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A question about electrostatics / Gauss's law

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose a conducting spherical shell carries a charge of 3.00 nC and that a charge of -2 nC is at the center of the sphere. If the distance from the center to the inner shell is 2.00 m, and the distance from the center to the outer shell is 2.40 m, find the electric field at:

    a.) r = 1.5 m from center
    b.) r = 2.2 m from center
    c.) r = 2.5 m from center
    d.) What is the charge distribution on the sphere?

    2. Relevant equations

    Using Gauss's law and the principle of electric flux, I was able to correctly derive the equation:

    (E)(r^2) = (Qinside)(ke)

    E=electric field
    ke = 9 x 10^9



    3. The attempt at a solution

    I was able to get parts (a) and (c) just be using the equation and plugging in the charges that I knew.

    The answer for (b) = 0
    the answer for (d) = 2.00 nC on the inner surface, 1.00 nC on the outer surface.


    I have worked for like 2 days on this problem and couldn't figure it out. Please help! Thanks in advance.
     
  2. jcsd
  3. May 18, 2009 #2

    LowlyPion

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    Well the answer for b) is 0. It is a conductor and charge within a conductor will lay at the surface.

    http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gausur.html#c2

    In d) since the field is 0, and the Gaussian surface within the conductor yields 0, then doesn't that mean that the net charge within is 0? If the center has -2nC, then won't the inner surface of the conductor necessarily have +2nC? and if 2 of them are on the inner surface ... what must be left of the 3nC that it is charged with to be lounging about on the outer surface?
     
  4. May 18, 2009 #3
    I see. But why would 2 of them be on the inner surface?
     
  5. May 18, 2009 #4

    LowlyPion

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    What's at the center?

    -2nC.

    What will the closed Gaussian surface within the middle of the conductor be? 0.

    If the net of the charge inside the Gaussian must be 0, then 2nC must reside on the inner surface.
     
  6. May 18, 2009 #5
    ah ok. thanks a lot!
     
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