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Homework Help: Magnitude of a electric field

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A solid insulating sphere of radius a has a uniform charge density p and a total charge Q. Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are b and c. (a) Find the magnitude of the electric field in the regions r<a, a<r<b, b<r<c, and r<c. (b) Determine the induced charge per unit area on the inner and outer surfaces of the hollow sphere.

    2. Relevant equations

    Gauss's Law= S E dA= Q/E_0

    3. The attempt at a solution

    S E dA= Q/E_0
    = E(4pir^2)= Q/E_0

    I am doing this right so far? if so not sure what to do next, if not, not sure what to do.
  2. jcsd
  3. Sep 27, 2009 #2


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    You are doing fine so far. Next you need to apply Gauss' Law by choosing different gaussian surfaces in each of the regions of interest

    First at r<a
    Second at a<r<b
    Third at b<r<c
    Fourth at r>c (You say r<c, but I assume you meant r>c)

    Apply Gauss' Law as you have stated it. Make sure that you calculate the charge enclosed by each surface correctly and remember that inside a conductor the electric field is zero.
  4. Sep 27, 2009 #3
    ok this is what I have so far...

    for r>a

    S E dA= Q/E_0
    = E(4pir^2)= Q/E_0
    E=Q/4piE_0r^2= kQ/r^2

    for a<r<b

    S E dA = E S dA= E(4pir^2)= q(internal)/E_0
    E= (Q/(4/3)pia^3)r/3(1/4pik)= kr(Q/a^3)

    for r>c
    would it be the same as the 1st one?

    for b<r<c
    would it be 0?

    Did I do any of these right? and for the wrong ones what do I need to do?
  5. Sep 27, 2009 #4


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    I assume you mean r < a. In this case not all the charge Q is enclosed by the Gaussian surface. Only a fraction of it. Can you figure out what that fraction is?

    Actually this is the answer to the previous part and the answer to the previous part is the answer to this one. Can you see why?

    It would be the same as in the region a<r<b (corrected).
    Yes it would.
    See above.
  6. Sep 27, 2009 #5
    Thanks....I think I just copied it down wrong in here from what I wrote down.
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