Question

Five point charges, each of value +q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value -q placed at the centre of the hexagon?

Answer 1

Among the five forces on five charges,2pairs(four) of forces are cancelled with each other since they are equal in magnitude and opposite in direction.

The resultant force is due to a single force given below.

$F=\frac{1}{4\pi {ϵ}_{\circ }}\times \frac{q\times q}{l^2}$

$F=\frac{1}{4\pi {ϵ}_{\circ }}\times \frac{q^2}{l^2}$

(attractive in nature)