1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Millikan oildrop experiment: archimedes' principle

  1. Aug 25, 2012 #1

    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!
  2. jcsd
  3. Aug 25, 2012 #2


    User Avatar

    Staff: Mentor

    It is ρ (Greek letter rho) not p.

    ρ is a density of the oil.
  4. Aug 25, 2012 #3


    User Avatar
    Science Advisor

    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.
  5. Aug 25, 2012 #4
    Ah, yes, I understand now! Many thanks to both of you! :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook