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Milikan's Oil Drop

  1. May 10, 2008 #1
    Milikan's oild drop experiment is a famous experiment to dertermine the charge on an electron. In order to perform the calculations it is necessary to solve the equatuion that describes the speed of small drops of oil falling under gravity. This equation is:

    dv/dt = 9.8 - Cv/D^2

    Where C= 3.1*10^-6 m^2/s and D is the diameter in metres. The oil drop is assumed to start from rest.

    Solve the differential equation and hence show that the eventual (terminal) velocity in terms of D is v(t)=9.8D62/C

    I have tried several times to integrate this question without success, I think I am making the mistake right at the start. Any help would be greatly appreciated.
  2. jcsd
  3. May 11, 2008 #2
    Can you show your work? That way we'll be able to spot your error.
  4. May 11, 2008 #3

    dv/dt = 9.8 - Cv/D^2
    1/9.8 - Cv/D^2 dv = dt
    then i integrate both sides

    1/(-C/D^2) ln (9.8- Cv/D^2) = t + c
    - D^2/C (ln 9.8 - Cv/D^2) = t+c
    ln (9.8 - Cv/D^2) = -Ct/D^2 + c
    9.8 - Cv/D^2 = e^(-Ct/D^2 + c)

    and here i just get a little lost... am I on the right track?
  5. May 11, 2008 #4


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    Hi shad0w0f3vil! :smile:

    Yes, the integration looks fine. :smile:

    Just plug t = ∞ into it, and solve.

    Alternatively, you could get the terminal velocity just by looking at the original dv/dt = 9.8 - Cv/D². :rolleyes:
  6. May 11, 2008 #5
    ok I will have a better look at it tomorrow and get back to you
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