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Gauss's law help

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A hollow spherical conducting shell has a uniformly distributed total surface charge density of -15[tex]\mu[/tex]C/m^2. It's outer surface has a radius R1=0.25m. Its inner radius is R2=0.15 m. A point charge, Q= -6.0[tex]\mu[/tex]C, is placed at the center of the spherical charge. Determine the charge on the inner and outer surfaces of the shell, and show electric field lines.

    2. Relevant equations
    closed integral E*dA= E(4[tex]\pi[/tex]r^2)= Q/[tex]\epsilon[/tex]

    3. The attempt at a solution
    since the hollow shell surfaces have a distinguishable inner and outer radius how do I calculate the area? Wouldn't I have to calculate the volume? Please can someone just tell me what the inner and outer charges are. Sorry I am new at using the math format.
  2. jcsd
  3. Oct 28, 2009 #2

    Doc Al

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    Staff: Mentor

    What's the surface area of a sphere?
    No, you're given a surface charge density, not a volume charge density.
    To find how the charge distributes once that point charge is inserted, use Gauss's law.

    Hint: Before that point charge is inserted, where does all the charge on the shell reside?
  4. Oct 28, 2009 #3
    Would the charge distributes be the charge density per area plus the the point charge in the center?
  5. Oct 28, 2009 #4

    Doc Al

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    Staff: Mentor

    I don't understand the question.

    The total charge on the conducting sphere is fixed. When the point charge is placed inside, the charge on the sphere redistributes itself between its two surfaces.
  6. Oct 28, 2009 #5
    So the net charge on the inner surface would become positive 6 micro coulombs?
  7. Oct 28, 2009 #6

    Andrew Mason

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    Science Advisor
    Homework Helper

    Yes, but you need to be able to explain it using Gauss' law.

    Take a Gaussian sphere whose surface is inside the conducting sphere shell - ie has a radius greater than the inside surface of the shell but less than the outside surface. What is the field through through this sphere's surface? (ie in the interior of the conducting shell). Apply Gauss' law to that Gaussian sphere to determine the charge enclosed by it. The charges enclosed are the point charge at the centre and the charge on the inner surface of the conducting shell.

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