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 flal of the proton is

Options

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

Answer

C

Explanation

Force on a charged particle in the presence of an electric field is given as
`F=qE`………(i)
Where, q is the charge on the charged particle and E is the electric field.
From Newton's second law of motion, force on a particle with mass m is given as
F=ma..............(ii)
From Eqs. (i) and (ii), we get
F=ma=qE`rArr a=(qE)/m`..........(iii)
Now, consider that a particle falls from rest through a vertical distance h. Therefore, u=0 andthe second equation of motion becomes
`s=ut+1/2at^(2)`
or `h= 0xxt+1/2at^(2)`
`=1/2 xx (qE)/(m)t^(2)` from Eq. (iii)
`rArr t^(2)=(2hm)/(qE)` or `t=sqrt(2hm)/(qE)`
since, the particle given in the question is electron and proton, and the quantitiy `sqrt(2h)/(qE)` (here, `q_(p) = q_(e) = e)` for both of them is constant. Thus, we can write
`t=ksqrt(m)`
Where, `k=sqrt(2h)/(qE)` or `t propto sqrt(m)`
As, mass of proton `m_(p) gtgt` mass of electron `(m_(e)`. Thus,the time of fall of an electron would be smaller than the time of fall of a protons.