Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Bound charges of a cube

  1. Jul 16, 2008 #1
    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 length a. 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).
  2. jcsd
  3. Jul 16, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi SonOfOle! :smile:

    (use \cdot instead of \bullet :wink:)

    [tex]\nabla \cdot \vec{r}\ =\ \nabla \cdot (x,y,z)\ =[/tex] … ? :smile:
    Last edited: Jul 16, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook