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

Explanation

If we place a charge `q` at the sixth vertex of the regular hexagon, then the net force on the charge `-q` placed at the center of hexagon will be zero due to symmetry.
The force on charge `q` placed on the sixth vertex balances the net force on charge `-q` due to the other live charges placed at the five `-q` due to charge `q` placed at the sixth vertex will be.
`F=(1 qxxq)/(4piepsilon_(0)L^(2))`

where `l` is the distnce of the center of hexagon from any vertex (directed from `O` to `C`). The magnitude of force on the point charge of value `-q` coulomb placed at the center of the hexagon is `(1//4pispsilon_(0)kq^(2)//L^(2))` directed from `O` to `F`.