Question

an electron falls from rest through a vertical distance h in a uniform are vertically upward directly electric field E the direction of electric field is now reversed keeping its magnitude the same a proton is allowed to flall from rest in it through the same vertical distance h the times of fall of electron in comparison to the time of fall of proton is

Answer 1

If I am right , this question is asked in NEET 2018 .
It's really easy equation . How , Let see ,

A rough daigram is shown in figure.
As you know, electron is charged particle , it experiences elctric force due to present in electric field E .
so, the force act on electron , F = eE
F = ma
eE = ma ⇒a = eE/m , here m is mass of electron.
because electron starts to move in rest , ∴ u = 0
Now, use S = ut + 1/2 at²
h = 0 + 1/2 (eE/m)t²
t = √{2hm/eE}

Similarly in case of proton,
t' = √{2hM/pE} , here M is mass of proton
You also know, magnitude of charge on proton equals electron.
so, e = p
∴ t' = √{2hM/eE}

Here it is clear that ,
t/t' = √{m/M}
But mass of proton >> mass of electron
∴ M >> m
∴ t/t' < 1
t < t'

Hence, time taken by electron is smaller than time taken by proton.

Method 2:- as you know, acceleration on electric field , a = qE/m
so, time taken to cover h distance = √{2hm/qE}
If h , E ,q are constant then, t $\propto$ √m
Hence, Heavier charged particle take longer time.
So, electron take less time than proton.

Answer 2

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