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Millikan Experiment

  1. Feb 19, 2009 #1
    1. The problem statement, all variables and given/known data
    You are given twelve cans (one of which is empty) which are filled with a certain number of ball bearings. Using nothing more than a balance, calculate the mass of a single bearing

    2. Relevant equations
    Total Mass=Mo+n(mo)
    Where total mass=mass of ball bearings
    Mo=smallest mass
    n=number of bearings
    mo= mass of each bearing

    3. The attempt at a solution
    I realize the equation above is a little confusing (formatting in PF is not easy). The way I attempted this problem was by first trying to organize the mass of the twelve cans (after subtracting the mass of the empty can, of course). I found the smallest difference between two masses, which was 2.3 g. From this point, however, I'm not sure what to do. I could possible divide each mass by 2.3 g, giving me the theoretical number of ball bearings. But I don't see how I could verify my answer. Any help is appreciated. Thank you.
  2. jcsd
  3. Feb 19, 2009 #2


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    Homework Helper

    You probably have the answer with 2.3 g. Do check to make sure all the can masses (less empty can mass) are multiples of 2.3. If not, try half of 2.3, then a third of 2.3 until it works.

    Millikan continued looking for smaller charges for 20 years before publishing your results!
  4. Feb 19, 2009 #3
    The problem I was having, however, was dealing with error. Assuming the mass is 2.3 g, then these cans have a lot of balls in them (up to 50). 2.3 g does not go into each mass evenly, but that doesn't mean I should look for a smaller number (right?). Obviously, 0.1 g would be a factor of each but that's not correct.
  5. Feb 19, 2009 #4


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    Yes, some judgment about accuracy is certainly needed! Just thinking ...
    The numbers are given to one decimal place, so an accuracy of plus or minus 0.05 g on the given measurements is suggested. If one of the cans measures 25 g, we could note that 24.95/2.3 = 10.8 and 25.05/2.3 = 10.9, so 2.3 g is not a possible marble mass. But the 2.3 g comes from subtracting two numbers that are each +/- .05, so it could be out by up to .1 g and 25/11 = 2.27 g is a possible marble mass.

    Maybe playing around like this with all the given masses would turn up a marble mass that works to within the error in measurement. Curious, if the accuracy of measurement is really +/- .05 then the LARGEST can mass yields the finest suggested marble mass because the error gets divided by a large number of marbles. Usually you get a % error - but I don't think you do with mass balances.
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