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(C) The potential at the centre $O$ of the hexagon due to a charge $q$ at each vertex is the sum of the potentials due to each individual charge.Since

Vedclass · Accepted Answer

$5.4 	imes 10^5 	ext{ volt}$

Vedclass · Answer

(C) The potential at the centre $O$ of the hexagon due to a charge $q$ at each vertex is the sum of the potentials due to each individual charge.Since the distance from the centre to each vertex of a regular hexagon is equal to the side length $r = 10 	ext{ cm} = 0.1 	ext{ m}$,the potential $V$ at the centre is given by:$V = 6 	imes \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} ight)$Given $q = 1 \mu	ext{C} = 1 	imes 10^{-6} 	ext{ C}$,$r = 0.1 	ext{ m}$,and $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$:$V = 6 	imes \left( 9 	imes 10^9 	imes \frac{1 	imes 10^{-6}}{0.1} ight)$$V = 6 	imes 9 	imes 10^9 	imes 10^{-5}$$V = 54 	imes 10^4 = 5.4 	imes 10^5 	ext{ volt}$.

$A$ regular hexagon of side $10 \text{ cm}$ has a charge of $1 \mu\text{C}$ at each of its vertices. The potential at the centre of the hexagon is $\left[\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI unit}\right]$.

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