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

$-\frac{Q}{4}$

Since, q is at the centre of two charges Q and Q, net force on it is zero, whatever the magnitude and sign of charge on it. For the equilibrium of Q, q should attract it. Simultaneously these attractions and repulsions should be equal.

$\frac{1}{4 \pi \varepsilon_0} \frac{Q Q}{r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{Q q}{(r / 2)^2}$

or    $q=\frac{Q}{4}$

∴ correct option is (b)