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Millikan Oil drop experiment

  1. Nov 13, 2016 #1
    Millikan Oil drop experiment.
    For my current lab, we are recreating the milian oil drop experiment to measure the charge of an electron. However, we are using 1-micron diameter latex spheres in place of oil drops.

    I am having difficulty deriving an equation for the speed of the drop. Only the linear part of air resistance is taken into account. Without an electric field the particle takes about 15 seconds to fall a distance of 15mm. With the current applied, different spheres travel at different velocities dependent on their charge. And the same sphere moves faster when the field is applied in the direction of gravity.

    Attempt at solution:
    v1 is rising against gravity
    v1 = [qE - mg] / (6*pi*eta*r)
    where eta is viscosity of air
    v2 is when field is reversed and aids gravity
    v2 = [qE + mg] / (6*pi*eta*r)

    ---> How do I find qE if I know the voltage difference between the two plates is 50V. What about 100V or 150V?

    ---> How do I measure the viscosity of air to use in the equation for v1 and v2?

    ---> Does the charge “q” on each sphere have to be an integer multiple of e, the charge of a single electron?
    Last edited: Nov 13, 2016
  2. jcsd
  3. Nov 13, 2016 #2


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    How is the electric field related to the voltage across the capacitor plates? Remember, the field between the plates may be assumed to be uniform.
    Use your zero field data.
  4. Nov 13, 2016 #3
    So apparently the gap between the plates d is 15 mm. I assume the dimensions of the plates are larger than 15 mm. In that case the electric field near the center of the plates can be taken to be E = V/d across the full width of the gap.
  5. Nov 13, 2016 #4


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