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Homework Help: Electric Field with a Solid Spherical Conductor

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data

    A spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity

    A. points generally toward the outer surface of the conductor.
    B. points generally away from the outer surface of the conductor.
    C. is zero.

    2. Relevant equations


    3. The attempt at a solution

    I'm thinking it should be C. A conductor has to have a net charge of 0 inside, so if you place a positive charge inside of the cavity, then the negative charges will be outside of the cavity. Therefore, the electric field could flow from the positive to the negative and I think the net electric field in the cavity would be zero.

    Any confirmation or denial would be great.
    Thanks in advanced.
  2. jcsd
  3. Sep 2, 2013 #2

    Simon Bridge

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    ... according to the problem statement, where is the charge?

    Have you heard of Gausses Law?
  4. Sep 2, 2013 #3

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  5. Sep 2, 2013 #4

    Simon Bridge

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    I see - the "positive test charge" in the definition of the electric force has to be so small that it does not significantly displace the charges whose field you are testing. i.e. the test charge cannot attract negative charges to it.

    There is actually an easier way to think about it - if you make a closed surface anywhere, then the electric field lines through that surface depend on the total charge enclosed by the surface. If there is no charge inside the surface, then there is no electric field.

    Even easier - what is the electric field inside a charged conductor?
  6. Sep 2, 2013 #5
    Ok, so if I were to create a Gaussian surface just around the cavity - not the whole system, there is no charge in the cavity, so there is no electric field inside the cavity, correct?
  7. Sep 2, 2013 #6

    Simon Bridge

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    There you go - wasn't that simple!

    When the conductor was charged, all the positive charge got evenly distributed about the outside skin.
    The electric field everywhere inside is zero - does not matter what cavities there are.
  8. Sep 3, 2013 #7
    Now I am very confused. My professor talked about this question in lecture today. He said that the answer should be B - points generally away from the outer surface of the conductor. I understand that the electric field lines from the positive charges should be pointed outwards, but why would that matter in the cavity? I thought the cavity should have no electric field.

    He said that since there is not symmetry you cannot just draw a Gaussian surface around the cavity because the center of cavity is not the center of the conductor.

    Why should this matter? Why can't you arbitrarily draw the Gaussian surface over the cavity, which would show that there is no net charge inside the cavity, therefore no electric field?
    Last edited: Sep 3, 2013
  9. Sep 3, 2013 #8

    Simon Bridge

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    Did anyone ask him about that?

    Well ... if you charge two plates, there is an electric field between them even though there is no charge between them.

    Darn - I'm afraid I've mislead you here.
    Technically gausses law says that the net flux through a surface depends on the charges enclosed.
    If all the charges are outside, then the same amount of flux enters as leaves so the net flux through the surface is zero. :(

    Maybe I'm slipping ... I'd have thought that the charge would be uniformly distributed about the outer surface of the conductor. This should mean that the field will cancel for any position inside the sphere.

    Even though the cavity is not centered in the sphere, you can still draw a gaussian surface centered on the sphere that completely encloses the cavity. There are no charges enclosed, therefore the the field through the surface is zero. This leaves the possibility of an uneven charge distribution outside the cavity - which is ruled out by the continuity relation.

    Also see:
    Last edited: Sep 4, 2013
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