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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 as compared to the time of fall of the proton is

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

The correct option is B smaller

For electron:-

meg+Fe=mea

a=meg+eEme

ae=g+emeE

Now, h=12aet2

te=  2hg+eEme

For proton:
mpg+Fe=mpa

ap=g+eEmp
Now, h=12×apt2
tp=  2hg+eEmp
mp>me
te<tp


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