Two positive charges $q_1$ and $q_2$ are located at points 1 and 2 with radius vectors ${\overrightarrow{r}}_1$ and ${\overrightarrow{r}}_2$. Find a negative charge $q_3$ and the radius vector ${\overrightarrow{r}}_3$ of a point 3 at which it has to be placed for the force acting on each of the three charges to be equal to zero.

Before we start the problem, we must keep in mind that the radius vector of any point P is actually the position vector (P V) of the point P which is the vector drawn from the origin to the point P.

So, let the origin be at O (see figure)

For equilibrium of $-q_3$, we have

$$\begin{gathered} {\overrightarrow{F}}_{31}+{\overrightarrow{F}}_{32}=\overrightarrow{0} \\ \Rightarrow \frac{q_1{q}_3}{4\pi {\epsilon }_0{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right|}^3}\left({\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right)+\frac{q_2{q}_3}{4\pi \\ {\epsilon }_0{\left|{\overrightarrow{r}}_2-{\overrightarrow{r}}_3\right|}^3}\left({\overrightarrow{r}}_2-{\overrightarrow{r}}_3\right)=\overrightarrow{0} \end{gathered}$$

Also, from figure, we see that

$$\begin{gathered} {\hat{r}}_{13}=-{\hat{r}}_{23} \\ \Rightarrow \frac{{\overrightarrow{r}}_1-{\overrightarrow{r}}_3}{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right|}=-\frac{{\overrightarrow{r}}_2-{\overrightarrow{r}}_3}{\left|{\overrightarrow{r}}_2-{\overrightarrow{r}}_3\right|} \end{gathered}$$

Put (2) in (1), we get

$\frac{q_1}{{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right|}^2}=\frac{q_2}{{\left|{\overrightarrow{r}}_2-{\overrightarrow{r}}_3\right|}^3}$

$\Rightarrow \pm \sqrt{q_1}\left({\overrightarrow{r}}_2-{\overrightarrow{r}}_3\right)=\sqrt{q_2}\left({\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right)$

Taking positive sign, we get

$$\begin{gathered} \sqrt{q_1}{\overrightarrow{r}}_2-\sqrt{q_2}{\overrightarrow{r}}_1=\left(\sqrt{q_1}-\sqrt{q_2}\right){\overrightarrow{r}}_3 \\ \Rightarrow {\overrightarrow{r}}_3=\frac{\sqrt{q_1}{\overrightarrow{r}}_2-\sqrt{q_2}{\overrightarrow{r}}_1}{\sqrt{q_1}-\sqrt{q_2}} \end{gathered}$$

This must be REJECTED as it will lie outside $q_1$ and $q_2$

Taking negative sign, we get

$\sqrt{q_1}{\overrightarrow{r}}_2+\sqrt{q_2}{\overrightarrow{r}}_1=\left(\sqrt{q_1}+\sqrt{q_2}\right){\overrightarrow{r}}_3$

$\Rightarrow {\overrightarrow{r}}_3$=$\frac{q1}{r\to 2}$ + $\sqrt{q2}{\overrightarrow{r}}_1$$\sqrt{q_1}$+$\sqrt{q_2}$

Clearly, this position vector or radius vector will lie on the line joining $q_1$ and $q_2,$ between them.

So, ${\overrightarrow{r}}_3=\frac{\sqrt{q_1}{\overrightarrow{r}}_2+\sqrt{q_2}{\overrightarrow{r}}_1}{\sqrt{q_1}+\sqrt{q_2}}$

For $q_1$ to be in equilibrium, let charge placed on 3 be $q_3$, then

$$\begin{gathered} {\overrightarrow{F}}_{12}+{\overrightarrow{F}}_{13}=\overrightarrow{0} \\ \Rightarrow \frac{q_1{q}_2}{4\pi {\epsilon }_0{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_2\right|}^3}\left({\overrightarrow{r}}_1-{\overrightarrow{r}}_2\right)+\frac{q_1{q}_3}{4\pi {\epsilon }_0{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right|}^3}\left({\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right)=\overrightarrow{0} \end{gathered}$$

While writing the above expression, we have not taken any nature of $q_3.$The answer will have its sign attached to $q_3.$ So, as of now $q_3$ is assumed to be positive and then lets see what happens. Also, as per the assumption made

$$\begin{gathered} {\hat{r}}_{12}={\hat{r}}_{13} \\ \Rightarrow \frac{q_2}{{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_2\right|}^2}=-\frac{q_3}{{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right|}^2} \left\{\because \frac{{\overrightarrow{r}}_1-{\overrightarrow{r}}_2}{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_2\right|}=\frac{{\overrightarrow{r}}_1-{\overrightarrow{r}}_3}{\left|{\overrightarrow{r}}_1-{\overrightarrow{r}}_3\right|}\right\} \\ \Rightarrow q_3=-q_2{\left|\frac{{\overrightarrow{r}}_1-{\overrightarrow{r}}_3}{{\overrightarrow{r}}_1-{\overrightarrow{r}}_2}\right|}^2 \\ \text{ Now, }{\overrightarrow{r}}_1-{\overrightarrow{r}}_3={\overrightarrow{r}}_1-\left(\frac{\sqrt{q_1}{\overrightarrow{r}}_2+\sqrt{q_2}{\overrightarrow{r}}_1}{\sqrt{q_1}+\sqrt{q_2}}\right) \\ \Rightarrow {\overrightarrow{r}}_1-{\overrightarrow{r}}_3=\frac{\left.\sqrt{q_1}{\overrightarrow{r}}_1+\sqrt{q_2}\right)}{\sqrt{r_1}-\sqrt{q_1}+\sqrt{q_1}{\overrightarrow{r}}_2-\sqrt{q_2}} \\ \Rightarrow {\overrightarrow{r}}_1-{\overrightarrow{r}}_3=\frac{\sqrt{q_1}\left({\overrightarrow{r}}_1-{\overrightarrow{r}}_2\right)}{\sqrt{q_1}+\sqrt{q_2}} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\left|\frac{{\overrightarrow{r}}_1-{\overrightarrow{r}}_3}{{\overrightarrow{r}}_1-{\overrightarrow{r}}_2}\right|}^2=\frac{q_1}{{\left(\sqrt{q_1}+\sqrt{q_2}\right)}^2} \\ \text{ Using (3), we get } \\ \Rightarrow q_3=-\frac{q_1{q}_2}{{\left(\sqrt{q_1}+\sqrt{q_2}\right)}^2} \end{gathered}$$