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Millikan's experiment with a twist

  1. Mar 18, 2006 #1
    The Millikan's oil drop experiment is doen with the electric field horizontally, rather than vertically, giving the charged droplets an acceleration in the horizontal direction. The result is that the droplet falls in a straight line which makes an angle theta with the vertical. Show that sin(theta)= qE/(b*v), where v is the terminal velocity. (bv is the drag force upward).

    What are the forces acting on the oil? Just the horizontal force, right? So qE=ma, how do I get it to look like sin(theta)= qE/(b*v)?

    Last edited: Mar 18, 2006
  2. jcsd
  3. Mar 18, 2006 #2
    What is b? Also wouldn't gravity also be acting on the oil drop since it's falling?
  4. Mar 18, 2006 #3
    bv is the upward force, and b=6*pi*n*a, where a is the radius of the oil drop and n is the coefficient of viscosity of the air.
    So there is also a velocity downward?
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