Milliken Oil Drop Experiment. Help Please.

  • Thread starter shikagami
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Can some one help me with this Milliken oil drop problem.

Data from a Milliken oil drop experiment are given below. Calculate the value of the charge on an electron from this data:

Voltage needed to suspend drop (V) Drop diameter (microns)
103------ 1.1
44.7----- 1.2
127------ 1.6
175------ 1.5
11.3----- 0.9

Given:
Plate separation: 0.5cm
Density of oil: 0.95 g/mL
Gravitational constant: 9.80 m/s^2

Ok... this is what I did: q=mg/E
1. I calculated for electric field first since: E= kV/m
2. One microns is equal to 10^-6 m
3. I changed volts into kV.
4. I took the average of all 5 of the electric fields and got 67896.71718 kV/m
5. Now I solved for the mass by cubing the plate separation and cancelled it with the mL... to get grams then I changed grams into kilograms to get 1.1875 x 10^-4 kg.
6. And finally I plug everything back into the equation: q=mg/E and get the answer of 2.706 x 10^-8 coulombs.

Please tell me if I did it right.
 

Answers and Replies

  • #2
Gokul43201
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No, you made mistakes because you have not properly understood the concept.

1. The force on a charge q due to an electric field E is qE. The electric field between a pair of plates at a voltage difference V, is given by V/d, where d is the distance between the plates (and has nothing to do with the size of the drop).

2. The gravitational force on the oil-drop is its weight, mg. m is the mass of the drop which is the product of volume and density. You are given the density, and you can calculate the volume of each drop assuming it is a perfect sphere.

3. For the drop to be suspended between the plates, the upward and downward forces must match. So qE = mg or q = mg/E. Having found m of each drop and the corresponding E acting on it, you can find the value of q on each drop. Do not average all the values; find for each case separately.

4. Use the different values of q to find the answer.
 

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