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Millikan oil drop experiment charge determination

  1. Oct 13, 2016 #1
    Howdy y'all!

    If you could help with the following question, my physics class and I would be extremely grateful.

    A charged oil droplet is suspended motionless between two parallel plates (d=0.01m) that are held at a potential difference V. Periodically, the charge on the droplet changes as in the original oil drop experiment. Each time the charge changes, V is adjusted so that the droplet remains motionless. Here is a table of recorded values of the voltage:

    i. 350 V

    ii. 408.3 V

    iii. 490 V

    iv. 612.5 V

    From the data above, determine the charge on the dorplet for case (i) above. What assumptions do you need to make? (Hint: the ratio of voltages = ?)

    Thanks a ton, we've been stumped.

    I'm gonna be honest here, me and 2 friends have been working on this for about 4 hours, and we don't really have any substantial work to show. Any help would be great. Thanks.
     
  2. jcsd
  3. Oct 13, 2016 #2

    Charles Link

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    I can give you a hint, but I haven't solved it myself yet: ## 350* \, Q_1=408.3* \, Q_2=490* \, Q_3=612.5* \, Q_4 ##. ## Q_4<Q_3<Q_2<Q_1 ##. Find some ## Q_o ## so that ## Q_4=n_4 \, Q_o ##, ## Q_3=n_3 \, Q_o ##, etc., ## n_4, n_3,... ## integers (hopefully small ones). Sorry, I edited a couple of times because I read it incorrectly.
     
    Last edited: Oct 13, 2016
  4. Oct 13, 2016 #3

    Okay, this makes sense. However, where can we go from here? Is there anyway to solve for the integers?
     
  5. Oct 13, 2016 #4

    Charles Link

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    I have it, but I'm not allowed to give the solution. I can give you a hint though. The smallest number, ## Q_4 ## is greater than 3. Another hint is the numbers are exact enough, that I think the data is probably simply constructed by the professor as a good learning exercise. One additional hint=let ## Q_4=n_4 ## (Ignore the ## Q_o ## part mentioned previously.) Please let us know if you figured out the answer.
     
    Last edited: Oct 13, 2016
  6. Oct 13, 2016 #5
    Okay, so we talked it out a little bit. So does this mean that the answer is 7e=1.12x10^-18 coulumbs?
     
  7. Oct 13, 2016 #6

    Charles Link

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    Yes. One additional question for you=what did you get for the other 3 integers? And were the calculations almost exact?
     
  8. Oct 13, 2016 #7
    We got all the integers as 7,6,5, and 4. Thank you so much! We just screamed in excitement and relief.
     
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