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Millikan's oil drop experiment

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Revered members,
    Kindly see my attachment.

    2. Relevant equations



    3. The attempt at a solution
    When the charged oil drop is balanced, electric force = weight of oil drop.
    Eq = mg
    Eq = F + U, where F is the viscous force and U is the upthrust force. Since the oil drop attains terminal velocity mg = F + U.
    I know viscous force always act upwards. But i can't figure out the direction of viscous force when oil drop is balanced. Please help, members
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 2, 2011 #2

    NascentOxygen

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    Staff: Mentor

    If there is no relative movement (between a body and the medium) there will be zero viscous friction force.
     
  4. Oct 2, 2011 #3
    Viscous force is directly proportional to velocity.
    Viscous force only arises when there is movement of the oil drop (eg. when the electric field is removed and the drop falls. Here viscous force will continue to increase until upthrust + viscous force = weight. Then, resultant force being zero, the drop will fall with terminal velocity)
    Also, viscous force always opposes direction of motion.
     
    Last edited: Oct 2, 2011
  5. Oct 2, 2011 #4
    Thanks a lot NascentOxygen and Physics S16 for your replies.
    What about the direction of viscous force when terminal velocity is achieved? Will it be upwards?
    I think upwards from the statement of Physics S16 that viscous force always opposes direction of motion
     
  6. Oct 2, 2011 #5

    NascentOxygen

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    Staff: Mentor

    Yes, always exactly opposite to the direction of [relative] motion.
     
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