Two negative charges of unit magnitude and a positive charge q are placed along a straight line. At what position and for what value of q will the system be in equilibrium? Check whether it is a stable or neutral equilibrium. 

 from the two charges P and Q respectively, as shown in the figure below:

For equilibrium of q, the forces on it exerted by P and Q must be co-linear, equal and opposite.

Force on q by P 

       ${\mathrm{F}}_{\mathrm{qP}} = \frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0 {{\mathrm{r}}_1}^2} \mathrm{towards} \mathrm{p}$

Force on q by Q 

      ${\mathrm{F}}_{\mathrm{qQ}} = \frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{{\mathrm{r}}_2}^2}$ towards Q

$\therefore$               $\left|{\mathrm{F}}_{\mathrm{qP}}\right| = \left|{\mathrm{F}}_{\mathrm{qQ}}\right| \mathrm{or} \frac{\mathrm{q}}{{{\mathrm{r}}_1}^2} = \frac{\mathrm{q}}{{{\mathrm{r}}_2}^2}$
$\therefore$                     ${\mathrm{r}}_1 ={\mathrm{r}}_2 = \mathrm{r}$

Hence charge q should be equidistant from P and Q.

For the system to be in equilibrium, the charges P and Q must also be in equilibrium.
Now,
$$\begin{gathered} {\mathrm{F}}_{\mathrm{pq}} = \mathrm{force} \mathrm{on} \mathrm{P} \mathrm{by} \mathrm{q} = \frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} (\mathrm{towards} \mathrm{q}) \\ {\mathrm{F}}_{\mathrm{PQ}} = \mathrm{force} \mathrm{on} \mathrm{Q} \mathrm{by} \mathrm{P} =\frac{1}{4{\mathrm{\pi \epsilon }}_0(2\mathrm{r}{)}^2}(\mathrm{away} \mathrm{from} \mathrm{P} \mathrm{and} \mathrm{away} \mathrm{from} \mathrm{q}) \end{gathered}$$

Since F_pq and F_PQ are oppositely directed along the same line, we have, for equilibrium,
           
               $\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} = \frac{1}{4{\mathrm{\pi \epsilon }}_0\left(4{\mathrm{r}}^2\right)}$
or                   $\mathrm{q} = \frac{1}{4}$
Similarly for the equilibrium of Q, we would get
$\mathrm{q} = \frac{1}{4}.$ 
Thus $\mathrm{q} = \frac{1}{4}$ in magnitude of either charge P or Q.

Stability: A slight displacement of q towards P increases the magnitude of F_qP and decreases the magnitude of F_qQ. Consequently, the displacement of q is increased. Thus the three charges are no longer in equilibrium. Hence the original equilibrium is unstable for displacement along the axis on which the charges are located. For a displacement of q along a direction normal to the line PQ, the resultant of the two forces of attraction F_qp and F_qQ will bring the charge q back to its original position. Thus the equilibrium is stable for displacement in the vertical direction.