Sphere, dielectrics and bound charge

Question

I have this situation. A conducting metal sphere of radius $a$ and charge $Q$, surrounded by a dieletric shell with inner radius $a$ and outern radius $b$.

If I try to evaluate the induced charge density $\rho_b$, in the region $a<r<b$ is it correct to say that I should get these results? $$ \vec{E}(a<r<b)=\vec{D}/\epsilon=\frac{Q}{4\pi\epsilon}\frac{\hat{r}}{r^2} $$ So, considering thath $\vec{P}=\epsilon_0\chi\vec{E}$ $$ \rho_b = -\vec{\nabla} \cdot \vec{P} = -\frac{\epsilon_0\chi Q}{4\pi\epsilon}\vec{\nabla}\cdot\bigg(\frac{\hat{r}}{r^2}\bigg)= \frac{\epsilon_0\chi Q}{2\pi\epsilon}\frac{1}{r^3} $$ My question is here because I found the solution for $\rho_b$ to be 0 on a textbook. Am I missing something?

Answer 1

You made a mistake at the last step. It is wrong to say $$\nabla\cdot \left(\frac{\hat{r}}{r^2}\right)=-2\frac{1}{r^3}$$

Divergence of inverse square radial vector field is trickier than simple 1D differentiation.

You need to use the fact that

$$\nabla\cdot \left(\frac{\hat{r}}{r^2}\right)=\delta^3(0),$$ where $\delta^3(0)$ is the Dirac delta function centered at the origin (i.e. infinity at origin, and zero everywhere else). As the dielectric layer does not extends to the origin, so $\rho_b=0$.