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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
- a)smaller
- b)5 times greater
- c)10 times greater
- d)equal
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
An electron falls from rest through a vertical distance h in a unifor...
Explanation:
When an electron falls through a uniform and vertically upward directed electric field E, it experiences a force in the opposite direction to that of the electric field. This is because the electric charge on the electron is negative, and hence, the direction of the electric force is opposite to that of the electric field. Therefore, the electron is accelerated in the downward direction.
On reversing the direction of the electric field, the force on the electron also gets reversed, and it is now accelerated in the upward direction. The magnitude of the electric force on the electron remains the same in both cases.
Similarly, when a proton falls through the same electric field, it experiences a force in the same direction as that of the electric field. This is because the electric charge on the proton is positive, and hence, the direction of the electric force is the same as that of the electric field. Therefore, the proton is accelerated in the upward direction.
Comparison of the time of fall of the electron and proton:
Let t1 be the time taken by the electron to fall through the vertical distance h, and t2 be the time taken by the proton to fall through the same vertical distance h.
Using the equations of motion, we can write:
For the electron:
h = (1/2)gt1^2 + (1/2)at1^2
where g is the acceleration due to gravity and a is the acceleration due to the electric force on the electron.
a = qE/m
where q is the charge on the electron, E is the magnitude of the electric field, and m is the mass of the electron.
Substituting the values of a and g, we get:
h = (1/2)(9.8 + qE/m)t1^2
For the proton:
h = (1/2)gt2^2 + (1/2)at2^2
where a is the acceleration due to the electric force on the proton.
a = qE/m
where q is the charge on the proton, E is the magnitude of the electric field, and m is the mass of the proton.
Substituting the values of a and g, we get:
h = (1/2)(9.8 - qE/m)t2^2
Dividing the two equations, we get:
t1^2/t2^2 = (9.8 + qE/m)/(9.8 - qE/m)
As the magnitude of the electric field E is the same for both cases, and the mass of the electron is much smaller than that of the proton, we can assume that qE/m < />
Therefore, we can simplify the above equation as:
t1^2/t2^2 ≈ 1
Taking the square root of both sides, we get:
t1/t2 ≈ 1
Hence, the time of fall of the electron is approximately equal to the time of fall of the proton.
Therefore, the correct option is (a) smaller.
When an electron falls through a uniform and vertically upward directed electric field E, it experiences a force in the opposite direction to that of the electric field. This is because the electric charge on the electron is negative, and hence, the direction of the electric force is opposite to that of the electric field. Therefore, the electron is accelerated in the downward direction.
On reversing the direction of the electric field, the force on the electron also gets reversed, and it is now accelerated in the upward direction. The magnitude of the electric force on the electron remains the same in both cases.
Similarly, when a proton falls through the same electric field, it experiences a force in the same direction as that of the electric field. This is because the electric charge on the proton is positive, and hence, the direction of the electric force is the same as that of the electric field. Therefore, the proton is accelerated in the upward direction.
Comparison of the time of fall of the electron and proton:
Let t1 be the time taken by the electron to fall through the vertical distance h, and t2 be the time taken by the proton to fall through the same vertical distance h.
Using the equations of motion, we can write:
For the electron:
h = (1/2)gt1^2 + (1/2)at1^2
where g is the acceleration due to gravity and a is the acceleration due to the electric force on the electron.
a = qE/m
where q is the charge on the electron, E is the magnitude of the electric field, and m is the mass of the electron.
Substituting the values of a and g, we get:
h = (1/2)(9.8 + qE/m)t1^2
For the proton:
h = (1/2)gt2^2 + (1/2)at2^2
where a is the acceleration due to the electric force on the proton.
a = qE/m
where q is the charge on the proton, E is the magnitude of the electric field, and m is the mass of the proton.
Substituting the values of a and g, we get:
h = (1/2)(9.8 - qE/m)t2^2
Dividing the two equations, we get:
t1^2/t2^2 = (9.8 + qE/m)/(9.8 - qE/m)
As the magnitude of the electric field E is the same for both cases, and the mass of the electron is much smaller than that of the proton, we can assume that qE/m < />
Therefore, we can simplify the above equation as:
t1^2/t2^2 ≈ 1
Taking the square root of both sides, we get:
t1/t2 ≈ 1
Hence, the time of fall of the electron is approximately equal to the time of fall of the proton.
Therefore, the correct option is (a) smaller.
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Community Answer
An electron falls from rest through a vertical distance h in a unifor...
Force experienced by a charged particle in an electric field, F = qE
As F = ma
As electron and proton both fall from same height at rest. Then initial velocity =0
From
∴ 
∴ 
∴ Electron has smaller mass so it will take smaller time.
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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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer?
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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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer?.
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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice 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 isa)smallerb)5 times greaterc)10 times greaterd)equalCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice NEET tests.
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