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(C) For a regular hexagon with side $a$,the distance from each vertex to the center $O$ is $r = a$.$1$. Electric Potential $(V)$: The potential at the

Vedclass · Accepted Answer

$E=0$ and $V=0$

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(C) For a regular hexagon with side $a$,the distance from each vertex to the center $O$ is $r = a$.$1$. Electric Potential $(V)$: The potential at the center $O$ is the algebraic sum of potentials due to individual charges: $V = \sum \frac{k q_i}{r_i} = \frac{k}{a} (q - q + q - q + q - q) = 0$.$2$. Electric Field $(E)$: The electric field at the center $O$ is the vector sum of fields due to individual charges. Each pair of opposite charges (e.g.,$q$ and $-q$) produces a resultant field at the center directed towards the negative charge. Specifically,a charge $q$ at one vertex and $-q$ at the opposite vertex create a net field of magnitude $\frac{2kq}{a^2}$ directed towards the negative charge. Since there are three such pairs oriented at $120^\circ$ to each other,their vector sum is zero. Thus,$E = 0$.

Six point charges are placed at the vertices of a regular hexagon of side $a$ as shown. If $E$ represents the electric field and $V$ represents the electric potential at the center $O$,then:

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