Sample Calculation on Milikan's Data Analysis

AI Thread Summary
The discussion revolves around calculating the charge on an oil drop using Millikan's data analysis. The electric field strength is determined to be 1.2 x 10^5 N/C, and the volume of the oil drop is calculated as 1.7456 x 10^-12 m^3, leading to a mass of 1.4838 x 10^-19 kg. The participant is attempting to equate the electric force (Fe) to the gravitational force (Fg) to find the charge (Q) on the drop. There is confusion regarding the correct values to use for Fg, specifically whether to use the radius of the drop or double that distance. The participant ultimately realizes that Fg can be calculated using the formula Fg = mg.
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Homework Statement


Find the charge on the oil drop in Coulombs, and the number of electrons it is "missing".
Vplates= 2400 V
Dplates = 0.020 m
Oil Density = 850 kg/m3
Oil Drop Radius = .000051 m (from Stokes’ Law)



Homework Equations


E = V/d
Volume = 4/3πr3
m= density x volume


The Attempt at a Solution


\Sigma = (2400 V)/(0.020 m) = 1.2 x 105
Volume = 4/3 π(0.000051)3 = 1.7456 x 10-12 m3
m = (850 kg/m3)(1.7456 x 10-12 m3) = 1.4838 x 10-19 kg
And this is where I got to... now I know that Fe = Fg and I was thinking I could find Fg, but am stuck :( I would appreciate it if you could help me out!
 
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Sorry about the electric field equations.. they came out weird haha
 
How much is the electric force on the charged oil drop in terms of its charge? And how much is the force of gravity if the mass is m?

ehild
 
I know that Fe = Q\Sigma so if I find Fg I can solve for Q. However, I am confused as to what values I use for Fg. I know the gravitational constant and the two masses (which are equal?), but am confused about the distance between them. Should I be using the radius of the drop or twice the radius?
 
Sorry about that.. Fe = QE
 
okay I realized I just need to use Fg = mg :)
 
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