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E-Field in Hollow Polarized Dielectric

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine the electric field intensity at the center of a small spherical cavity cut out of a large block of dielectric in which a polarization P exists.

    3. The attempt at a solution
    charge densitysurface = P(dot)n --> charge densitysurface = P(dot)r

    This is about as far as I've gotten... My problem is I've never dealt with a non-uniform charge distribution before. My first goal was to find the surface charge around the sphere, but the polarization makes it such that it's non-uniform. I've assumed that the dielectric is polarized in the positive-z direction, which puts a concentration of negative charges at the top of the sphere and a concentration of positive charges at the bottom. My professor recommended using a sin() function to integrate over the sphere (positive->neutral->negative->neutral->repeat), but I don't really know how to fit this into a surface integral. Any thoughts?

    Last edited: Sep 23, 2008
  2. jcsd
  3. Sep 24, 2008 #2
    The material is a dialectric. I don't think it's to be supposed that there is surface charge.
  4. Sep 26, 2008 #3
    There actually i s surface charge - picture a surface of thin dielectric in the xy-plane. If you polarize the whole thing with an e-field in the z-direction, the dipoles at the top of the plane will have their positive charges pointing up (giving a positive surface charge). Likewise, the bottom will have a negative surface charge. This case is tricky, because it's a sphere, so there's a smooth transition around the surface that goes sinusoidally. The professor has actually solved the problem since I first posted the problem, so I'll post a solution later for anybody who's interested.
  5. Oct 6, 2010 #4
    I'd love to see that solution.
  6. Oct 6, 2010 #5
    Haha - so, I've graduated since I originally posted this problem a couple years ago, but I actually just pulled out a box of old schoolwork to sort through. If I can find the solution, I'll post it.
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