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Millikan Oil Drop

  1. Oct 21, 2007 #1

    In a Millikan experiment the distance of rise or fall of a droplet is .60 cm and the average time of fall (ie field off) is 21.0 s. With the field turned on, the observed successive rise times are 46.0, 15.5, 28.1, 12.9, 45.3 and 20.0 s
    a) Prove that charge is quantized.
    b) If the oil density p=858 kg / m^3 and the viscosity of air is 1.83 x 10^-5 kg /m*s, find the radius, volume and mass of the drop used in this experiment.

    2. Relevant equations

    qE = mg + EV

    f = 6 pi E r v = mg

    3. The attempt at a solution

    I started this by drawing the force diagram for no field. The charge makes no difference here and you assume that the drop falls at terminal velocity through the whole .6 cm. So:

    v = .6cm / 21 s = .029 cm/s

    In order to prove that charge is quantized, must show that the ratio of charges are approximately small whole numbers. So need to find the charge of each drop from their rise times. This is where I start to struggle. Clearly the larger the negative charge, the faster the drop will rise, resulting in a shorter rise time.

    For the second part, I think I can use vel = 2 r^2 p g /9 (visc. air)
    but then why are there several different rise times for a given oil drop? Is this because it is taking on different numbers of electrons each time? If it is, then the velocity should change, so then the radius would change, but that doesn't make sense. I've looked up this experiment all over the web for further info, but few descriptions mention air viscosity or calculating mass from different rise times.

    Any hints or explanations would be helpful. Thank you!
  2. jcsd
  3. Oct 22, 2007 #2


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    Science Advisor

    Use the drop time to get r.
    Use the rise times to get q, knowing r which is constant for the same drop.
    Use the DIFFERENCE in q from different rises, to get the small numbers.

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