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 coulomb placed at the center of the hexagon is

If there had been a sixth charge +q at the remaining vertex of hexagon, force due to all the six charges on −q at O will be zero.Now if/is the force due to the sixth charge and F due to the remaining five charges, then F→+f→=0,  i.e.,  F→=−f→|F→|=|f→|=14πε0q×qL2=14πε0(qL)2