An electron falls from rest through a vertical distance h in a uniform and vertically upwards 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 proton is

Options

A. smaller
B. 5 times greater
C. 10 times greater
D. equal

Answer

A

Explanation

Force experienced by a charged particle in an electric field, F = qE
As F = ma
`:. m a = qE rArr a = (qE)/(m)` ….(i)
As electron and proton both all fall from same height at rest. Then initial velocity = 0
from `s = u t + (1)/(2) a t^(2) " "(because u = 0)`
`:. h = (1)/(2) a t^(2) rArr h = (1)/(2)(qE)/(m)t^(2)` [Using (i)]
`:. t = sqrt((2hm)/(qE)) rArr t prop sqrt(m)` as .q. is same for electron and proton.