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Physics
Classical Physics
Electromagnetism
How could electric susceptibilbility depend on position?
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[QUOTE="vanhees71, post: 6639752, member: 260864"] It's a bit "nebulous". I guess what he considers is the effective (bound) surface charge of a homogeneous and isotropic dielectric. [edit: corrected in view of #3] The bound-charge density within the dielectric is $$\rho=-\vec{\nabla} \cdot \vec{P},$$ which is 0, for ##\vec{P}=\text{const}##, within the dielectric. Trivially it's also 0 outside the dielectric, where is vacuum, i.e., no charges at all. At the surface you have, however bound-surface charges, which you get by taking the "surface divergence". Let ##\vec{n}## be the surface-normal unity vector pointing out of the material. Then with a Gaussian pillbox with two sides parallel to the boundary of the dielectric, you get $$\sigma=-\mathrm{Div} \vec{P}=\vec{n} \cdot \vec{P}.$$ [/QUOTE]
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Physics
Classical Physics
Electromagnetism
How could electric susceptibilbility depend on position?
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