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Quick Gauss' Law Answer Check

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A sphere of radius a has a charge density that varies with distance from its center d(r) = do*(r/a)**2.

    Express the electric field as a function of distance from the center of the sphere r, a, do, and the permitivity of free space ez for each of the falling regions:

    0<r<a


    2. Relevant equations

    [itex]\Phi[/itex]*dA= Q/[itex]\epsilon[/itex]o

    3. The attempt at a solution

    Using Gauss's Law I have E*4 [itex]\pi[/itex] r^2 = (4/5)*[itex]\pi[/itex]*do*r^3/ [itex]\epsilon[/itex]o

    so E = (1/5)*do*r/[itex]\epsilon[/itex]o

    But this isn't the right answer...help?
     
  2. jcsd
  3. Feb 14, 2012 #2

    ehild

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    You made a mistake, E*4 [itex]\pi[/itex] r^2 = (4/5)*[itex]\pi[/itex]*do*r^5/ [itex]\epsilon[/itex]o correctly.


    ehild
     
  4. Feb 14, 2012 #3
    Oh ok, I see that now. How come it's r^5 instead of r^3? I'm just having trouble with the actual picture of the sphere I guess.
     
  5. Feb 14, 2012 #4

    ehild

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    Show your work in detail.


    ehild
     
  6. Feb 15, 2012 #5
    I would, but I barely have an idea of the problem. The professor is just rushing through examples without covering the concept and barely touching the points.

    I can see though that Q = (4/5)*π*do*r^5
    but I don't understand why...I thought charge didn't have to do with area (as it looks to be because of the pi) and that the charge in this problem was d?

    Sorry for the complete lack of understanding, just trying to figure this out..
     
  7. Feb 15, 2012 #6

    ehild

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    Gauss' Law states that the integral of E around a closed surface is equal to the enclosed charge divided by epsilon-nought.

    You need the charge enclosed in a sphere of radius r<a. The charge density depends on the distance from the centre, ρ. The charge Δq confined in a shell is Δq=volume of the shell *charge density d(ρ).
    The volume of the shell of radius ρ and thickness Δρ is 4πρ^2Δρ. To get the whole charge, Δq should be integrated from ρ=0 to ρ=r.

    [tex]q=\int_0^r{(d_0\frac{ρ^2}{a^2})4πρ^2dρ}[/tex]

    ehild
     
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