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(D) The given figure shows six equal charges,$q$,at the vertices of a regular hexagon.Charge,$q = 5 \; \mu C = 5 	imes 10^{-6} \; C$Side of the hexag

Vedclass · Accepted Answer

$2.7 	imes 10^{6} \; V$

Vedclass · Answer

(D) The given figure shows six equal charges,$q$,at the vertices of a regular hexagon.Charge,$q = 5 \; \mu C = 5 	imes 10^{-6} \; C$Side of the hexagon,$l = 10 \; cm = 0.1 \; m$In a regular hexagon,the distance of each vertex from the centre $O$ is equal to the side length of the hexagon. Therefore,the distance $d = 10 \; cm = 0.1 \; m$.The electric potential $V$ at the centre $O$ due to a single charge $q$ at distance $d$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d}$.Since there are six identical charges at the vertices,the total potential at the centre is the sum of the potentials due to each charge:$V_{total} = 6 	imes \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d} ight)$Using $\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \; N m^{2} C^{-2}$:$V_{total} = 6 	imes \frac{9 	imes 10^{9} 	imes 5 	imes 10^{-6}}{0.1}$$V_{total} = \frac{270 	imes 10^{3}}{0.1} = 2700 	imes 10^{3} = 2.7 	imes 10^{6} \; V$Therefore,the potential at the centre of the hexagon is $2.7 	imes 10^{6} \; V$.

$A$ 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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