Six charges are placed around a regular hexagon of side length 𝑎 as shown in the figure. Five of them have charge 𝑞, and the remaining one has charge 𝑥. The perpendicular from each charge to the nearest hexagon side passes through the center O of the hexagon and is bisected by the side. Which of the following statement(s) is(are) correct in SI units? (A) When 𝑥 = 𝑞, the magnitude of the electric field at O is zero. (B) When 𝑥 = −𝑞, the magnitude of the electric field at O is � �� �� ��. (C) When 𝑥 = 2𝑞, the potential at O is �� �√����� . (D) When 𝑥 = −3𝑞, the potential at O is − �� �√����� .

 We analyze the problem step by step to determine which statements are correct.

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Step 1: Symmetry of the system

• The hexagon is regular, and five charges are $q$, while one charge is $x$.

• The perpendicular from each charge to the nearest side passes through the center $O$ and is bisected by the side. This implies that the charges are symmetrically placed around the hexagon.

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Step 2: Electric field at the center $O$

The electric field at the center $O$ due to a charge $q$ at a distance $r$ is:

$$E=\frac{kq}{r^2},$$

where $k=\frac{1}{4\pi {ϵ}_0}$.

(A) When $x=q$, the magnitude of the electric field at $O$ is zero.

• If $x=q$, all six charges are equal, and due to the symmetry of the hexagon, the electric fields at $O$ cancel out.

• Statement (A) is correct.

(B) When $x=-q$, the magnitude of the electric field at $O$ is $\frac{kq}{a^2}$.

• If $x=-q$, the system has five charges $q$ and one charge $-q$. The electric fields due to the five $q$ charges cancel out, leaving only the field due to the $-q$ charge.

• The distance from each charge to $O$ is $a$ (since the hexagon is regular). Thus, the electric field at $O$ is:

  $$E=\frac{kq}{a^2}\mathrm{.}$$

• Statement (B) is correct.

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Step 3: Potential at the center $O$

The potential at the center $O$ due to a charge $q$ at a distance $r$ is:

$$V=\frac{kq}{r}\mathrm{.}$$

(C) When $x=2q$, the potential at $O$ is $\frac{6kq}{a\sqrt{3}}$.

• If $x=2q$, the total potential at $O$ is the sum of the potentials due to the five $q$ charges and the one $2q$ charge.

• The distance from each charge to $O$ is $a$. Thus, the potential at $O$ is:

  $$V=5\cdot \frac{kq}{a}+\frac{k(2q)}{a}=\frac{7kq}{a}\mathrm{.}$$

• The given potential $\frac{6kq}{a\sqrt{3}}$ is incorrect.

• Statement (C) is incorrect.

(D) When $x=-3q$, the potential at $O$ is $-\frac{6kq}{a\sqrt{3}}$.

• If $x=-3q$, the total potential at $O$ is the sum of the potentials due to the five $q$ charges and the one $-3q$ charge.

• The distance from each charge to $O$ is $a$. Thus, the potential at $O$ is:

  $$V=5\cdot \frac{kq}{a}+\frac{k(-3q)}{a}=\frac{2kq}{a}\mathrm{.}$$

• The given potential $-\frac{6kq}{a\sqrt{3}}$ is incorrect.

• Statement (D) is incorrect.

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Final Answer:

Only statements (A) and (B) are correct.