_{1}$

\begin{align*} |q_{1}| &= |q_{2}| \left(\dfrac{r_{1}^{2}}{r_{2}^{2}}\right)\\ &= |-3.0 \,\text{nC}| \left(\dfrac{(2.0 \,\text{cm})^{2}}{(4.0 \,\text{cm})^{2}}\right)\\ &= \boxed{+ 0.75 \,\text{nC}} \end{align*}

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