A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices.

The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.

Where,

Charge, $q=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}$

Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm

Distance of each vertex from centre O, d = 10 cm

Electric potential at point O,

$V=\frac{6 \times q}{4 \pi \epsilon_{0} d}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{C}^{-2} \mathrm{~m}^{-2}$

$\therefore V=\frac{6 \times 9 \times 10^{9} \times 5 \times 10^{-6}}{0.1}$

$=2.7 \times 10^{6} \mathrm{~V}$

Therefore, the potential at the centre of the hexagon is $2.7 \times 10^{6} \mathrm{~V}$.