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Millikan Oil Drop Experiment

  • Thread starter Daniel Luo
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  • #1
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Homework Statement



When an oil drop falls freely, the velocity first increases, but afterwards it quickly reaches a constant speed, since the air resistance becomes equal to the weight of the oil drop. The air resistance is given by:

Fair=6[itex]\pi[/itex]r[itex]\eta[/itex]v,

where [itex]\eta[/itex] is the viscosity of air with a value of 1.759*10-5 N s/m2, Oil's density is 918.7 kg/m3. In Millikan's experiment, the speed of the oil drop is 5.449*10-4 m/s.

- Find the radius and mass of the oil drop.

Afterwards, electricity was sent through the plates, and the speed of the oil drop increases by 5.746*10-4 m/s. The upward electric force acting on the oil drop is given by:

FE=qE,

where E is the electric field strength: E = 3.178*105 N/C.

Find the electric charge of the oil drop, q.

Homework Equations



For no. 1:

Weight = Air resistance
mg = Fair=6[itex]\pi[/itex]r[itex]\eta[/itex]v,
[itex]\rho[/itex]Vg=Fair=6[itex]\pi[/itex]r[itex]\eta[/itex]v,

No. 2:

FE=qE
mg = FE + Fair

The Attempt at a Solution



All right. What I did in no. 1 was to say that the weight was equal to the air resistance. Assuming the drop to be spherical, I said:

[itex]\rho[/itex](4/3[itex]\pi[/itex]r3)g = 6[itex]\pi[/itex]r[itex]\eta[/itex]v. Then I solved for r:

r = 2.19*10-6 m

And I found m:

m = [itex]\rho[/itex](4/3[itex]\pi[/itex]r3) = 4.04*10-14 kg (This seems to be too light even for a raindrop?)


For no. 2: I said that the weight was equal to the sum of air resistance and the electric force:

mg = 6[itex]\pi[/itex]r[itex]\eta[/itex]v + qE

Solving for q gave: q = -1.31*10-18 C.

However, this answer seems wrong, as Millikan tried to find the charge of the electron e from this experiment, which is 1.602*10-19 C.

Have I done something wrong? In case I did, what was my mistake?

Thanks a lot!
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
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Well a raindrop is water and may have a radius around 1-2mm.
The oil drop has a radius of around 3 microns ... about 1000x smaller.
Millican needed a microscope to see them.

Technically, Millican found the electron charge-to-mass ratio from this experiment.
To find out how much charge a single electron has, he needed to know how many electrons were responsible for the charge.

How many excess electrons are there in your oil drop?
 
  • #3
haruspex
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Technically, Millican found the electron charge-to-mass ratio from this experiment.
No, I think that was obtained through a much earlier experiment by someone else. The point of Millikan's experiment was to find the absolute charge. (The mass of the electrons would have been immaterial in the oil drop.). The technique was to measure the charge on many oil drops and find the GCD.
 
  • #4
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How do I find out? Do I use the mass, molar mass and Avogadro's number?
 
  • #5
haruspex
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How do I find out? Do I use the mass, molar mass and Avogadro's number?
You have found the charge, what's the problem? If you like, you can estimate the number of excess electrons in the drop.
 
  • #6
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It just doesn't seem right? I get the no. of electron in the drop to be 8 (rounded).
 
  • #7
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Am I not supposed to find the actual electric charge of an electron? Isn't that he point of the experiment?
 
  • #8
Simon Bridge
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No, I think that was obtained through a much earlier experiment by someone else.
Oh you are right - that was J. J. Thomson who did the q/m thingy.
I keep getting them mixed up.

You still can't do it with just one drop.

Daniel Luo said:
It just doesn't seem right? I get the no. of electron in the drop to be 8 (rounded).
1. you were not asked to find the charge of an electron.
2. I doubt you found the total number of electrons in the oil drop ... recall: neutral oil has electrons (-) and atomic nuclei (+) in equal amounts. Since your oil drop is negatively charged ...............
 
  • #9
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So I should just go ahead with my answer which corresponds to approx. the charge of 8 electrons?
 
  • #10
Simon Bridge
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According to what you wrote in post #1, you have only been asked to determine the radius, mass, and charge, for the whole drop. That's all. Nothing about determining anything for the electron nor anything about finding out how many electrons there are on the drop.

In order to discover the charge of an electron by Millican's method, you need to repeat the experiment for a great many different oil drops.
http://webphysics.davidson.edu/appl...ts/pqp_errata/cd_errata_fixes/section4_5.html
 
  • #11
ehild
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So I should just go ahead with my answer which corresponds to approx. the charge of 8 electrons?
That is quite a good result, the charge is not much more than 8 electrons. Usually, the oil drop gains more than one electrons in that experiment. Anyway, when I did it as a student during Laboratory Practice, I got usually half-integer results :biggrin:

ehild
 

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