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The Electric Field Due to a Thin Spherical Shell, Gauss's Law

  1. Oct 29, 2008 #1
    1. A thin spherical shell of radius a has a total charge Q distributed
    uniformly over its surface. Find the
    electric field at points outside the shell and inside the shell




    2. [tex]\int[/tex]E dA = Q/[tex]\epsilon[/tex]



    3. THis is an example in my textbook, and it claims that:

    The calculation for the field outside the shell is identical
    to that for a solid sphere (which yields E = KQ/r^2). If we
    construct a spherical gaussian surface of radius r >a concentric
    with the shell, the charge inside this surface
    is Q. Therefore, the field at a point outside the shell is equivalent
    to that due to a point charge Q located at the center.

    I dont see how it is equivalent to a point charge at the center, or how it is the same as the calculation for a sphere.

    The electric field of a sphere with total charge > 0 is directed directly outward from the sphere and thus will be perpendicular to a gaussian surface (as long as the centers are the same).
    But in this case of a ring, the electric field will not be perpendicular to the gaussian surface all the time, will it?

    Thanks
     
  2. jcsd
  3. Oct 29, 2008 #2
    "But in this case of a ring, the electric field will not be perpendicular to the gaussian surface all the time, will it?"

    A spherical shell is not a ring.
     
  4. Oct 29, 2008 #3
    yea...jsut noticed that...now I feel stupid.

    thanks
     
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