Five point charges, each +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.

Explanation

The forces due to the charges placed diagonally opposite at the vertices of hexagon, on the charge –q cancel in pairs. Hence net force is due to one charge only.
Net force, `|vec(F)|=(1)/( 4 pi epsilon_(0))(q^(2))/(l^(2))`