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Electric Fields in Millikan's experiment

  1. Dec 14, 2006 #1
    In Millikan's experiment, an oil drop of radius 0.9493 µm and density 0.876 g/cm3 is suspended in chamber C (Figure 22-14) when a downward-pointing electric field of 1.92 105 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign.)

    So, to do this problem, i know that you have to use (4/3)pi*r^3 to find the volume. Then you times the volume by the density (in meters). That finds the mass, which you then times by gravity. That should be the answer but its wrong.

    My work- Volume=3.5834e-18
    (3.5834e-18)(8.74e-7)= 3.139e-24
    (8.139e-24)*9.81=3.07944e-23
     
  2. jcsd
  3. Dec 14, 2006 #2

    Doc Al

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    Staff: Mentor

    What this will give you is the weight of the drop. That's a needed step in the solution, but what's next? What role does the electric field play in this problem? Hint: What forces act on the suspended oil drop?
     
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