1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    tiny-tim

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Bound charges of a cube
  1. Bound Charges (Replies: 15)

Loading...