A regular hexagon of side 12 cm has a charge of 6 μC at each of its vertices. What will be the potential at the centre of the hexagon?

$2.7 × 10^6 V$

The correct answer is Option (4) → $2.7 × 10^6 V$

For a regular hexagon, distance from centre to any vertex is equal to its side.

Given: side $r = 12\ \text{cm} = 0.12\ \text{m}$, charge at each vertex $q = 6\ \mu C = 6 \times 10^{-6}\ \text{C}$.

Potential at centre due to one charge: $V_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$

Total potential at centre (6 vertices): $V = 6V_1 = \frac{6}{4 \pi \varepsilon_0} \frac{q}{r}$

Substituting $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9$ :

$V = 9 \times 10^9 \times \frac{6 \times 6 \times 10^{-6}}{0.12}$

$V = 9 \times 10^9 \times 3 \times 10^{-4}$

$V = 2.7 \times 10^6\ \text{V}$

Therefore, $V = 2.7 \times 10^6\ \text{V}$