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Millikan oildrop experiment: archimedes' principle

  • Thread starter Looh
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  • #1
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Hello!

My physics teacher gave me an assignment to work out (theoretically) how Millikan's oildrop experiment works.

The simple principle of the experiment (as far as I know):
[itex]E = \frac{F}{Q}[/itex]
[itex]F = mg[/itex]
[itex]mg = qE \rightarrow q = \frac{mg}{E}[/itex]

However, after reading a bit on Wikipedia it seems as if there is more involved in order to receive an acceptable value of the particle.

Archimedes' principle:
[itex]F = pVg[/itex]

Volume of a sphere (the oildrop):
[itex]V = \frac{4πr^3}{3}[/itex]

The weight of the drop:
[itex]w = \frac{4πr^3}{3}(p - p_{air})g[/itex]

I don't understand why p is being subtracted by p_air, what density does p refer to?


I'm sorry if I'm ambiguous or outright wrong, I'm not particularly good at physics.
You can read more about the method I'm trying to understand here: http://en.wikipedia.org/wiki/Oil_drop_experiment#Method

Thanks in advance!
 

Answers and Replies

  • #2
Borek
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It is ρ (Greek letter rho) not p.

ρ is a density of the oil.
 
  • #3
HallsofIvy
Science Advisor
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In other words, [itex]\frac{4\pi r^3}{3}\rho[/itex], the volume of the oil drop times the density of oil, is the actual gravitational force on the oil drop. [itex]\frac{4\pi r^3}{3}\rho_{air}[/itex] is that same volume times the density of oil. Their difference is the "net weight", the actual weight of the oil minus the weight of the air it displaces.
 
  • #4
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It is ρ (Greek letter rho) not p.

ρ is a density of the oil.
In other words, [itex]\frac{4\pi r^3}{3}\rho[/itex], the volume of the oil drop times the density of oil, is the actual gravitational force on the oil drop. [itex]\frac{4\pi r^3}{3}\rho_{air}[/itex] is that same volume times the density of oil. Their difference is the "net weight", the actual weight of the oil minus the weight of the air it displaces.
Ah, yes, I understand now! Many thanks to both of you! :smile:
 

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