Induced charge density by non-uniform dipole density in dielectric?

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quasar987
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In chapter 10 section 3 volume 2 of the Feynman lectures on physics, there is a passage that I can't force myself to agree with. He says, talking about the polarization vector of dielectrics,

First consider a sheet of material in which there is a certain dipole moment per unit volume. Will there be on the average any charge density produced by this? Not if P is uniform. If the positive and negative charges being displaced relative to each other have the same average density, the fact that they are displaced does not produce any net charge inside the volume. On the other hand, if P were larger at one place and smaller at another, that would mean that more charge would be moved into some region than away from it; we would then expect to get a volume density of charge.
Can someone explain this perhaps differently?

I really don't see how a non-uniform polarisation vector across a dielectric have to do with charge density. If P is larger at one place, it is either because 1) E is greater there, or 2) the density of atoms N is greater there. But in either case, the total charge in a volume element remains that of the sum of the atoms, which is null for regular dielectric atoms regardeless of the number of them N in that volume or of the magnitude of the "dipole distance" [itex]\vec{d}[/itex] (as in [itex]\vec{p}=q\vec{d})[/itex].
 

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Meir Achuz
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Feynman can entertain, but avoids the nitty gritty.
Try this picture: One long molecule having +q at one end and -q at the other, so its dipole moment is qL. Then next to it, a molecule with +- 2 q at the ends. Sort of like this: -q-----+q -2q-----+2q.
This represents a P that is increasing to the right. If you look in the middle, there is a net charge of -q. This corresponds to
\rho_bound=-div P (in Gaussian units).
 

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