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 f is the force due to the sixth charge and F due to the remaining five charges, thenF→+f→=0,  i.e.,  F→=−f→⇒|F→|=|f→|=14πε0q×qL2=14πε0qL2