Charge of $5 μ C$ is at each of the vertices of a regular hexagon of side $10 c m$ . Determine the electric potential at the centre of the regular hexagon.

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Solution

For Hexagonal shape
$O A = O B - O C = O D = O E = O F = 10 c m = 0.1 m$
Potential at O
$V_{O} = V_{A} + V_{B} + V_{C} + V_{D} + V_{E} + V_{F} = 6 V_{A}$
Due to equal of charge and distance
$\begin{matrix} V_{A} & = V_{B} = V_{C} = V_{D} = V_{E} = V_{F} ∴ V_{O} & = 6 \times \frac{1}{4 π ε_{0}} \frac{q}{O A} = 6 \times 9 \times 10^{9 \cdot} \times \frac{5 \times 10^{- 6}}{0.1} = 27 \times 10^{5} or V_{O} & = 2.7 \times 10^{6} volt \end{matrix}$

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