Ncert Exercises & Solutions

A regular hexagon with sides of length \(10~\text{cm}\) has a charge of \(5~\mu\text C\) at each of its vertices. What is the electric potential at the center of the hexagon? (Assume the potential at infinity is zero.)
1. \(2.7 \times 10^6~\text V\)
2. \(4.5 \times 10^6~\text V\)
3. \(9.0 \times 10^6~\text V\)
4. \(1.8 \times 10^7~\text V\)

The given figure shows six equal amounts of charges, q, at the vertices of a regular hexagon.

Where,
Charge,

q = 5 μ C = 5 \times 10^{- 6} C

Side of the hexagon,
1 = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d =10 cm
Electric potential at point O,

V = \frac{1}{4 π ε_{0}} \cdot \frac{6 \times q}{d}

Where,
Where, E = Permitivity of free space and

\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}

∴ V = \frac{9 \times 10^{9} \times 6 \times 5 \times 10^{- 6}}{0.1} = 2.7 \times 10^{6} V

Therefore, the potential at the centre of the hexagon is 2.7 x

10^{6}

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