# 2.2) A regular hexagon of side 10 cm has a charge $5\mu C$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Answers (1)

The electric potential at O due to one charge,

$V_1 = \frac{q}{4\pi\epsilon_0 r}$

q = 5 × 10-6 C

r = distance between charge and O = 10 cm = 0.1 m

Using the superposition principle, each charge at corners contribute in the same direction to the total electric potential at the point O.

$V = 6\times\frac{q}{4\pi\epsilon_0 r}$

$\implies V = 6\times\frac{9\times10^9 Nm^2C^{-2}\times5\times10^{-6}C}{0.1m}$

= $2.7 \times 10^6 V$

Therefore the required potential at the centre is $2.7 \times 10^6 V$

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