Question

if five identical charged particles of charge q each are kept at vertices of , regular hexagon,then force acting on charge '-q' kept at centre of hexagon is :(Here r is the distance of vertex from centre of hexagon

Answer 1

Answer:

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=

4πϵ

Answer 2

Answer:

-K(q^(2))/(r^(2))

Explanation:

Since the total force at the centre will be due to 5 charges out of them 2 pairs will cancel out eachother's forces and the only one charge that remains will exert a force on the central charge who's magnitude will be $Kq^{2}/r^{2}$