A regular hexagon of side 6 cm has a charge | vedclass

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(B) In a regular hexagon,the distance from the centre to each vertex is equal to the side length of the hexagon. Given side $a = 6 \ cm = 0.06 \ m$. S

Vedclass · Accepted Answer

$1.8 	imes 10^6 \ V$

Vedclass · Answer

(B) In a regular hexagon,the distance from the centre to each vertex is equal to the side length of the hexagon. Given side $a = 6 \ cm = 0.06 \ m$. Since there are $6$ vertices,each with a charge $q = 2 \ \mu C = 2 	imes 10^{-6} \ C$,the total potential $V$ at the centre is the sum of the potentials due to each charge. $V = 6 	imes \left( \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{a} ight)$ Substituting the values: $V = 6 	imes 9 	imes 10^9 	imes \frac{2 	imes 10^{-6}}{0.06}$ $V = 54 	imes 10^9 	imes \frac{2 	imes 10^{-6}}{6 	imes 10^{-2}}$ $V = 54 	imes 10^9 	imes \frac{1}{3} 	imes 10^{-4}$ $V = 18 	imes 10^5 \ V = 1.8 	imes 10^6 \ V$.

$A$ regular hexagon of side $6 \ cm$ has a charge of $2 \ \mu C$ at each of its vertices. What is the potential at the centre of the hexagon? $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right]$

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