Question

Five point charges, each of value $+q$ are placed on five vertices of a regular hexagon of side $L$. The magnitude of the force on a point charge of value $-q$ placed at the centre of the hexagon is $\frac{n{q}^2}{4\pi {ϵ}_0{L}^2}$. Then the value of n?

Answer 1

Among the five forces due to five charges, 2 pair (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. Also since it is a hexagon, distance between the centre and corner is equal to side length of the hexagon.
$F=\frac{1}{4\pi {ϵ}_0}\times \frac{q\times q}{L^2}$

$F=\frac{1}{4\pi {ϵ}_0}\times \frac{q^2}{L^2}$ (attarcative in nature)

On comparing with $\frac{n{q}^2}{4\pi {ϵ}_0{L}^2}$:
We get, $n=1$