I How could electric susceptibilbility depend on position?

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The discussion centers on the interpretation of Griffiths' statement regarding electric susceptibility and its dependence on position within a dielectric. It clarifies that the bound-charge density is zero when the polarization vector, P, is constant throughout the dielectric. However, at the surface of the dielectric, bound-surface charges arise, calculated using the surface divergence of P. The conversation highlights that if P were zero, the displacement field D would not align with expected values, indicating a discrepancy. Ultimately, the susceptibility varies across the boundary, reflecting different values on either side.
Ahmed1029
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In the statement encircled, what does Griffiths actually mean?
Screenshot_2022-06-07-15-17-53-45_e2d5b3f32b79de1d45acd1fad96fbb0f.jpg
 
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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
$$\sigma=-\mathrm{Div} \vec{P}=\vec{n} \cdot \vec{P}.$$
 
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vanhees71 said:
which is, for P→=0, within the dielectric
You probably mean ##\vec{P}=\text{constant}## inside the dielectric because if it was zero then ##\vec{D}=\epsilon_0\vec{E}+\vec{P}=\epsilon_0\vec{E}## inside the dielectric which doesn't look right...
 
At the boundary the susceptibility has different values on the two sides of the boundary.
 
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