anonymous
  • anonymous
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
Physics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
= 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
  • IrishBoy123
for 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
  • IrishBoy123
brilliant 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Michele_Laino
  • Michele_Laino
yes! The Millikan experiment is very important!
Abhisar
  • Abhisar
Yes, 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.