Six charges are placed at the vertices of a regular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure as a function of distance x from point O is (assume x >>a)

Two opposite charges situated at opposite vertices forms a dipole and point lies on bisector line of dipole for one dipole.E→=−Kp→x3Direction of field due to these three dipoles is in a horizontal plane at distance x from plane of hexagon and parallel to it, mutually having angle 60° with each other.E0=E+E=2E=2Q(2a)4πε0⋅x3=Qaπε0x3