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Question:

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field 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?

Equal

5times greater

10times greater

Smaller

Solution:

For electron:-
-Meg + Fe = Mea
a = (meg + eE)/me
Now, h = 1/2 aet2
t1 = √(2h/(g + eE/me))
For proton:-
-mpg + Fe = mpa
a = (g + eE/mp)
Now, h = 1/2 × apt2 ⇒ t2 = √(2h/(g + eE/mp))
∴ mp > me
So, t1 > t2