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JEE Main201910 Apr 2019Morning ShiftPhysicsElectrostaticsActual

A uniformly charged ring of radius 3 a and total charge q is placed in x ‐ y plane centred at origin. A point charge q is moving towards the ring along the z - axis and has speed v at z = 4 a . The minimum value of v such that it crosses the origin is:

Options

  1. A2 m 1 15 q 2 4 π ϵ 0 a 1 / 2
  2. B2 m 4 15 q 2 4 π ϵ 0 a 1 / 2
  3. C2 m 1 5 q 2 4 π ϵ 0 a 1 / 2
  4. D2 m 2 15 q 2 4 π ϵ 0 a 1 / 2

Correct answer

D. 2 m 2 15 q 2 4 π ϵ 0 a 1 / 2

Step-by-step solution

From the conservation of energy, 1 4 π ε 0 q 2 3 a 2 + 4 a 2 + 1 2 m v 2 = 1 4 π ε 0 q 2 3 a v = q 2 4 π ε 0 a × 2 15 × 2 m = 2 m 2 15 q 2 4 π ε 0 a 1 / 2

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