Calculate Charge on Electron from Milliken Oil Drop Data

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The discussion revolves around calculating the charge of an electron using data from the Milliken oil drop experiment. The user presents specific data points, including voltage and drop diameter, and outlines their calculation process, which involves determining the electric field and mass of the oil drop. They arrive at a charge value of 2.706 x 10^-8 coulombs but seek validation of their approach and results. Key aspects of their method include converting units and averaging electric fields. The user requests confirmation or guidance on their calculations.
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Milliken Oil Drop Problem anyone? Help please.

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 canceled 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.
 
Last edited:
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come on anyone please help me or at least tell me if I'm in the right direction.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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