A regular hexagon

Let O be the center of the hexagon. It contains the charges at all its 6 vertices, each charge = + 5 μC = 5 × 10^-6 C. The side of the hexagon is 10 cm = 0.1 m

It follows that the point O, when joined to the two ends of a side of the hexagon forms an equilateral triangle

Electric potential at O due to one charge,

Here, q = amount of charge = 5 × 10^-6 C

r = distance between charge and O = 0.1 m

ε₀ = absolute permittivity of free space

Since at the each of the hexagon, a charge of 5 μC (5 × 10^-6 C) is placed, total electric potential at the point O due to the charges at the six corners,

⇒

⇒

⇒ V₁ = 2.7 × 10⁶ V