A regular hexagon of side 10 cm has a charge 5μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Let O be the centre of the hexagon
It follows that the point O, when joined to the ends of a side of the hexagon forms an equilateral triangle.
∴AO=BO=CO=DO=EO=FO=10cm=0.1 m
Since at each comer of the hexagon, a charge of 5pc i.e., 5×10−6C is placed, total electric potential at point o due to the charges at the six comers,
V=6×(14π∈0×qr)
=6×9×109×5×10−60.1
=2.7×106V.