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Electromagnetism help!

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Use Gauss’s law to obtain a vector expression for the electric field produced by an
    infinte sheet of charge with a surface density of 1 C m−2 , confined to the xy plane. Hence
    show that the divergence of this electric field is zero for all points not in the xy plane.


    2. Relevant equations
    [tex] \int {E.dS} = {Q}/{epsilon} [/tex]
    del.E = rho/epsilon


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 14, 2009 #2

    lanedance

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    try picking a useful guassian surface
     
  4. Oct 14, 2009 #3
    i still dont really know where to go with it, any other hints you can give?
     
  5. Oct 14, 2009 #4

    gabbagabbahey

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    Hi andyfreesty1le, welcome to PF!:smile:

    First figure out what kind of symmetry the charge distribution (and hence the electric field it produces) possesses (e.g. spherical, cylindrical, planar, etc.). Then read your textbook to find examples of what kind of Gaussian surface you would use for problems with that type of symmetry.
     
  6. Oct 14, 2009 #5
    Ok, so according to my textbook, i would use cylindrical gaussian surface. It tells me how to find the electric field, but i cant see how to work out the electric field vector.
     
  7. Oct 14, 2009 #6

    gabbagabbahey

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    No, you use cylindrical Gaussian surfaces when the charge distribution is cylindrically symmetric. A plane does not possess cylindrical symmetry.
     
  8. Oct 14, 2009 #7

    lanedance

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    you could use a cylindrical surface, or a thin rectangular prism surface equivalently

    place it so the flat ends of either the cylinder or prism ar parallel to the plane

    consider the case when it contains the plane and when it doesn't, then relate the charge contained to the flux through the surface in each case. It may also help to consider wthen teh thisness of the prism shrink to infinitesimal
     
  9. Oct 16, 2009 #8
    You can use a cylinder. Or you can use a "box".

    Gauss's Law states that the charge inside a volume is equal to the electric field integrated over the surface of that volume.

    So you're looking for the electric field.

    First, imagine a box being intersected by the charged plane.

    Imagine the "height" of the box goes to zero, so we don't have to concern ourselves with the field coming out of the sides.

    If the box has dimensions [tex]x^2[/tex], what is the charge inside it?

    Hint: charge density is charge per area. Multiply that by the area enclosed.

    Now, integrate the electric field by the surface area of the volume.

    Then solve for E.
     
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