Energy of 2 spherical shells filled with dielectric

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Homework Statement


2 concentric conducting spherical shells, with radii a and 2a, have charge +Q and -Q respectively. The space between the shells is filled with a linear dielectric with permittivity ε(r) = (ε0*a)/(1.5*a - 0.5*r), which varies with distance r.
a) Use Gauss's law to determine the displacement field between the shells
b) Determine the bound surface and volume charge between the shells
c) Determine the total energy of the system
d) Determine the capacitance

Homework Equations


D = ε0E+P
D = εE

The Attempt at a Solution


I already did part a and got the following equation for D:
D = Q/4πr2 in the radial direction
For part b I rearranged the equation, D=εE to get an expression for E which was:
E=(Q(1.5a-0.5r))/4πr2ε0a in the radial direction
Then I used the equation D = ε0E+P to solve for P, giving me the following:
P=(Q/4πr2)(1 - ((1.5a - 0.5r)/a)) in the radial direction
Now that I had an expression for P I tried to find the surface and volume bound charge, this is where my confusion starts. For the volume bound charge I found the negative of the divergence of P and ended up getting an expression that depended on r.
Volume bound charge = Q/8πar2
It doesn't make sense to me that the volume bound charge depends on r so now I'm wondering if I did something wrong when I found the expression for P. Can anyone tell me if I've made a mistake somewhere? The other question that I have is regarding part c, is there an electric field outside the outer sphere because of the bound charges? The charge from the 2 spherical shells cancel out so Gauss's law should give 0 field outside the outer shell, but if the bound charge contributes to the field then it will be non-zero and I'll have to include it in my calculation of the total energy. Any help is appreciated.
 
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Why does it bother you that there is a volume charge density that varies with r?

Your work looks good except you might want to check the sign of the volume bound charge.

Regarding whether there is a field outside the outer conductor, have you accounted for all the charge? There is free charge on the conductors and there is bound volume charge within the dielectric. But there is also some additional charge that you have not yet taken into account.
 
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If you mean the surface bound charge on the outer surface when you say I haven't accounted for all the charge yet, then I already found that. Thanks for pointing out the sign of the volume bound charge, I found the spot where I missed a minus sign. Now it seems that the total volume bound charge and the surface bound charge on the outer surface cancel out so there is no electric field outside the sphere and I was able to complete the rest of the question.
 
OK. Good work.