What are the bound charges of a polarized dielectric cube with no free charges?

[SonOfOle]

Homework Statement

Consider a permanently polarized dielectric cube with the origin of the coordinates at the center of the cube. The cube has a side of length a. The permanent polarization of the dielectric is \vec{P} = c \vec{r}. The vector \vec{r} is the radius vector from the origin of the coordinates to the point \vec{r} (x,y,z). There are no free charges on the system. Compute the bound charges of the system.

Homework Equations

\sigma_b=\vec{P} \bullet \hat{n}
\rho_b=-\nabla \bullet \vec{P}

The Attempt at a Solution

We can see by symmetry that \sigma_b = 6 \sigma_{b, oneside}, and also that
c\vec{r} \bullet \hat{n}=a/2 for each face of the cube, so the net bound surface charge is \frac{3}{4} a^3 c.

Assuming the above checks out, the bound volume charge is where I run into difficulty.

I think something is wrong with the below set-up, but please let me know. Thanks.

\rho_b=-\nabla \bullet \vec{P}
-\nabla \bullet (c \vec{r})=-\nabla \bullet (c \sqrt{x^2+y^2+x^2}) = \frac{x+y+z}{\sqrt{x^2+y^2+x^2}} which you would then integrate over the volume for dxdydz (each going from -a/2 to +a/2).

 

[tiny-tim]

I think something is wrong with the below set-up, but please let me know. Thanks.

\rho_b=-\nabla \bullet \vec{P}
-\nabla \bullet (c \vec{r})=-\nabla \bullet (c \sqrt{x^2+y^2+x^2}) = \frac{x+y+z}{\sqrt{x^2+y^2+x^2}}

Hi SonOfOle!

(use \cdot instead of \bullet )

\nabla \cdot \vec{r}\ =\ \nabla \cdot (x,y,z)\ = … ?