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Electric field within a solid sphere

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data

    A nonconducting, solid sphere of radius a is placed at the center of a spherical conducting shell of inner radius b (> a) and outer radius c, as shown in the figure below. A charge +Q is distributed uniformly through the sphere, which thus carries a charge density ρ (C/m3). The outer shell carries a total charge -3Q.

    Find the electric field E(r) within the solid sphere, i.e., at a radius r < a

    2. Relevant equations
    ∫EdA=Qenclosed/Eo


    3. The attempt at a solution

    E*4∏r^2=(ρ*(4/3)*∏*a^3)/Eo

    I thought I knew what i was doing, but now i'm not quite so sure. I know what i have above is not the correct answer.
     
  2. jcsd
  3. Feb 13, 2013 #2

    BruceW

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    Homework Helper

    Think about how much charge you are enclosing by your Gaussian surface.
     
  4. Feb 13, 2013 #3
    well, i think that i am enclosing something less than Q. Since Q is uniform over the entire sphere. If i am taking only a portion of that sphere, then i think it should be less.
     
  5. Feb 13, 2013 #4

    BruceW

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    yep. well, technically, the charge density is uniform over the entire sphere. You know that a Gaussian sphere of radius a would enclose Q, so how much charge does a Gaussian sphere of radius r(<a) enclose?
     
  6. Feb 13, 2013 #5
    A ratio of what a enclosed. So like, Qr/a?
     
  7. Feb 13, 2013 #6

    jtbell

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    Staff: Mentor

    It's a ratio, but not that ratio. Suggestion: first find the charge density ρ in terms of Q and a.
     
  8. Feb 13, 2013 #7
    That's what i tried to start to do with :E*4∏r^2=(ρ*(4/3)*∏*a^3)/Eo
     
  9. Feb 13, 2013 #8

    jtbell

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    Staff: Mentor

    You have a sphere of radius a and total charge Q. Forget everything else about the problem for the moment. What is the charge density ρ?

    Once you have that, what is the charge on a sphere with the same charge density, but with radius r?
     
  10. Feb 15, 2013 #9
    charge density p is charge/volume. That's kinda what i was trying to do. Was i just making it more complicated than i needed to. Should i just find p and then multiply it by the new volume that i am trying to find?
     
  11. Feb 15, 2013 #10

    BruceW

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    yep. that will give you the charge enclosed by your Gaussian surface (since charge density is constant inside the sphere)
     
  12. Feb 15, 2013 #11
    ohhh okay. Well thank you all very much
     
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