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Millikan's Experiment (pt. 2)

  1. Nov 19, 2004 #1
    I finally know how he did it! Knowing the mass, voltage, and radius, he used:

    q = mgr / v to find the elementary charge. He averaged out all his findings to get the best result

    But...if the mass of the drop was known, how would one know how many electrons the oil drop had? For example, if all oil drops had no less than 3.2x10^-19C of charge, how would Millikan assume that e = 1.6x10^-19C? It's just kinda confusing.
  2. jcsd
  3. Nov 20, 2004 #2
    We you actually perform the experiment, you get values of q that are multiples of e.
  4. Nov 20, 2004 #3
    when u perform the experiment many times, u get different amounts of charge on the drop for each exp. comparing these different values, you should be able to find the lowest value (lets say it's 3.2 x 10^-19C). then with this lowest value u find, for each of the other readings, how many times are they multiples of this lowest value. u should be able to get clean clear quotients that are either whole numbers or simple fractions with denominators such 2, 4 and so on. if u get a fraction, say, 1.5 somewhere, then u know that 3.2 x 10^-19C cant be the fundamental charge cos there is a mulitiple of 0.5 that u cant account for. this extra 0.5 will suggest that the lowest charge is 0.5 x 3.2 x 10^-19C.

    extrapolating from here and doing the experiments many times over will get u the fundamental charge of 1.6 x 10^-19C.
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