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Milikin Drop experiment change of charge with E-Field

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data

    In my honours lab we are performing the Milikin drop experiment, and we've come into a slight problem, where when calculating the radius of the drops and the charge on the drops, the charge seems to vary with change in E-field.

    When performing the experiment we first needed to find a charge drop after ionizing a spray of drops by applying an E-field and seeing which drops seem to accelerate in the opposite direction of the field.

    When a drop is aquired we use the voltage applied to move the drop to the top and turn off the voltage and first allow the drop to free fall and record its motion downwards. This is what we called the free fall velocity which allows us to calculate the radius of the drop using equation (1)

    After a few trials of the free fall we move on to turning on the voltage and record a video of the down velocity and up velocity at the same voltage. Then we change the voltage and do the same, for a few times. Then by knowing our radius we are able to calulate the electro force qE. When plotting a graph of v vs E or v vs potential we are able to view if our data is reliable or not. If viewed reliable we would calculate the values in the following equations


    2. Relevant equations

    1. [itex]\frac{4\pia^{3}\rho g}{3}[/itex] = [itex]6\pi a\eta^{'}v[/itex]


    2. [itex]\frac{4\pia^{3}\rho g}{3} - qE[/itex] = [itex]6\pi a\eta^{'}v[/itex]

    [itex]\eta^{'}[/itex] = [itex]\frac{\eta}{1+\frac{b}{pa}}[/itex]

    b = 0.00617 torr-cm

    p = ambient air pressure

    a = radius

    eta = viscosity of air

    eta prime = effective viscosity of air

    rho = density of the oil

    q = charge of the drop which should be an interger value

    E = electric field

    g = gravity

    v = velocity of drop

    3. The attempt at a solution

    The attempt at the solution is in origin, so it's kind of hard to post, but it's more about why the drop seems to change charge with different E-fields, even though it is not colliding with other drops.

    Sorry it's kind of long, hope someone can help
     
  2. jcsd
  3. Oct 14, 2011 #2
    I forgot to say that we also found the voltage at which the drop balances in one spot
     
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