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Homework Help: E-field of non-conducting, nonuniform shell

  1. Jan 25, 2007 #1
    1. The problem statement, all variables and given/known data


    Right have a shell that has a volym charge distribution P=(G/r^2) and with inner radius a and outer radius b.

    Now what is the e-field at r<a, a<r<b, r>b


    2. Relevant equations

    gauss's law: surfaceint[E.da]=Qenclosed/epsilon0
    Qenc=volymint[PdV)



    3. The attempt at a solution

    Right this is what i tried.

    for r<a: there is no enclosed charge so E is zero at all points.

    for a<r<b:

    well first E is radially outwards and normal to the surface so the gauss integral just go to
    E4[pi]r^2

    Qenclosed=volymint[PdV] between r and a

    goes to -> volymint[(G/r^2)(4[pi]r^2)dr] -> 4[pi]G*volymint[dr] evulated at r and a

    -> 4[pi]G(r-a)

    and E4[pi]r^2=(4[pi]G(r-a))/epsilon0
    which then goes to -> E=(G(r-a))/(epsilon0*r^2)

    then for r>b:

    it's the same calc as for a<r<b but we end up with E=(G(b-a))/(epsilon0*r^2)


    now i'm wondering have I done something wrong and if I had to plot it on a graph where we have E/r then for the first part upto a E would be zero and for the part where r>b it would follow a 1/r^2 relation, but how do i figure out the realation for part a<r<b, my algebra seems to suck a bit.. =)

    but yeah.. does this look correct?
    and if so, can you help me or give me a hint hwo to work out the relation.
     
  2. jcsd
  3. Jan 26, 2007 #2
    I am having some difficulty reading your equation, so I would not comment on them...

    For r<a, E=0..
    For a<r<b, E grows linearly..
    For b<r, E = Q/r^2 whereas Q = the total charge in the shell..
    E is continuous at r = a and r = b..

    If your solution satisfy the conditions above, your solution should be correct....
     
  4. Jan 27, 2007 #3
    pretty sure you are wrong.. outside the spherical shell E will be proportinal to 1/r^2 as for any other spherical charge..

    but inside the shell it will be proportinal to (1/r)-(1/r^2) following my equations, becuase the shell doesn't have uniform charge..

    also note the shell is non-conducting..
     
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