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Terminal Velocity and Millikan's Oil Drop Experiment

  1. Sep 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose I drop a ball from high enough in the sky that, at some point, it reaches terminal velocity. At this point, the ball is no longer accelerating. It is travelling at a constant velocity. Does this mean that at terminal velocity, the force of gravity on the ball is equal to the force of air resistance (viscousity)? A net force of zero?


    If so, then what is going on in Millikan's Oil drop experiment? A particle drops between two plates. Suppose the plates are not electrically charged. Then at some point, the falling particle will reach terminal velocity, at which point the force of gravity on the particle will equal the force of air resistance.

    However, when the plates are then electrically charged, the particle is then decellerated (assuming the particle is negative and the top plate is positive). When the particle comes to a complete stop, the electrical force attracting the negative particle upward is exactly equal to the force of gravity pulling the particle downward. So once again, a net force of zero?

    So is it that as the electrical charge of the plates are increased, the total upward force on the particle is greater than the total downward force? And this would be why the particle is slowing down (or accelerating upward) until coming to a complete stop.

    I am trying to imagine to "quantify" the relationship that is occuring between the downward force of gravity, the upward force of air resistance, and the upward force of the positively charged top plate.

    Intuitively, if feels as though the downward force of gravity always remains the same (9.8m/s^2), the air resistance decreases while the electric charge of the plates increase. But if the force of air resistance decreased by the same quanity as the increase of the force of the charged plates, then the particle would remain falling at a constant speed. Therefore, I would guess that there is some sort of relationship between the upward forces as one force increases and the other decreases so as to slow the particle down to a complete stop.

    What would be that relationship between the upward forces of air resistance and electrical charge of the plate?



    2. Relevant equations



    3. The attempt at a solution

    Not homework, just self study.
     
  2. jcsd
  3. Sep 14, 2012 #2

    ehild

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    The air resistance depends on the speed of the particle, proportional to the square of the speed Fr=kv2
    http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object

    The applied upward electric field slows down the particles. They are decelerated, have upward acceleration, so an upward resultant force qE-mg- kv2>0 till the particle moves. If the particle is in rest, v=0 and qE=mg.

    ehild
     
  4. Sep 14, 2012 #3
    Thanks ehild, is this a fair assesment?...

    When the particle is in free fall, at first the downward force of gravity is greater than the upward force of air resistance so there is a positive net force in the downward direction resulting in acceleration of the particle downward. mg>kv^2

    Then as velocity increases, the upward force of air resistance increases until there is a net balance of upward and downward force on the particle, at which time the particle travels downward at a constant speed (zero acceleration). mg=kv^2

    Then if the charged plate is turned on, the upward force on the particle will be greater than the downward force, so the particle will accelerate upward (the speed of fall will slow down). mg<kv^2 + qE

    And then when the particle come to rest, the upward force of air resistance decreases to zero [kv^2 = 0], and once again there is a an upward and downward net force of zero on the particle, but now, instead of mg=kv^2, it's mg = qE.
     
  5. Sep 14, 2012 #4
    p.s. just wanting to confirm if that assesment is accurate so as to ensure I actually have an intuitive idea of what is going on, in terms of the physics, during that portion of the Millikan experiment.
     
  6. Sep 14, 2012 #5

    ehild

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    Your explanation is correct.

    ehild
     
  7. Sep 15, 2012 #6
    appreciate it ehild.
     
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