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In Millikan oil drop experiment a drop of charge Q and radius r is kept constant between two plates of potential difference of 800 volt. Then charge on other drop of radius 2 r which is kept constant with a potential difference of 3200 V is
- a)Q/2
- b)2Q
- c)4Q
- d)Q/4
Correct answer is option 'B'. Can you explain this answer?
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In Millikan oil drop experiment a drop of charge Q and radius r is kep...
Explanation:
Millikan oil drop experiment is used to determine the charge of an electron. The experiment is performed by suspending a tiny, charged droplet of oil between two parallel metal plates by balancing its weight with an electric force.
Given, a drop of charge Q and radius r is kept constant between two plates of potential difference of 800 V.
Formula: The electric force acting on the oil drop due to the electric field between the plates is given by F = QE, where E is the electric field between the plates.
Calculation:
Let the electric force acting on the oil drop be F1. Then, we have F1 = QE1, where E1 is the electric field between the plates.
Similarly, let the electric force acting on the second oil drop be F2. Then, we have F2 = QE2, where E2 is the electric field between the plates.
Since both oil drops are in equilibrium, we have F1 = F2. Therefore, we have QE1 = QE2.
Given, the second drop has a radius of 2r and is charged. Let the charge on the second drop be Q2.
Then, we can write Q2 = (4/3)πr3 ρ2, where ρ2 is the density of the oil.
Since the potential difference between the plates is 3200 V for the second drop, we can write E2 = V2/d = 3200/2d, where d is the distance between the plates.
Similarly, we can write E1 = V1/d = 800/d, where V1 is the potential difference between the plates for the first oil drop.
Now, using the formula QE1 = QE2, we have Q(800/d) = Q2(3200/2d).
Substituting Q2 = (4/3)π(2r)3ρ2, we get Q(800/d) = (4/3)π(2r)3ρ2(3200/2d).
Simplifying the expression, we get Q = 8Q2.
Therefore, the charge on the second drop is Q2 = Q/8 = Q/2^3 = Q/2.
Hence, the correct answer is option B, i.e., 2Q.
Millikan oil drop experiment is used to determine the charge of an electron. The experiment is performed by suspending a tiny, charged droplet of oil between two parallel metal plates by balancing its weight with an electric force.
Given, a drop of charge Q and radius r is kept constant between two plates of potential difference of 800 V.
Formula: The electric force acting on the oil drop due to the electric field between the plates is given by F = QE, where E is the electric field between the plates.
Calculation:
Let the electric force acting on the oil drop be F1. Then, we have F1 = QE1, where E1 is the electric field between the plates.
Similarly, let the electric force acting on the second oil drop be F2. Then, we have F2 = QE2, where E2 is the electric field between the plates.
Since both oil drops are in equilibrium, we have F1 = F2. Therefore, we have QE1 = QE2.
Given, the second drop has a radius of 2r and is charged. Let the charge on the second drop be Q2.
Then, we can write Q2 = (4/3)πr3 ρ2, where ρ2 is the density of the oil.
Since the potential difference between the plates is 3200 V for the second drop, we can write E2 = V2/d = 3200/2d, where d is the distance between the plates.
Similarly, we can write E1 = V1/d = 800/d, where V1 is the potential difference between the plates for the first oil drop.
Now, using the formula QE1 = QE2, we have Q(800/d) = Q2(3200/2d).
Substituting Q2 = (4/3)π(2r)3ρ2, we get Q(800/d) = (4/3)π(2r)3ρ2(3200/2d).
Simplifying the expression, we get Q = 8Q2.
Therefore, the charge on the second drop is Q2 = Q/8 = Q/2^3 = Q/2.
Hence, the correct answer is option B, i.e., 2Q.
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In Millikan oil drop experiment a drop of charge Q and radius r is kept constant between two plates of potential difference of 800 volt. Then charge on other drop of radius 2 r which is kept constant with a potential difference of 3200 V isa)Q/2b)2Qc)4Qd)Q/4Correct answer is option 'B'. Can you explain this answer?
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