Millikan oildrop experiment: archimedes' principle

[Looh]

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):
E = \frac{F}{Q}
F = mg
mg = qE \rightarrow q = \frac{mg}{E}

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:
F = pVg

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

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

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!

 

[Borek]

It is ρ (Greek letter rho) not p.

ρ is a density of the oil.

 

[HallsofIvy]

In other words, \frac{4\pi r^3}{3}\rho, the volume of the oil drop times the density of oil, is the actual gravitational force on the oil drop. \frac{4\pi r^3}{3}\rho_{air} 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.

 

[Looh]

It is ρ (Greek letter rho) not p.

ρ is a density of the oil.

In other words, \frac{4\pi r^3}{3}\rho, the volume of the oil drop times the density of oil, is the actual gravitational force on the oil drop. \frac{4\pi r^3}{3}\rho_{air} 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!