Question

Five point charges, each of charge +q are placed on five vertices of a regular hexagon of side 'l '. Find the magnitude of the resultant force on a charge -q placed at the centre of the hexagon.

Answer 1

Effective force = 0

as all the forces cancel out each other ∴ Net Force on charge = 0

Answer 2

All the five charges, each of charge +q are placed on 5 vertices of a hexagon. Now if put a charge -q at the center of the hexagon. Force felt by -q charge will be F_A, F_B, F_C, F_D and F_E by charges present on vertices A, B, C, D, and E respectively. As all the charges are the same and are separated by equal distances from -q charge, so magnitudes of all the forces felt by charge -q is the same.

Hence force F_A  will cancel the force F_D  &  force F_B  will cancel the force  F_E, as they are working in opposite directions. So the net force working on the charge -q will be only F_C.

`"F"_"net" = "F"_"C" = 1/(4pi ε_@) "q.q"/"l"^2`... (Towards C)

OR `| "F"_"net"| = 1/(4pi ε_@) "q.q"/"l"^2`