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Milikan's Oil Drop

  • #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.
 

Answers and Replies

  • #2
354
0
Can you show your work? That way we'll be able to spot your error.
 
  • #3
ok

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?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
249
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:
 
  • #5
ok I will have a better look at it tomorrow and get back to you
 

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