(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

[tex]\sigma_b=\vec{P} \bullet \hat{n}[/tex]

[tex]\rho_b=-\nabla \bullet \vec{P}[/tex]

3. The attempt at a solution

We can see by symmetry that [tex]\sigma_b = 6 \sigma_{b, oneside}[/tex], and also that

[tex]c\vec{r} \bullet \hat{n}=a/2[/tex] for each face of the cube, so the net bound surface charge is [tex]\frac{3}{4} a^3 c [/tex].

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.

[tex]\rho_b=-\nabla \bullet \vec{P}[/tex]

[tex]-\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}}[/tex] which you would then integrate over the volume for dxdydz (each going from -a/2 to +a/2).

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# Homework Help: Bound charges of a cube

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