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.
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The given figure shows six equal amount of charges, $q$, at the vertices of a regular hexagon.
Where, Charge, $q=5\, \mu\, C=5 \times 10^{-6} \,C$
Side of the hexagon, $l= AB = BC = CD = DE = EF = FA =10 \,cm$
Distance of each vertex from centre $O , d =10\, cm$
Electric potential at point $O$
$V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{6 \times q}{d}$
Where,
Where, $\varepsilon_{0}=$ Permittivity of free space and $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9}\, Nm ^{2} \,C ^{-2}$
$\therefore V=\frac{9 \times 10^{9} \times 6 \times 5 \times 10^{-6}}{0.1}=2.7 \times 10^{6}\, V$
Therefore, the potential at the centre of the hexagon is $2.7 \times 10^{6} \;V$
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