Six charges are placed around a regular hexag | vedclass

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(C) The distance of each charge from the center $O$ is $d = a \cos(30^{\circ}) = \frac{\sqrt{3}}{2}a$.$(A)$ When $x=q$,all six charges are equal and p

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$A, B$

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(C) The distance of each charge from the center $O$ is $d = a \cos(30^{\circ}) = \frac{\sqrt{3}}{2}a$.$(A)$ When $x=q$,all six charges are equal and placed symmetrically. By symmetry,the electric field at $O$ is zero. Statement $(A)$ is correct.$(B)$ When $x=-q$,the five charges $q$ produce a field equivalent to a charge $-q$ at the position of $x$ plus a charge $q$ at the position of $x$ (to complete the hexagon). The net field at $O$ is due to the charge $-q$ at $x$ and the additional $-q$ at $x$,resulting in a field of magnitude $E = \frac{1}{4\pi\epsilon_0} \frac{2q}{d^2} = \frac{2q}{4\pi\epsilon_0 (3a^2/4)} = \frac{2q}{3\pi\epsilon_0 a^2}$. This does not match the given value. Statement $(B)$ is incorrect.$(C)$ Potential $V = \frac{1}{4\pi\epsilon_0} \sum \frac{q_i}{d} = \frac{1}{4\pi\epsilon_0 d} (5q + x)$. For $x=2q$,$V = \frac{7q}{4\pi\epsilon_0 (\sqrt{3}a/2)} = \frac{7q}{2\sqrt{3}\pi\epsilon_0 a}$. This does not match. Statement $(C)$ is incorrect.$(D)$ For $x=-3q$,$V = \frac{1}{4\pi\epsilon_0 d} (5q - 3q) = \frac{2q}{4\pi\epsilon_0 (\sqrt{3}a/2)} = \frac{q}{\sqrt{3}\pi\epsilon_0 a}$. This does not match. Statement $(D)$ is incorrect.Re-evaluating the provided options,only $(A)$ is correct.

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