A charge $$q$$ is placed at the centre of the line joining two equal charges $$Q$$. Show that the system of three charges will be in equilibrium if $$q = -\dfrac {Q}{4}$$.
Charge $$q$$ is in equilibrium since charges $$A$$ and $$B$$ exert equal and opposite forces on it.
For equilibrium of charge $$Q$$ at $$B$$;
$$F_{BC} + F_{AB} = 0$$
$$\Rightarrow \dfrac {1}{4\pi \epsilon_{0}} \dfrac {qQ}{(l/2)^{2}} + \dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q . Q}{l^{2}} = 0$$
$$\Rightarrow \dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q}{l^{2}} (4q + Q) = 0 \Rightarrow q = -\dfrac {Q}{4}$$.