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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