E-field of non-conducting, nonuniform shell

1. Jan 25, 2007

Pixter

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. Jan 26, 2007

chanvincent

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....

3. Jan 27, 2007

Pixter

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..