A regular hexagon of side 10 cm has a charge 5 µC 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×(4π∈01×rq) =6×9×109×0.15×10−6 =2.7×106V.
