A regular hexagon of side 10 cm has a charge \(1 \mu \mathrm{C}\) at each of its vertices. The potential at the centre of hexagon is \(\left[\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right.\) SI unit \(]\)

Potential at the centre of hexagon
\(V_0 =V_A+V_B+V_C+V_D+V_E+V_F\)
\(=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{O A}+\frac{q}{O B}+\frac{q}{O C}+\frac{q}{O D}+\frac{q}{O E}+\frac{q}{O F}\right]\)
For a hexagon, distance from centre to vertices is equal to length of side.
\(\begin{aligned}
\therefore \mathrm{OA}=\mathrm{OB}=\mathrm{OC}=\mathrm{OD}=\mathrm{OE}=\mathrm{OF} & =10 \mathrm{~cm} \\
& =0.1 \mathrm{~m}
\end{aligned}\)
\(\therefore \mathrm{V}_\theta =\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{0.1}(6)\)
\(=9 \times 10^9 \times \frac{\left(1 \times 10^{-6}\right) 6}{0.1}\)
\(=5.4 \times 10^5 \mathrm{volt}\)