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Electroscope Charge Distribution

  1. Aug 28, 2012 #1

    I am helping out with a class and the students were given a question about an electroscope that is being charged by induction, and they have to label the charge distribution on a diagram. The diagram is drawn such that the conductive elements are seperated - indicating a force between them. The answer that is given is here:

    Electroscope Induction

    I have been getting the students thinking about the "sea" of electrons in the conducting rod and so some of them want to put positive charges on the top half and negative charges on the bottom half so that there is still a repulsion with net charge zero (no transfer/conduction). The more I think about it, the more I think they are not really wrong. I think the real answer lies inbetween that shown in the diagram, and that which I have described. Clearly if you bring a HIGHLY negative plate to the top, the students would be pretty much correct.

    Does any one wish to clarify this point(whether this is right or wrong), or add to the discussion?

  2. jcsd
  3. Aug 29, 2012 #2

    Simon Bridge

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  4. Aug 29, 2012 #3
    The student had something more like this:

    The Upper "V" was positive, and the lower was negative. In otherwords half the rod was positive, the other was negative. This allowed the repulsion to still be correct.
  5. Aug 29, 2012 #4

    Simon Bridge

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    Oh you mean the top part of the electroscope vane was positive and the lower negative?
    That's fair enough - how would you go about determining the difference?

    In practice you don't get a sharp change from positive to negative along a conductor like that. Can you find a charge distribution consistent with this setup that does not have the repulsion?

    The question just wants the students to demonstrate that they understand the principles qualitatively.
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