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anonymous
 one year ago
In his book, The Electron, Millikan reports an early series of measurements he performed on a single oil drop. The motion of the drop was timed with a stopwatch over a vertical distance of 10.21 mm. With the electric field switched off, the time of the fall was t1 = 11.880 s. With the electric field switched on, Millikan found that the time of rise of the drop over the same distance varied on successive trials because the drop sometimes had more electric charge, sometimes less. The following table gives the time of rise for several trials:
t2 = 80.708 s
= 140.565 s
= 79.600 s
anonymous
 one year ago
In his book, The Electron, Millikan reports an early series of measurements he performed on a single oil drop. The motion of the drop was timed with a stopwatch over a vertical distance of 10.21 mm. With the electric field switched off, the time of the fall was t1 = 11.880 s. With the electric field switched on, Millikan found that the time of rise of the drop over the same distance varied on successive trials because the drop sometimes had more electric charge, sometimes less. The following table gives the time of rise for several trials: t2 = 80.708 s = 140.565 s = 79.600 s

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0= 137.308 s = 34.638 s = 500.1 s = 19.69 s =42.302 s Other relevant data: Plate Distance = 16 mm Volts = 5085 V Oil Density = 0.9199 g/cm^3 Air Viscositu = 1.824*10^(4) g/s*cm (a) Calculate the mass of the oil drop. (b) For each value of t2, calculate the electric charge on the drop. (c) Check that all these values of the electric charge are (nearly) integral multiples of the fundamental electric charge e = 1.66*10^(19) C. I think I can answer all parts of the problem if I knew how to get the radius of the oil drop. I’m pretty sure it has something to do with the density and the formula for the volume of a sphere, however I can’t figure it out. Can anyone help?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3for r, use this [\(v_1 \) is terminal velocity for initial drop] : \[r^2 = \frac{9 \eta v_1}{2 g (\rho  \rho _{air})}. \,\] which is taken from here: https://gyazo.com/3badd0e6cdac923f6bde345c5776d9a9

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3brilliant question btw! this is, i think, the seminal paper. https://www.aip.org/history/gap/PDF/millikan.pdf you should get some medals just for asking this question!! @Michele_Laino @Astrophysics @arindameducationusc @Robert136 @Abhisar

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0yes! The Millikan experiment is very important!

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Yes, this is the experiment he used to find the e/m ratio of electron. c: Here is a fun simulation on it. http://g.web.umkc.edu/gounevt/Animations/Animations211/MillikanOilDropExp.swf
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