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Millikan Oil Drop Experiment Question

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A 1.50 x 10^-14 kg oil drop accelerates downward at a rate of 1.80 m/s^2 when placed between two horizontal plates that are 9.40 ch apart. The potential difference between the two plates is 980 V. Determine the magnitude of the charge on the oil drop.

    m = 1.50 x 10^-14 kg
    a = 1.80 m/s^2 (down)
    d = 9.40 cm = 9.40 x 10^-2 m
    V = 980 V


    2. Relevant equations
    Fnet = Fg + Fe
    Fnet = ma
    Fe = Eq
    Fg = mg
    E = V/d


    3. The attempt at a solution
    Fnet = Fg + Fe
    Fe = Fnet - Fg
    = ma - mg
    = ((1.50 x 10^-14 kg)*1.80 m/s^2)) - ((1.50 x 10^-14 kg)(9.81 m/s^2))
    = -1.2015 x 10^-13 N

    E = V/d
    = (980 V) / (9.40 x 10^-2 m)
    = 10425.53191 V/m

    Fe = Eq
    q = Fe / E
    = (-1.2015 x 10^-13 N) / (10425.53191 V/m)
    = -1.1524592 x 10^-17 C

    The charge on the oil drop is approx. -1.15 x 10^-17 C ?????????
     
  2. jcsd
  3. Jan 10, 2013 #2

    TSny

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    Gold Member

    Looks ok to me. Does the charge correspond to an integer number of electrons on the drop?

    {Accurate analysis of the Millikan Oil Drop Experiment would take into account the viscosity of the air and the buoyant force acting on the drop, and involves measuring terminal velocities rather than accelerations. See http://en.wikipedia.org/wiki/Oil_drop_experiment . But what you did looks correct based on the information given.}
     
  4. Jan 10, 2013 #3
    If I divide that charge by 1.60 x 10^-19 C I get 72.0207 electrons. Does it have to be a whole number?
     
  5. Jan 10, 2013 #4

    TSny

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    Yes! At least if the data is good! That's one reason the experiment is considered a classic. Millikan found the net charge of the drops to be "quantized" as integer multiples of a fundamental amount of charge e. 72.02 is close to an integer. The discrepancy can be chalked up as "experimental uncertainty in the data".

    [EDIT: Millikan reported net charges on the drops ranging from 1*e to 136*e. See One of Millikan's Original Papers ]
     
    Last edited: Jan 10, 2013
  6. Jan 15, 2013 #5
    Okay, I understand! Thank you for your help!
     
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