Is the Total Charge of Polarization in a Non-Homogeneous Dielectric Zero?

[Granger]

Homework Statement

Consider an infinite environment with electrical permittivity non-homogeneous $$\epsilon=\epsilon_0(1+a/r)$$ a being a positive constant. A conducting sphere of radius R and charge Q is put on that environment, centered at r=0. Determine the electric field $$E$$, the electrical potential $$V$$, the volume density of polarization charge $$\rho'$$ and the surface density of polarization charge $$\sigma'$$. Prove that the total charge of polarization in the dielectric is zero.

Homework Equations

The Attempt at a Solution

So I'm having a lot of trouble with the last part of the question.
I got for the the volume density of polarization charge $\rho'$ and the surface density of polarization charge $$\sigma'$$:

$$\rho'= \frac{Qa}{4\pi r^2(r+a)^2}$$
for $$r>R$$

$$\sigma'= \frac{-Qa}{4\pi R^2(R+a)}$$
for $$r=R$$

Now I now that the total charge will be the sum of the total charge in surface with the total charge in volume. Total charge in surface is easy.
$$Q'_{surface}=4\pi R^2\sigma'= \frac{-Qa}{(R+a)}$$

However I'm not quite sure on how to compute the total volume density. I thought it might be an integral from R to infinity of

$$\frac{4}{3} \pi r^3 \frac{Qa}{4 \pi r^2 (r+a)^2}=\frac{Qar}{3(r+a)^2}$$

but the integral of that expression on those boundaries diverge.
What should I do then?
Any suggestions?

 

[kuruman]

You cannot multiply the density by the volume the way you did to get a "charge" at r and then evaluate as r goes to ∞. You need to consider a shell that has charge dq = ρ(r) dV, then integrate. What's dV for a shell of radius r and thickness dr?

 

[Granger]

Oh! The volume of the shell is $$dV=4\pi r^2 dr$$. This way I get to the total charge equaling zero. Thank you very much!