Calculate Charge on Electron from Milliken Oil Drop Data

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SUMMARY

The discussion focuses on calculating the charge of an electron using data from the Milliken oil drop experiment. The user applied the formula q=mg/E, where they first calculated the electric field (E) using the voltage and plate separation. After converting units and averaging the electric fields, they determined the mass of the oil drop and ultimately calculated the charge as 2.706 x 10^-8 coulombs. The calculations and methodology presented are correct and align with established physics principles.

PREREQUISITES
  • Understanding of the Milliken oil drop experiment
  • Familiarity with electric field calculations (E = kV/m)
  • Knowledge of unit conversions (microns to meters, grams to kilograms)
  • Basic physics concepts including mass (m) and gravitational force (g)
NEXT STEPS
  • Review the principles of the Milliken oil drop experiment
  • Learn about electric field strength calculations in physics
  • Study unit conversion techniques for scientific measurements
  • Explore the significance of charge quantization in atomic physics
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Students in physics, educators teaching electromagnetism, and researchers interested in experimental physics methodologies will benefit from this discussion.

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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.
 

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