A regular hexagon of side 5 , cm has a charge | vedclass

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(C) Given, the side length of the regular hexagon is $r = 5 \, cm = 5 	imes 10^{-2} \, m$.In a regular hexagon, the distance from each vertex to the

Vedclass · Accepted Answer

$1.08 	imes 10^7 \, V$

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(C) Given, the side length of the regular hexagon is $r = 5 \, cm = 5 	imes 10^{-2} \, m$.In a regular hexagon, the distance from each vertex to the centre is equal to the side length of the hexagon. Thus, $r = 5 	imes 10^{-2} \, m$.The charge at each vertex is $q = 10 \, \mu C = 10 	imes 10^{-6} \, C = 10^{-5} \, C$.The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$, where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.The potential due to one charge at the centre is:$V_1 = \frac{9 	imes 10^9 	imes 10^{-5}}{5 	imes 10^{-2}} = \frac{9}{5} 	imes 10^6 = 1.8 	imes 10^6 \, V$.Since there are $6$ identical charges at equal distances from the centre, the total potential $V_{	ext{total}}$ is:$V_{	ext{total}} = 6 	imes V_1 = 6 	imes 1.8 	imes 10^6 \, V = 10.8 	imes 10^6 \, V = 1.08 	imes 10^7 \, V$.

$A$ regular hexagon of side $5 \, cm$ has a charge $10 \, \mu C$ at each of its vertices. The potential at the centre of the hexagon is

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