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Electric Field inside a Sphere (Gauss' Law)

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A sphere has outer radius 15 and inner radius 5.

    Between r= 5 and 15, the sphere is solid and contains a total charge of 20 C.

    At r < 5, the sphere is hollow.

    Calculate the electric field at r = 8.

    2. Relevant equations

    Gauss' Law --> EA = (q)/epsilon zero

    3. The attempt at a solution
    What I'm confused about is the hollow sphere part. But ignoring that, here's how I would do the problem.

    The volume of the entire sphere is (4/3)pi(15^3) = 14137.16694 m^3.
    The volume of the hollow part is (4/3)pi(5^3)= 523.598 m^3.

    Therefore, the volume of just the solid part is 14137 - 523 = 13613.568 m^3.

    The charge density per volume is then 20 / 13613 = 0.001469.

    To calculate the electric field at r = 8, I need to find the volume of such a sphere.
    Same procedure as above, take the volume minus the empty part to find 1621.06.

    Multiply this by the charge density... 1621.06 * 0.001469 to find 2.38133 C.

    This is the internal charge.

    Take a gaussian surface of radius 8. It will have surface area (4pi)(8^2)
    E = (internal charge)/ (epsilon zero)(area)

    E= 2.38 / (eps. zero)(804.24) = huge number, 3.34 x 10 ^ 8.

    What's confusing me is whether it's ok to take the surface area of a sphere with radius 8, because of that hollow part. Do I need to subtract the hollow part's surface area or something?
     
    Last edited: Feb 25, 2009
  2. jcsd
  3. Feb 25, 2009 #2

    LowlyPion

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    Homework Helper

    With the Gaussian surface you are determining the charge inside. The surface of course is the spherical surface at 8. No other surface need apply.

    As to the rest of it, that looks OK. Subtract out the hollow volume to determine charge density and then determine the volume at 8 subtracting out the void again.
     
  4. Feb 25, 2009 #3
    So it doesn't really matter how the charge is distributed inside the Gaussian sphere, then? The charge can be all on the surface or all at the center, etc. and the procedure is still the same?
     
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