Electric Field and Potential of Charges around Regular Hexagon

To determine the correct statements, we first analyze the geometry of the charge configuration around the regular hexagon.

1. Distance of each charge from the center $O$

For a regular hexagon of side length $a$, the perpendicular distance from the center $O$ to any of its sides (the inradius $r$) is given by: $r = a \cos(30^\circ) = \frac{\sqrt{3}}{2}a$

As given in the problem and shown in the diagram, the perpendicular segment from each charge to the nearest side passes through $O$ and is bisected by that side. This implies that the distance from the side to the charge is equal to the distance from $O$ to the side, which is $r$.

Therefore, the total distance $R$ of each charge from the center $O$ is: $R = r + r = 2r = 2\left(\frac{\sqrt{3}}{2}a\right) = \sqrt{3}a$

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2. Analysis of Options

Option A:

When $x = q$, six identical charges $q$ are positioned symmetrically around $O$ at equal angular intervals of $60^\circ$ at a uniform distance $R = \sqrt{3}a$.

By central symmetry, the electric fields produced by diametrically opposite charges are equal in magnitude and opposite in direction, canceling each other out completely: $E_{\text{net}} = 0$

Thus, Option A is correct.

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Option B:

When $x = -q$, five charges are $+q$ and one charge is $-q$. By the principle of superposition, this distribution can be represented as:

1. A symmetric distribution of six identical $+q$ charges at distance $R$ from $O$ (which gives a net electric field of zero at $O$).
2. An additional charge of $-2q$ located at the position of $x$.

The magnitude of the net electric field at $O$ is therefore due to the effective charge of $-2q$: $E_{\text{net}} = \frac{1}{4\pi \epsilon_o} \frac{|-2q|}{R^2}$

Substituting $R = \sqrt{3}a$, so $R^2 = 3a^2$: $E_{\text{net}} = \frac{1}{4\pi \epsilon_o} \frac{2q}{3a^2} = \frac{q}{6\pi \epsilon_o a^2}$

Thus, Option B is correct.

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Option C:

The electric potential at the center $O$ is the scalar sum of the potentials due to all six charges: $V_O = \frac{1}{4\pi \epsilon_o R} \sum_{i=1}^6 q_i = \frac{1}{4\pi \epsilon_o (\sqrt{3}a)} (5q + x)$

For $x = 2q$: $V_O = \frac{1}{4\pi \epsilon_o \sqrt{3}a} (5q + 2q) = \frac{7q}{4\sqrt{3}\pi \epsilon_o a}$

Thus, Option C is correct.

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Option D:

For $x = -3q$, using the formula for potential at $O$: $V_O = \frac{1}{4\pi \epsilon_o \sqrt{3}a} (5q - 3q) = \frac{2q}{4\sqrt{3}\pi \epsilon_o a} = \frac{q}{2\sqrt{3}\pi \epsilon_o a}$

This is not equal to $-\frac{3q}{4\sqrt{3}\pi \epsilon_o a}$.

Thus, Option D is incorrect.

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Conclusion

The correct options are A, B, and C.