Given data:
Charge at the center = q
Equal charges = Q
The system of the three charges will be in equilibrium.

Explanation:
The charge q placed at the center will experience an equal and opposite force from the charges Q. This is because of the Coulomb's law. According to Coulomb's law, the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Now let's assume that the distance between the charges Q and the center charge q is ‘r’. So the electrostatic force between the charges will be:

F = k(Qq/r^2)

Where k is the Coulomb's constant.

As the charges Q are at equal distance from the center charge q, they will experience equal and opposite force. Therefore, the net force acting on the system will be zero.

∴ k(Qq/r^2) + k(Qq/r^2) = 0

On solving the above equation, we get:

q = -Q/2

This means that for the system to be in equilibrium, the charge q should be negative and equal to half the magnitude of the charges Q.

However, this negative charge cannot exist at the center of the charges Q as it will violate the principle of superposition. Therefore, the charge q must be positive and equal to Q/4. This will create a stable equilibrium configuration for the system.

Hence, the correct option is B) Q/4.

Solution:

Given: Three charges q, Q, Q are placed at the vertices of an equilateral triangle.

To find: Charge q required to be placed at the center of the triangle so that the system remains in equilibrium.

Approach: For the system to be in equilibrium, the net force acting on each charge should be zero. We will use Coulomb's law to find the force acting on the center charge q due to the other two charges Q.

Let's assume that the side of the equilateral triangle is of length ‘a’.

The distance between the center charge q and the charges Q is ‘a/2’.

Using Coulomb's law, the force acting on the center charge q due to each of the charges Q is given by:

F = k (qQ / r²), where k = 1/4πε₀ is the Coulomb's constant and r = a/2 is the distance between the charges.

The direction of the forces will be along the lines joining the charges.

Since the charges Q are at equal distance from q, the forces due to them will be equal in magnitude and opposite in direction. Therefore, the net force due to them will be zero.

Hence, the net force acting on the center charge q is due to the charge Q only.

The forces due to the charges Q are making an angle of 120° with each other. Therefore, their horizontal components cancel out and the net force acting on the center charge q is along the vertical direction.

So, the net force acting on the center charge q is given by:

Fnet = 2F sin(60°) = (2kqQ/a²) x (sqrt(3)/2)

For equilibrium, Fnet = 0. Therefore,

q = (a²Q) / (4kQ sin(60°))

q = (a²Q) / (4kQ x sqrt(3)/2)

q = (a²Q) / (2kQ x sqrt(3))

q = (1/2) x (a²Q) / (4πε₀ x a² x sqrt(3))

q = Q / (4πε₀ x sqrt(3))

q = Q/4

Hence, the required charge q is Q/4.