Six charges each equal to +q are placed at the corners of regular hexagon of side a. The electric potential at the point where the diagonals intersect is ‘V’ and the electric field at that point is E. Then

The distance of the point of intersection of diagonals = a
Potential due to each charge $=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{q}}{\mathrm{a}}$
$\therefore$Total potential $=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{6\mathrm{q}}{\mathrm{a}}$
The net electric field at that point of intersection of diagonals is zero because the electric field at this point due to equal charges at opposite corners will cancel each other in pairs so, 2 & 3 are correct.