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JEE Main 2022 (28 Jul, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2022 (28 Jul, Shift 2)

Q22JEE Main 2022 · 28 JUL · SHIFT 2Medium

Motion of Charges in an Electric Field

A uniform electric field E=(8 m/e) V/mE = (8\,m/e)\text{ V/m} is created between two parallel plates of length 1 m1\text{ m} as shown in figure, (where m=m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s2\text{ m/s}. The angle of the deviation (θ)(\theta) of the path of the electron as it comes out of the field will be _________.

Figure

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Answer: (b)

  1. Acceleration across the plates: a=eEm=8 m/s2a = \frac{eE}{m} = 8\text{ m/s}^2.
  2. Time inside: t=Lu=12=0.5 st = \frac{L}{u} = \frac{1}{2} = 0.5\text{ s}.
  3. vy=at=4 m/sv_y = at = 4\text{ m/s}, vx=2 m/sv_x = 2\text{ m/s}, so tan⁡θ=42=2\tan\theta = \frac{4}{2} = 2.
  4. So the answer is (b) tan⁡−1(2)\tan^{-1}(2).
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