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?
Answer & explanation
Correct answer: option 4
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, $V = 2.7 \times 10^6\ \text{V}$