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Homework Help: Millikan's oil drop experiment homework

  1. Jul 7, 2006 #1
    Hi everyone I'm stuck with a few questions hopefully I can get some assistance.

    1) A student repeating Millikan's oil drop experiment to measure electronic charge used an incorrectly calibrated voltmeter and obtained the following values for charges on the drops:

    Charges on drop / 10^-19C

    Using these results what value should he take for the electronic charge, in C?

    2) If one of the slits of the standard Young's double slit demonstration of interference in light is painted over so that it only transmit only half the light intensity, why does the dark lines increase in density and bright lines decrease in density?

    3) One junction of X of a thermocouple is placed in melting ice at 273K and the other junction Y in steam at 373K. The e.m.f measured is 1.0mV. Now Y is transferred to a bath whose temp is 398K. Assume variation of e.m.f with temp diff is linear, the e.m.f recored will be?

    4) See the attachment. The resistance between A and C is 8ohm. What is resistance R?

    For 1, I know charges exist in multiples that they are discrete. I assumed all the values given to be from a drop since they stated on oil drop. How then can I find the charge?

    For 2, I know bright lines will decrease in intensity as at the bright lines, the total intensity is lower compared to 2 equal intense waves constructive interfering. But why does the dark fringe increase in intensity? Does it mean it becomes darker? Why so?

    3) 1.0mV = k(373-273)
    E = k(398-273)
    Solving, I get the value of E to be 0.125. Answer given was 1.25.

    4) I don't know how to start at all. Using kirchoff's law doesn't seem to provide anything. Any advice? Answer 4ohms.

    Sorry for the long post hope someone could lend a helping hand! thanks!

    Attached Files:

  2. jcsd
  3. Jul 7, 2006 #2


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    You must find a charge such that all the numbers you listed are *integer* mulltiples ofthat fundamental charge (there is actually an infinite number of possibilities but usually one consider the largest possible value such that all measurements are integre multiple of thatvalue).

    The first thing to try for for the fundamental charge is obviously the smallest measurement, 3.9. But of course, that does not work since the next value, 5.2 is not a multiple of 3.9.

    So 3.9 itself is a multiple of the fundamental value.

    Obviously, if all those numbers are multiple of the fundamental charge, their *difference* will be a multiple of the fundamental charge. So the next thing to try is to take in your list th two values the closest together and use the *difference* between the two as the fundamental charge. Is it then possible to write all your values as multiple of that difference? If so, that will be your fundamental charge (again, that's not the only solution but that will be the largest possible value for a possible fundamental charge).

  4. Jul 8, 2006 #3
    Thanks Pat I got it..

    Any help for the rest of the questions? Much appreciated thanks!
  5. Jul 8, 2006 #4
    Suppose the amplitudes are x and y. Then as y decreases x+y decreases but x-y increases. And since Intensity is proportional to Amplitude^2, so Intensity decreases in first case and increases in 2nd.

    And check your calculation for 3.
    Last edited: Jul 8, 2006
  6. Jul 9, 2006 #5
    4)You don't need to use KVL for this. Think that you are connecting an imaginary battery between points A and C. You will find that the resistancce of 1 ohm can be removed from the circuit (why ?).
    Now you should be able to easily apply the laws for series and parallel combinations to get your answer.
    Do you follow ?

  7. Jul 10, 2006 #6
    Thanks for the help.

    For Q3, I checked and I still got 0.125V. Did you managed to get the given answer?

    I'm not very good in electricity hopefully you can help me out further. It can be removed because B would not be connected so it'll be an open circuit there so no current flows?

    I've got another question here regarding electricity as well (see attachment). The question is what is the reading of the volmeter? I drew the loops in myself and the currents I1 I2 and I3. But how do I go around solving the voltmeter reading?

    Many thanks again in advance.

    Attached Files:

    • 1.GIF
      File size:
      2.3 KB
  8. Jul 10, 2006 #7
    Correct. Hope you got the answer :wink:
  9. Jul 10, 2006 #8
    Thanks Arunbg. Any idea on how is it possible to solve for the 3 currents in the above circuit? Or is there any easier or more conventional way of solving this question?
  10. Jul 10, 2006 #9
    The only sure way is KVL . Use it for the two small loops and remember
    However, I'm not sure about the voltmeter. Is its resistance given or is it considered to be an ideal voltmeter (resistance tending to infinity) ?

    3) I suggest you check the units.

    Last edited: Jul 10, 2006
  11. Jul 10, 2006 #10

    Doc Al

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    I don't understand that last diagram. Is that circled V supposed to be a voltmeter? Stuck in the middle of the circuit?
  12. Jul 11, 2006 #11
    Oops I apologise for not being clear! The question is what is the reading on the voltmeter? The question didn't say anything about the resistance of the voltmeter. I still can't solve it as even if I had found the currents, how do I find the voltmeter reading?
  13. Jul 11, 2006 #12

    Doc Al

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    Sounds like the only thing you can do is treat it as an ideal voltmeter (with infinite resistance), as arunbg pointed out. In that case, what's the value of I2?

    You just need to find the total voltage drop across its terminals. Hint: Add up the voltage drops around the loop.
  14. Jul 11, 2006 #13
    In order to do that, the currents must still be solved in order to find the drop in potentials right? I have

    6 = 5(I3) + 10(I1) and
    2 = 5(I3)

    I solved for I1 to be 2/5A. Is this correct? Sounds weird as if it is so, I3 is also 2/5. I suppose I2 doesn't come into the picture as there is no potential drop for current I2?

  15. Jul 11, 2006 #14

    Doc Al

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    The way I interpret this odd voltmeter placement implies that I2 = 0. So I1 = I3.
  16. Jul 11, 2006 #15
    That should give the potential drop for the 5ohm resistor to be 2V so the voltmeter reading should be 2? In principle, does the voltmeter in this circuit measures the potential drop across the 5ohm resitor ? What if the potential drop there is just say to be 1.5V?
  17. Jul 11, 2006 #16

    Doc Al

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    Don't forget the 2V battery--that's also between the voltmeter terminals.

    Find the cumulative voltage drop from one side of the voltmeter to the other.
  18. Jul 11, 2006 #17
    Sorry I'm still a bit confused here. Should it read 0 or 4 volts? The 5ohm resistor has a 2V drop, and the battery supply is 2V so it should read 0? However the current through the 5ohm resistor is not totally due to the 2V source since I3 passes through the resistor which part of the current is from the 6V source?
  19. Jul 12, 2006 #18
    The voltage dropped across 5 ohm is only due to the 6V source . However when measuring with the voltmeter, the 2V source comes into play .
    Remember that for drops to cancel out each other, the drops should b in the opposite sense, ie don't aid each other .
    If you still have trouble, you can think of the 5 ohm resistance as a 2V battery, with terminals marked as per convention and use KVL (trivially).

  20. Jul 12, 2006 #19
    After all the help I'm given I hope I'm not getting it wrong aleady. It should read 0V?
  21. Jul 12, 2006 #20

    Doc Al

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    Exactly! And given this "cute" result, now we can see the point of the problem. :wink:

    (Note: All the current passing through the resistors is due to the 6V battery.)
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