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It seems to me that the volumetric density of polarisation charge in a linear homogeneous isotropic dielectric in an external field is always zero, but I find this rather surprising.

Consider such dielectric, then if the total field is $E = E_0 + E_s + E_v$ being $E_0$ an external polarising field and $E_s$ and $E_v$ the fields produced by the polarisation charges on the surface and volume respectively of the dielectric. We should have, inside the dielectric

$$ \nabla \cdot E = \nabla \cdot E_v = 4\pi \rho_v $$ Since $P= \chi E$ and $\rho_v = -\nabla \cdot P$ we have, taking into account

$$ E_v = -\nabla \int \frac{\rho(x')}{|x-x'|}d^3 x' $$ that

$$ -\rho_v = 4 \pi \chi \rho_v $$ from where $\rho_v = 0$ and also $\nabla \cdot P=0 $

Is this correct or am I missing some caveat? I have looked over some books but never found mention to this strange (and signifcant at an elementary level I think) fact.

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  • $\begingroup$ Why surprising? A uniform external field can't produce charge in the bulk of any neutral, isolated material, whether conducting or dielectric. All the charge will be at the surface. In fact, this is even more true for a dielectric. All the positive and negative charges are tightly bound. The field can displace them slightly into dipoles, but at the macroscopic level there is still no net charge in the volume. $\endgroup$ – user27118 May 21 '15 at 18:03
  • $\begingroup$ I don't think it is evident at all, in fact it is probably not true for a non-homgeneous dielectric, where $\epsilon(x)$, let alone a non-isotropic material. I believe this is something special of linear homogeneous isotropic dielectrics. $\endgroup$ – Rogelio Molina May 21 '15 at 18:06
  • $\begingroup$ Why? What would be the microscopic origin of a charge in the volume? $\endgroup$ – user27118 May 21 '15 at 18:22
  • $\begingroup$ The origin would be a non-vanishing $\nabla \cdot P$. Dipoles need not be homogeneously placed within the substance, hence local charge densities might arise in those cases. It is a standard fact that the local charge density in such case is given by $\rho= - \nabla \cdot P$ $\endgroup$ – Rogelio Molina May 21 '15 at 18:38
  • $\begingroup$ I see, but certainly they can't for an isotropic, uniform material. This is given in Jackson (compare 4.39 to 4.33). I do not think the linearity condition is necessary. It would be interested to know if isotropy is. $\endgroup$ – user27118 May 21 '15 at 20:02

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