An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E . The direction of electric fie - Physics MCQ NEET (UG) 2018 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

1. Distance h = \frac{1}{2} a t^{2} ⟹ t = \frac{2 h}{a} .2. Acceleration in an electric field a = \frac{qE}{m} .3. So, t = \frac{2 hm}{qE} .4. Since h, q, and E are the same for both the electron and the proton, t \propto m .5. The mass of an electron is much smaller than the mass of a proton. Therefore, the time of fall for the electron will be smaller.

Try it yourself before checking the explanation above.

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:

10 times greater

5 times greater

Smaller

Equal

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