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In the Landau's book "Electrodynamics of Continuous Media" he derives the relation between the polarisation $\vec P$ and the dipole moment $\vec p$. Starting with the definition of the dipole moment:

$\int \vec r \rho dV = -\int \vec r (\nabla \cdot \vec P) dV = \oint \vec r(d\vec f \vec P) + \int(\vec P \nabla) \vec r dV$ where $\rho$ is the charge density, $d\vec f$ is the surface element. Landau says the integral over the surface vanishes: $\oint \vec r(d\vec f \vec P) = 0$.

Q: why does the integral over the surface equal $0$? In a dielectric, in the electric field there are only surface charges. Thus I would conclude that $\oint \vec r(d\vec f \vec P)$ gives the total dipole moment because $d\vec f \cdot \vec P$ is the infinitesimally small charge at some point on the surface .

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Probably, surface induced charge density has been a' priori excluded by the author while replacing rho=divergence of polarization vector, because this relationship could be established if we consider rho=volume charge density only, and no surface charge density. This result emerges if we try to find the electrostatic potential at an external field coordinate due to a in general non- uniformly polarized dielectric. Also, you never mentioned whether the dielectric in question is linear or not. For linear dielectric media, for less intense electrostatic fields, the polarization in medium is directly proportional to the field. The fact that in the presence of external electrostatic fields only the induced surface charge densities exist is incorrect...that is true only if the polarization in the medium is spatially uniform, else not. The exact dependence of the polarization of the medium on the applied external field is governed solely by the homogeneity, isotropy and linearity/non-linearity of the medium.

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L&L intend the integral to be over the entirety of 3-space, not only the interior of the polarized body. The surface integal is zero because ${\bf P}$ is zero at infinity

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Landau integrates over "...a surface that encloses the body but nowhere enters it", so that P $=0$ everywhere along the surface.

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