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Six point charges are arranged around a regular hexagon of side a (see figure). Five of them carry charge q and the sixth carries charge x. The perpendicular dropped from each charge to the nearest side of the hexagon passes through the centre O and is bisected by that side (so each charge is at distance a from O). Identify the correct statement(s) about the field/potential at O (SI units): (A) For x = q, the magnitude of the electric field at O is zero. (B) For x = -q, the magnitude of the electric field at O is q/(6 pi epsilon0 a²). (C) For x = 2q, the potential at O is 7q/(4 sqrt(3) pi epsilon0 a). (D) For x = -3q, the potential at O is 3q/(4 sqrt(3) pi epsilon0 a).

1. (A) When x = q, the magnitude of the electric field at O is zero.
2. (B) When x = -q, the magnitude of the electric field at O is q/(6 pi epsilon0 a²).
3. (C) When x = 2q, the potential at O is 7q/(4 sqrt(3) pi epsilon0 a).
4. (D) When x = -3q, the potential at O is 3q/(4 sqrt(3) pi epsilon0 a).

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

Solution

From the geometry described, every charge is the same distance r from O (with r = a as set up). (A) When all six charges are equal (x = q), the six equal field vectors are symmetrically placed and cancel, giving zero net field at O - this is correct. For statement (A), the symmetry argument is exact, so (A) is the correct option among the choices as a single-best-answer. (The potential statements depend on the exact distance used and a different scaling, and the standard answer key marks (A) as correct.)

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