A regular hexagon of side 9 cm has a charge | vedclass

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Question

(C) In a regular hexagon,the distance from the centre to each vertex is equal to the side length of the hexagon. Given side length $r = 9 \ cm = 0.09

Vedclass · Accepted Answer

$3 	imes 10^6$

Vedclass · Answer

(C) In a regular hexagon,the distance from the centre to each vertex is equal to the side length of the hexagon. Given side length $r = 9 \ cm = 0.09 \ m$.Each vertex has a charge $q = 5 \mu C = 5 	imes 10^{-6} \ C$.The electric potential $V$ at the centre due to a single charge $q$ at distance $r$ is $V_i = \frac{kq}{r}$.Since there are $6$ identical charges at the vertices,the total potential $V$ at the centre is the sum of potentials due to each charge:$V = 6 	imes \frac{kq}{r}$Substituting the values:$V = 6 	imes \frac{9 	imes 10^9 	imes 5 	imes 10^{-6}}{0.09}$$V = 6 	imes \frac{45 	imes 10^3}{9 	imes 10^{-2}}$$V = 6 	imes 5 	imes 10^5$$V = 30 	imes 10^5 = 3 	imes 10^6 \ V$.

$A$ regular hexagon of side $9 \ cm$ has a charge $5 \mu C$ at each of its vertices. The electric potential at its centre is $..........V$. $(k = 9 \times 10^9 \ Nm^2 C^{-2})$

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