A point charge 'Q' is placed at the centre of the line joining two equal charges '+q' and '+q'. The value of 'Q' when the system is in equilibrium is

Let the two equal charges $+q$ be placed at a distance $r$ from each other.

The charge $Q$ is placed at the centre, so its distance from each $+q$ charge is $\dfrac{r}{2}$.

For the system to be in equilibrium, the net force on each charge must be zero.

The net force on the central charge $Q$ is zero for any value of $Q$ due to symmetry.

For the charge $+q$ to be in equilibrium, the force exerted by the other $+q$ charge must be equal and opposite to the force exerted by $Q$.

The force on $+q$ due to the other $+q$ is $F_1 = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2}$.

The force on $+q$ due to $Q$ is $F_2 = \dfrac{1}{4\pi\epsilon_0} \dfrac{qQ}{(\dfrac{r}{2})^2} = \dfrac{1}{4\pi\epsilon_0} \dfrac{4qQ}{r^2}$.

Equating the sum of forces to zero:

$F_1 + F_2 = 0$

$\dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2} + \dfrac{1}{4\pi\epsilon_0} \dfrac{4qQ}{r^2} = 0$

$q^2 + 4qQ = 0$

$Q = \dfrac{-q}{4}$

Answer: $\dfrac{-q}{4}$