# Understanding surface and volume polarization charges

Can someone please give me an intuition of this formulae? I am unable to visualize the formulae in equation 5.26.

Specific question - How does a dielectric when kept in an external electric field gives rise to both surface and volume charge density?

My take: In dielectrics, electrons are bound to the nucleus and hence cannot penetrate the outer surface of the dielectric at any cost. When a dielectric material is placed in an external electric field, positive charges (nuclei?) tend to drift from their equilibrium position in the direction of the applied external field, while negative charges (electrons?) tend to drift in the opposite direction thus creating an electric dipole.

Now since these charges which drift either way cannot penetrate the outer surface of the dielectric they are bound to get accumulated at the outer surfaces thus generating surface charge density.

What I am having trouble understanding is the process of volume charge generation within the dielectric. Won't there be equal positive and equal negative charges inside the volume?

• Can you ask a specific question? and post the formula? We don't like to guess what your thinking. Apr 26, 2017 at 4:11
• Is it okay now ? Apr 26, 2017 at 4:17
• Yep, looks good Apr 26, 2017 at 4:18

## 2 Answers

Dielectric polarization happens from polar molecules. The work done by an external field behaves like a charge, but unlike a electrons in a material, these charges cannot move. So if you had some dielectric material, you'd want to be able to find out what an external electric field does to it.

$P_{pv}$ is the bound charge, its locked up inside of the polar molecules, they aren't going to escape the material and move down a wire.

$P_{ps}$ is the surface charge. There could be charges on the surface, these need to be accounted for.

You add them both up and you know how much charge you have ( and what kind of useful things you can do with it like store energy)

• Dielectric polarization also happens from non-polar molecules. Apr 26, 2017 at 5:39
• And I have a doubt that how does movement of dilelectric molecules in presence of electric field gives rise to surface and volume charge densities. And your answer is not pertinent. Apr 26, 2017 at 5:40

My guess is that your formula is applicable to non-uniform (non homogeneous and/or anisotropic) dielectric materials.

If the material is homogeneous and isotropic I do not see any way a volumetric polarization charge could develop, so your difficulty in understanding why is - from my point of view - justified.

If the material is not homogeneous, you can see how charge can develop by considering it composed of multiple portions of homogeneous materials. Basically, the surface charge at each interface is what constitutes the volume charge in the chunk of material as a whole.

For anisotropic materials, I pass :-) but you might want to have a look at Purcell's "Electricity and Magnetism", 3rd edition, section 10.11 "The field of a charge in a dielectric medium and Gauss's Law". Figure 10.28 at p. 498 gives you a hint on how a single free charge placed inside an otherwise isotropic dielectric gives rise to a polarization with positive divergence.

Anyway, here is some comfort coming from the Physics Stack Exchange: Density of polarization charge is zero always for linear isotropic homogeneous dielectrics?

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

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.

They also give a reference:

[...] certainly [local charge densities] can't [arise] 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.