Solution:

Given, two equal charges Q are placed at a distance d from each other.

A charge q is placed at the center of the line joining the two charges.

To find: the value of q for which the system of the three charges will be in equilibrium.

Let us assume that the charges Q are positive and q is negative.

We know that the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Hence, the magnitude of the electrostatic force between the two charges Q is given by:

F = kQ²/d²

where k is the Coulomb's constant.

Now, let us consider the forces acting on the charge q.

The force due to the charge Q on the left is towards the right and the force due to the charge Q on the right is towards the left. These two forces cancel out each other as they are equal in magnitude and opposite in direction.

Hence, the net force on the charge q is zero.

For equilibrium, the net force on the charge q should be zero.

Hence, the force due to the charges Q should be balanced by the force due to the charge q.

The force due to the charge q on the left charge Q is towards the left and the force due to the charge q on the right charge Q is towards the right.

Hence, the magnitude of the force due to the charge q on each charge Q is given by:

F' = kqQ/(d/2)²

where d/2 is the distance between the charge q and each charge Q.

Since the net force on the charge q is zero, we have:

2F' = F

kqQ/(d/2)² = kQ²/d²

Simplifying, we get:

q = -Q/4

Hence, the value of q for which the system of the three charges will be in equilibrium is -Q/4.