Three point charges 'q' each are placed at the corners of an equilateral triangle of side length 'a' . Find the value of charge Q to be placed at the centre of the triangle for which the system remains in tge state of equilibrium.

Draw a triangle with +q charge on each vertex. Now the point charge places at the top most vertex (let's call it A, and the other two vertices B and C) experiences a repulsive force due to the fact that both the other charges are +ve as well. Thus it will experience two repulsive forces, and the net force (vector addition) can be calculated. The angle between the two forces is 60 as the triangle is equilateral. Thus Fnet = (F² + F² + 2F²cos(60))^(½) gives us Fnet = √3(F).

F is k(q)(q)/l²

Now due to the fact that this force is repulsive and it is directed away from the triangle, the only way the charge will remain stationary is if the charge Q to be placed at the centroid is -ve. By using Pythagoras theorem and simple properties of centroid we get the distance between the charge +q at A and the charge -Q at the centroid to be l/√3 .

Now , force between +q and -Q

F2 = k(q)(-Q)/(l/√3)1

Since the forces must balance each other in order for the system to remain stationary

Fnet = F2

Solving, we get Q= -q/√3