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JEE Main 2020 (4 Sep, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2020 (4 Sep, Shift 2)

Q35JEE Main 2020 · 4 SEP · SHIFT 2Difficult

Motion of Charges in an Electric Field

A particle of charge q and mass m is subjected to an electric field

E = E0 (1 – aax2) in the x-direction, where aa and E0 are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

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Answer: (c)

  1. Work–energy: KE=∫0xqE0(1−ax2) dx=qE0(x−ax33)KE = \int_0^x qE_0(1 - ax^2)\,dx = qE_0\left(x - \frac{ax^3}{3}\right)
  2. KE=0KE = 0 again when 1−ax23=01 - \frac{ax^2}{3} = 0.
  3. x=3ax = \sqrt{\frac{3}{a}}
  4. So the answer is (c).
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