Polarization charge density of homogeneous dielectric

  • #1

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Hi everyone,
there's something that I can't comprehend: when a homogeneous is in a conservative and non-uniform in module electric field polarization expression is given by P0χE. Supposing the most general situation there's: divPp where ρp is the polarization charge density in the dielectric. When my textbook introduces the electric induction vector D0E + P it says that divD=0 if in the volume (the dielectric material) you are going to integrate the divergence there are no free charges and it makes absolutely sense. Troubles have come in my mind when it states that this means divPp=0 too because "every electrostatic field makes the variation end up with 0". I mean what's the characteristic of electrostatic conservative non-uniform (in module) fields to make divP=0? I think I missed some logics here. I mean, why does it jump to the conclusion that polarization charges are only located on the surfaces of the dielectric material?
 

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  • #2
Charles Link
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In general ## \nabla \cdot P=\rho_p ## is not equal to zero in a dielectric material when an electric field is applied. There are a couple of geometries where a uniform polarization ## P ## in the material results from an applied uniform external electric field , because the polarization charge forms in such a manner on the surface of the solid, that its electric field, ## E_p ##, (inside the material), added to the applied electric field ## E_o ##, results in a uniform ## E_i ## in the material and thereby a uniform ## P ##. A uniform ## P ## will have zero divergence,(and thereby ## \rho_p=0 ##), and surface polarization charge density is ## \sigma_p =P \cdot \hat{n} ##. ## \\ ##This self-consistent uniform ## P ## case happens in the case of a dielectric slab, where the electric field from the surface polarization charges ## E_p=-\frac{P}{\epsilon_o} ## , and also for a dielectric sphere, where ## E_p=-\frac{P}{3 \epsilon_o} ##. For most geometric shapes, in an applied electric field that is uniform, the resulting ## P ## is not uniform, and the resulting electric field, along with the resulting polarization will be quite complex.## \\ ## One other case where a simple self-consistent solution occurs is a cylinder turned sideways. For that case ## E_p=-\frac{P}{2 \epsilon_o} ##. These simple cases can be readily solved by writing ## E_i=E_o+E_p ##, and ## P=\epsilon_o \chi E_i ##. Since ## E_p=-\frac{D P}{\epsilon_o } ##, where ## D ## is the geometric factor for a particular geometry, it is a simple matter of solving two equations for the two unknowns: ## E_i ## and ## P ##. ## \\ ## For most cases, there is no constant ## D ##, like there is for the 3 cases mentioned above. (## D=1 ## for a slab, ## D=\frac{1}{3} ## for a sphere, and ## D=\frac{1}{2} ## for a cylinder that is turned sideways). ## \\ ## Additional note: Outside the material, ## E_p =0 ## for the dielectric slab. For the sphere and cylinder geometries, ## E_p ## outside the material has a somewhat complex form, and the solution for those cases is presented in advanced E&M courses.
 
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  • #3
Thank you so much!
 
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  • #4
Charles Link
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One typo/correction: The first line of post 2 should read ## -\nabla \cdot P=\rho_p ## with a minus sign in the equation.
 
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