In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a potential difference 2400 V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 V. What is the charge on the second drop?Q4Q2Q3Q2
Question
In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a potential difference 2400 V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 V. What is the charge on the second drop?Q4Q2Q3Q2
Solution
In Millikan's oil drop experiment, the force due to gravity on the oil drop is balanced by the electric force. The force due to gravity is given by Fg = mg = 4/3 * π * r³ * ρ * g, where r is the radius of the drop, ρ is the density of the oil, and g is the acceleration due to gravity. The electric force is given by Fe = EQ, where E is the electric field and Q is the charge on the drop.
In the first case, we have 4/3 * π * r³ * ρ * g = EQ1, where Q1 is the charge on the first drop and E = V/d, where V is the potential difference and d is the distance between the plates.
In the second case, the radius of the drop is halved, so the force due to gravity is 4/3 * π * (r/2)³ * ρ * g = 1/8 * 4/3 * π * r³ * ρ * g. The potential difference is also reduced to 1/4 of the original, so E = V/4d. Therefore, we have 1/8 * 4/3 * π * r³ * ρ * g = EQ2/4, where Q2 is the charge on the second drop.
Solving the above equation for Q2, we get Q2 = Q1/2. Therefore, the charge on the second drop is half the charge on the first drop.
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