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Electric Flux and Field

  1. Feb 3, 2007 #1
    1. The problem statement, all variables and given/known data

    A cylindrical piece of insulating material is placed in an external electric field, as shown in the figure. The net electric flux passing through the surface of the cylinder is

    a.positive.

    b.negative.

    c.zero.

    Figure : http://i131.photobucket.com/albums/p289/SoaringCrane/Tip23_fig11.gif


    2. Relevant equations

    Possibly: electric flux = E*A*cos theta

    3. The attempt at a solution

    I’m really unsure about this question. Would the net flux be a. positive? The flux on the ends would be 0 because the E and dA vectors are perpendicular to each other while the flux on the wall near the middle will be positive since E and dA are parallel to each other???


    1. The problem statement, all variables and given/known data

    The electric field at the surface of a conductor


    a.is parallel to the surface.

    b.depends only on the total charge on the conductor.

    c.depends only on the area of the conductor.

    d.depends only on the curvature of the surface.

    e.depends on the area and curvature of the conductor and on its charge.



    2. Relevant equations

    This is really a conceptual question. Possibly electric flux = integral[E*dA] = Q_enclosed/epsilon_0

    3. The attempt at a solution

    Is the correct choice e. depends on the area and curvature of the conductor and on its charge? I know that the electric field intensity increases as the curvature increases. The electric field for an infinitely long sheet is based on the surface charge density, so area and charge (from Gauss’ law) are other factors?

    Thanks.
     
  2. jcsd
  3. Feb 3, 2007 #2

    Dick

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    You are trying to work way too hard for the first problem. What IS Gauss' Law?
    For the second one, yes it's conceptual. You know it depends on curvature and surface charge density. And you do know a) is false, right? (Why?) So everything is pointing at e) alright.
     
    Last edited: Feb 3, 2007
  4. Feb 3, 2007 #3
    Gauss' law states that the net electric flux for a closed surface is equal to the total electric charge over epsilon_0.
     
  5. Feb 3, 2007 #4

    Dick

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    You mean "total charge enclosed in the surface". Right?
     
    Last edited: Feb 3, 2007
  6. Feb 3, 2007 #5
    Yes, Q_enclosed/epsilon_0. What am I doing incorrectly for the first question?
     
  7. Feb 3, 2007 #6

    Dick

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    You aren't doing anything wrong! But how much charge is enclosed in the cylinder?
     
  8. Feb 3, 2007 #7
    It does not say how much charge is inside. (Is there even any charge inside??)
     
  9. Feb 3, 2007 #8

    Dick

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    Exactly! Judging by the phrase "external electric field" and looking at the picture I would say there is no charge inside. Hence?
     
  10. Feb 3, 2007 #9
    The net electric flux is 0.
     
  11. Feb 3, 2007 #10

    Dick

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    Yes. That would be correct.
     
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