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Homework Help: Confused by the Question (oil drop charge)

  1. Mar 14, 2007 #1
    This is a short question, from my understanding, to determine the charge of an object we can use the equation [tex]q=\frac{mg}{E}[/tex] , where m is mass of object and E is voltage of electrical field. Im asked to find the charge on a drop of oil suspended in the field. I was given both the mass and the applied voltage of the field, but i was also given distance between the field and also told that the lower plate is "at a lower potential". How do the other pieces of information contribute to the charge since they are not factors in the equation? i also have to find number of excess of deficit electrons, how do i do that? (perhaps by divide by charge of electron?)

    the givens were oil drop mass: 2.6 x 10^15 kg, plate separation = .005 m ,potential difference between plate = 270 V , and it also says the lower plate is at a lower potential.

    to determine E, i used E = v / d , where v is voltage, and d is plate separation, then i used the above equation to determine q.
     
    Last edited: Mar 15, 2007
  2. jcsd
  3. Mar 15, 2007 #2
    This is Millikan's Oil Drop Experiment - where the oil drop is 'suspended' between parallel plates. Normally the electric field is: E=N/C, ie force per coulomb of charge. In parallel plates the field is basically constant, and can instead be calculated by E=V/d. So if you have the distance and the voltage that gives you the electric field E. Stick that into your first equation and hopefully you'll get an integer multiple of electron charge :D
     
  4. Mar 15, 2007 #3
    i calculated the E using the given voltage distance between plates, but the q i got was [tex]-4.7\cdot10^{-19}[/tex] . i used -9.8 for my g value.

    When i divide this number by the charge of an electron , [tex]e=1.60\cdot10^{-19}[/tex] , i get 2.93 !

    What went wrong?
     
    Last edited: Mar 15, 2007
  5. Mar 15, 2007 #4

    andrevdh

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    Well, we don't have the given values, so we can't tell you where you went wrong.
     
  6. Mar 15, 2007 #5
    okay, i added the givens.
     
    Last edited: Mar 15, 2007
  7. Mar 15, 2007 #6
    the negative value of the charge would indicated an excess of electrons, so i thought dividing by the charge of an electron would give me the number of excess electrons, not sure if thats right however.
     
  8. Mar 15, 2007 #7
    still need help one this one...
     
  9. Mar 16, 2007 #8
    An electron is negatively charged so I don't think it's a problem getting a negative value for the charge. As for the answer, 2.93 is as close to 3 as makes no difference (considering how small a mass you're talking about) so I'd say thats all ok...You're supposed to get integer multiples of the electron charge so it's fine.

    I don't know about the excess electrons sorry.
     
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