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JEE Main202229 Jun 2022Morning ShiftPhysicsElectrostaticsActual

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 10 5 N C - 1 . If the charge on the particle is 40 μ C and the initial velocity is 200 m s - 1 , how much distance it will travel before coming to the rest momentarily

Options

  1. A0 . 5   m
  2. B1   m
  3. C5   m
  4. D10   m

Correct answer

A. 0 . 5   m

Step-by-step solution

Force experienced by a particle with charge q in an electric field F = q E . Now, the acceleration produced is given as a = F m = q E m = 40 × 10 - 6 × 10 5 1 × 10 - 4 = 4 × 10 4   m     s - 2 (As the particle is projected against the electric field, hence it is decelerated) Using, v 2 = u 2 - 2 a s We have, 0 2 = u 2 - 2 a s s = v 2 2 a = 200 2 2 × 4 × 10 4 = 0 . 5   m

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