Two positive and equal charges are fixed at certain distance. A third charge (small) is placed in between the line joining of the two charges and it experiences zero net force due to the other two.

The correct option is A The equilibrium is stable if small charge is positive

At equilibrium, the forces $F_1$ and $F_2$ are equal.

So, to check the nature of equilibrium, let us displace the particle slightly towards right from the equilibrium position. Let, the particle is diplaced by $x$.


Now, from the coulomb's law,

$F_1=\frac{KQq}{{(\frac{r}{2}+x)}^2}$

$F_2=\frac{KQq}{{(\frac{r}{2}-x)}^2}$

Since, $F_2>F_1$ so, $F_{\text{net}}$ is towards mean point so its a stable equilibrium if positive charge is placed.

If negative charge $-q$ is placed between the charges and is displaced slightly towards right , the net force on it will be away from origin i.e. towards point $2$ and the particle will not return to its equilibrium position , hence unstable equilibrium.

Hence, option (a) is correct answer.