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Forces acting on charged oil drops

  1. Sep 6, 2006 #1
    Hi guys. I'm a new member, just joined today. I've been looking for the answer to the following question for hours, but am unable to do so. Please try to help ASAP! Thanx a million. The question is:

    When a small sphere moves through the air with a low speed v it experiences a resitive force given by kv where k is a constant. If the oil drop carries a charge of magnitude q, show that when the potential difference between the plates is V2 the speed v with which the drop moves upwards is given by

    v=q/kd x (V2-V1)
     
  2. jcsd
  3. Sep 6, 2006 #2

    Doc Al

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    Welcome to PF!

    The rule here is that to get help you must show your work. :smile:

    But here's a hint: For the sphere to move at constant speed, what must be the net force on it?
     
  4. Sep 6, 2006 #3

    Andrew Mason

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    What is V1? (are we to assume it is the potential for which v = 0?)

    AM
     
    Last edited: Sep 6, 2006
  5. Sep 6, 2006 #4

    Doc Al

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    I assume that you meant to write the potential difference as V2 - V1, not V2.
     
  6. Sep 6, 2006 #5

    Andrew Mason

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    I think V1 has to be the voltage for which v = 0. Otherwise you would have to know the mass of the oil drop.

    AM
     
  7. Sep 7, 2006 #6

    Doc Al

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    Not sure what you mean. What determines the terminal velocity is the strength of the electric field, which depends on the potential difference. (Assuming that we can ignore gravity, the mass is not needed.)
     
  8. Sep 7, 2006 #7

    Andrew Mason

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    I don't think we can ignore gravity. In order to keep the oil drop still, you will need to apply an electrical force to the oil drop to balance gravity, so some potential difference to the plates is needed. That is V1. The speed will be determined by the electric force (qE=qV/d) in excess of that.

    AM
     
  9. Sep 7, 2006 #8

    Doc Al

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    OK. I'll buy that. :smile:
     
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