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Millikan oil drop problem

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data

    A 1.50x10-14 kg oil drop accelerates downwards at a rate of 1.80 m/s2 when placed between two horizontal plates that are 9.40 cm apart. The potential difference between the two plates is 980 V. Determine the magnitude of the charge on the oil drop.

    2. Relevant equations
    |E|=V/d
    Fnet=ma
    Fg=mg
    Fe=q|E|

    3. The attempt at a solution

    So first I find the strength of the electric field.

    |E|=V/d
    =(980 V)/(0.094 m)
    |E|=1.04x104V/m

    Then I find the net force.
    Fnet=ma
    =(1.50x10-14kg)(-1.80 m/s2)
    =-2.70x10-14N

    Then I find the force of gravity on the drop.
    Fg=mg
    =(1.50x10-14kg)(-9.81m/s2)
    =-1.47x10-13N

    Here is where I think I am going wrong. I try to find the electric force.
    Fnet=Fg+Fe
    Fe=fnet-Fg
    =(-2.70x10-14N)-(-1.4715x10-13N)
    =1.20x10-13N

    Then I find the charge.
    Fe=q|E|
    q=Fe/|E|
    =(1.2015x10-13N)/(1.04x104N/C
    =1.16x10-17 C

    Now this answer is obviously wrong. Where am I messing up?
     
  2. jcsd
  3. Nov 15, 2011 #2

    gneill

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    Staff: Mentor

    What makes you think that your answer is "obviously wrong"?
     
  4. Nov 15, 2011 #3
    Doesn't the charge have to be an integer multiple of the elementary charge (1.60x10-19C) ?
     
  5. Nov 15, 2011 #4

    gneill

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    Staff: Mentor

    Sure. But what experiment has perfect accuracy?

    Divide by the fundamental charge and see how close you come to a round number.
     
  6. Nov 15, 2011 #5
    Okay, so then my original answer is correct?
     
  7. Nov 15, 2011 #6
    well if you will ever do the actual Millikan experiment you will find that you will not exactly get 1.6x10^-19, but around 1.6x10^-19 ± some error x10^-17
     
  8. Nov 15, 2011 #7

    gneill

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    Staff: Mentor

    Your method and numbers look fine to me.
     
  9. Nov 15, 2011 #8
    Great, thanks for the help!
     
  10. Feb 24, 2012 #9
    What happen If we place negative charge between parallel opposite charge plate ?
    answer is negative charge attract toward positive plate...

    Then why Oil drop is suspended btw opposite plates in Millikan's oil drop experiment??
     
  11. Feb 24, 2012 #10

    gneill

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    Staff: Mentor

    What forces are operating on the oil drop? Draw a Free Body Diagram.
     
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