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I have a question regarding the bound charges in electrostatics, I think I am a bit confused, on one side I have read that bound charges in a capacitor with a dielectric inside the plates are on the surface of the dielectric material. On the other side, in other books bound charges refer to the charges of molecules which cannot move as it happens in metals (free electrons).

So, if I put a solid dielectric inside a capacitor the bound charges will be only on the surface? what about the charges on the molecules which compose the solid?

If I take for instance a glass of water, are there bound charges?

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armando, You lack the basics of stucture and composition of materials. What do You mean with this : ""as it happens metals"" ? – Georg Oct 6 '11 at 19:34
There is nothing lacking in this question. He means that in a metal charges are delocalized. – Ron Maimon Oct 6 '11 at 20:33
well, I meant, in metals the charges are free. But I am a bit confused, in some books bound charges refer to charges which cannot move in solids and solutions. In other books bound charges refer to the charges on the surface of dielectrics. – armando Oct 6 '11 at 20:46
up vote 1 down vote accepted

The charges in the middle cancel, leaving only sheets of charge on the surface, and where the electric field changes.

An easy classical model of a dielectric is a collection of small conducting spheres in a non dielectric insulator. The average charge in any macroscopic region is zero, but the polarization isn't.

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if for instance I want to compute the total bound charge in a solid/solution which charges should I take into account? Lets take NaCl+water solution, the free charges are those of Na and Cl ions because they can move freely. The bound charges are those of water? – armando Oct 6 '11 at 20:50
Yes-- but its not so simple to compute. The charges are the electrons and nuclei, and they make a dipole in an E field. – Ron Maimon Oct 6 '11 at 21:59

Bound charges are only on the surface of a dielectric if the dielectric is homogeneous and isotropic and free from free charge. Bound charge and polarization vector satisfy the equation

$$\rho= -\nabla\cdot \vec{P},$$ $$\sigma= \vec{P}\cdot\vec{n},$$ where $\rho$ is bound charge density and $\sigma$ is bound charge surface density, $\vec{P}$ polarization vector and $\vec{n}$ unit normal vector pointing outside.

In normal dielectrics, we have $\vec{D}=\epsilon\vec{E}=\epsilon_0\vec{E}+\vec{P}$, and wherever there is no free charge, $\nabla\cdot\vec{D}=0$, so $\nabla\cdot\vec{P}=0$, so $\rho=0$.

But if $\epsilon$ varies according to location, or is a tensor instead of scalar, bound charges will arise even in the middle of dielectrics.

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