JEE Advanced 2022 Paper 1 Q13 Physics Electrostatics & Circuits Electric Field & Potential Hard

JEE Advanced 2022 Paper 1 · Q13 · Electric Field & Potential

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?

[Figure: Regular hexagon of side length a centred at O. Six point charges placed outside the hexagon, each opposite the midpoint of a side, at perpendicular distance equal to the apothem (so the distance from each charge to O equals 2·apothem = a√3). Five charges = q, one = x.]

  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/(6πε₀a²).
  3. C. When x = 2q, the potential at O is 7q/(4√3 πε₀ a).
  4. D. When x = −3q, the potential at O is −3q/(4√3 πε₀ a).
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

The geometry: each charge is at distance r = a√3 from O (since the perpendicular from the charge to the nearest side has its midpoint on the side, making the charge-to-side distance = apothem = a√3/2, and charge-to-O distance = a√3/2 + a√3/2 = a√3). (A) With six equal q symmetrically placed, the field at O cancels by symmetry ⇒ E = 0 ✓. (B) With one charge replaced by −q, the net field is twice the field of one q at the unique charge's location: E = 2·k·q/r² = 2·q/(4πε₀·3a²) = q/(6πε₀a²) ✓. (C) Potential V = sum of kq_i/r_i with all r_i = a√3: V = (5q + 2q)/(4πε₀·a√3) = 7q/(4√3 πε₀ a) ✓. (D) With x = −3q: V = (5q − 3q)/(4πε₀·a√3) = 2q/(4√3 πε₀ a) = q/(2√3 πε₀ a), NOT −3q/(4√3 πε₀ a). ✗. Units check for (B) and (C): [C]/([F/m]·[m²]) = [V/m] ✓ for field; [C]/([F/m]·[m]) = [V] ✓ for potential.

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