Q111 / 180 · NEET 2018 · Physics

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field  E. The direction of electric filed is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

  • A
    10 times greater
  • B
    5 times greater
  • Correct answer
    C
    Smaller
  • D
    Equal

Solution

 Acceleration of charge q of mass m in uniform electric field is given as a=qEm.
Applying second equation of motion, S=ut+12at2.

Initially mass is at rest, so u=0.

h=0+12 qEmt2⇒t=2hmqE⇒t∝m 

as mass of proton is heavier than electron, hence electron will take less time.

Hence, the correct answer is option: 'C'.

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