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

  1. Aug 22, 2006 #1

    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,

    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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  3. Aug 23, 2006 #2

    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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