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Question regarding Millikan's oil drop experiment.

  1. Sep 12, 2012 #1
    My chemistry teacher recently explained the Millikan oil drop experiment to us. However, he rushes his lectures to the point where it is almost impossible to ask him questions.

    What I was wondering in regards to this experiment is why is the positive plate used in the experiment? I understand that the negative plate at the bottom is used to repel the negative electrons that were bound to the oil drops. However, my teacher's description of the experiment did not define the purpose of the positive plate above where the oil drops were suspended.

    Can someone please explain to me the purpose of the positive plate and its impact on the test results? His explanation was confusing, because it seemed like the intent of the experiment was to quantize the negative charge, and it would seem like the repulsion from the negative plate would accomplish the task and require less calculation than the addition of the positive plate. I would just like to gain a better understanding of all the details of the experiment, and I haven't been able to find a good website that explained it either.
     
  2. jcsd
  3. Sep 12, 2012 #2
    Because it creates a greater voltage. Without the positive plate the charged oil droplets would just be repelled by the negative plate but with the positive plate, the charged oil droplets are also attracted to the positive plate so theres a force pulling them upwards. I'm just speculating here though, you're right, getting rid of the positive plate would be eliminating an extra variable which would make the calculations easier. Its possible that without the positive plate, there wouldn't be a great enough change in the speed at which the droplets fall.

    Look at this video:

    the speed at which the charged droplets fall is inversely proportional to the voltage applied. With just a negative plate, the voltage won't be that great at all so you probably won't see a significant change in the speed at which the droplets fall. With a positive plate, you greatly increase the voltage.
     
    Last edited by a moderator: Sep 25, 2014
  4. Sep 12, 2012 #3

    Borek

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    Staff: Mentor

    I am not sure eliminating positive plate will make the calculations easier. When you have both plates you can expect electric field between them to be uniform, which is not necessarily the case with a single plate.
     
  5. Sep 12, 2012 #4
    I conjecture that the positive plate is needed to eliminate "edge effects." If there was no positive plate, the electric field lines coming from the negative plate would start to curve at distances comparable to the diameter of the plate. However, the positive plate would ensure that the electric field lines would be straight line segments between the negative and positive plates.
    Near the surface of the negative plate, the electric field lines will always be normal to the surface of the plate. This can be shown using the boundary conditions on a conducting surface. So the field lines are effectively straight near the negative plate. However, far from the negative plate the electric field lines will start to spread unless they are tethered by a positive plate.
    Near the surface of the positive plate, the electric field lines will again be normal to the surface of the plate. So the electric field lines will be straight line segments near the surface of the plate.
    If the two plates are parallel, the electric field lines will go straight from one surface to the other without curving. There will be a slight spreading near the edge of the plates. However, the spreading will be much larger if there wasn't matching between the plates.
    The electric field lines behave a little like ropes. Ropes need tension in order to be straight. Think of the positive plate as a tether for electric field lines that are attached to the negative plate. The positive plate provides the "tension" that straightens out the electric field lines.
     
  6. Sep 12, 2012 #5
    Well thanks to the two responses before this one, and especially thank you for such a thorough response here. It seems like my professor was trying to simplify the whole situation to keep from having to explain much of what you had here. I really enjoy knowing more details though, so thanks so much for making sense of this for me.
     
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