A charge ‘q’ is placed at the centre of the line joining two equal charges ‘Q’. The system of the three charges will be in equilibrium if ‘q’ is equal to

The situation is as shown in the figure.

Let two equal charges Q each placed at points A and B at a distance r apart. C is the centre of A B where charge q is placed. For equilibrium, net force on charge Q=0

$$\begin{gathered} \therefore \frac{1}{4\pi {\epsilon }_0}\frac{QQ}{r^2}+\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{(r/2{)}^2}=0 \\ \frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{r^2}=-\frac{1}{4\pi {\epsilon }_0}\frac{4Qq}{r^2} \\ \text{ or } Q=-4q \\ \text{ or } q=-\frac{Q}{4} \end{gathered}$$