2
$\begingroup$

I've stumbled about the following problem that I do not understand:

The polarization density P is defined such that its divergence gives the charge density due to bound (immobile) charges in a material. Now in a perfect conductor ("metal"), the electric field inside will be 0, hence there also can not be any (local) polarization. The static relative permittivity would therefore be undefined (0 divided by 0). On the other hand, one often reads that the relative permittivity of a metal is infinite, because it is interpreted as the factor by which an external electric field is reduced inside the material, and the electric displacement field D is supposed to be constant and finite. But then we would get P=D, hence a nonzero polarization density inside the material, even though there are no localized dipoles. So I guess my question essentially boils down to: Is there an unambiguous definition of either P or D or the permittivity or alternatively the "bound charge density" that would resolve my confusion? Sometimes I have the feeling that those quantities do not really correspond to observables, but are a matter of interpretation instead (but I never really found it clearly explained like that). But then on the other hand, refractive index (which is related to permittivity and permeability) is an observable, which makes me think that there must be an unambiguous definition.

You see I am thoroughly confused, and I would appreciate any clarification.

$\endgroup$

2 Answers 2

1
$\begingroup$

The question here is how to extend to concept of polarization from dielectrics to conductors.

Polarization in dielectrics involves in definition only those charged particles which belong to electrically neutral "internal set", i.e. a neutral molecule or a neutral crystal lattice piece. If there are other charged particles that cannot be included in the set because it would break electric neutrality, these particles are not part of the set that the definition of $\mathbf P$ is based on. This is because electric dipole moment of a system is unambiguous = independent of reference point in space only if the system is electrically neutral.

Macroscopic manifestation of these "internal set" charges is electric field coming from the polarized elements, and on surfaces where polarization has discontinuity (such as on boundary of the body) there is non-vanishing macroscopic electric charge due to polarization - so called "bound charge" - with density $\rho_{bound}$. However, being bound or mobile in time is not important feature here; the important feature is that the macroscopic charge is due to polarized state of the charged particles forming the neutral dielectric only, and that this charge is fully described by the polarization field via $\rho_{bound} = - \nabla \cdot \mathbf P$.

If dielectric body gets electrically charged, e.g. gets or loses electrons via friction, resulting macroscopic charge density is no longer described fully by $\mathbf P$ (which determines only part of it). The so-called free charge density $\rho_{free}$ has to be introduced via

$$ \rho_{free} = \rho_{total} - \rho_{bound}. $$

The name "free charge density" here does not mean the charge there is free to move like in conductor! In a dielectric body, this free charge will mostly stay in place unless interaction with other bodies or slow diffusion will remove it. We introduce free charge density not because of this charge's freedom to move in/on the body, but because it is due to additional charged particles foreign to the dielectric body, not accounted for by polarization $\mathbf P$.

In conductors we can define polarization in the same spirit; based on those charges which are part of electrically neutral internal set.

However, it is not clear whether conduction charges should be included into the neutral set or whether conduction charges together with corresponding part of opposite static charges bound to the conductor lattice should be excluded and viewed as additional "free charge". These two options give different values for polarization inside conductors.

If conduction charges are chosen to be part of the internal set, then things are as follows. Consider electrically neutral block of metal in external electric field of capacitor plates. The block will have some induced charge on its surfaces, but inside, electric charge density will vanish. The surface charge with density $\sigma$ is due to charges that belong to the neutral set, so polarization inside just below the surface is non-zero and obeys the relation

$$ \sigma = \mathbf P\cdot \mathbf n. $$

As the external field gets stronger, magnitude of density of induced surface charge gets greater, so magnitude of polarization inside has to increase as well. Since electric field inside vanishes, dielectric constant comes out as infinite. This viewpoint is strange in the sense that inside of the metal block, the external field is screened out, total electric field vanishes, and there is no local property showing for the polarized state. There are no stretched neutral molecules inside. It is all due to charge separation happening mainly on the surface. This weird state of matters does not happen in dielectrics because there is no macroscopic separation of charge on the surface due to lack of conduction.

On the other hand, if conduction charges are chosen not to be part of the internal set, then the induced surface charge is not viewed as due to polarization inside, but is instead viewed as additional "free" charge. Polarization inside therefore vanishes or is at most very weak = negligible, just like the electric field. However we have to include additional quantities - free charge density $\rho_{free}$ - to describe the system. This is a more intuitive picture where polarization describes just the polarized state of neutral non-conducting subsystems in the metal, and conduction/surface phenomena have separate description.

$\endgroup$
0
$\begingroup$

The displacement vector $D$ really doesn't have a physical meaning, it is just a mathematical aid for solving problems in electromagnetism by making Maxwell's equations simpler.

The DEFINITION of $D$ is $\vec{D}=\varepsilon_0 \vec{E} + \vec{P}$, and it is defined like that so we could write $\nabla \cdot D = \rho_{free}$, meaning we can solve problems with D when we have no knowledge of the bound charges in the material.

For the polarization density, you must understand the concept of dipole moment - it is a vector that is handy in describing the potential/electric field from a configuration of charges (most of the time, when there is $0$ net charge). The electric potential due to a some volume $dV$ with polarization density $\vec{P}$ is $\frac{(\vec{P}dV)\cdot \hat{r}}{4\pi \varepsilon_0 r^2}$.

In an ideal metal, there is no polarization density and no electric field, only free charges on the boundary of the metal.


Moreover, there is no such thing as bound charges, it is also a fictitious concept, created as a mathematical aid. It can be shown that you can ignore the electric potential and fields due to polarization density if instead you invent these "bound charges". What really is going on is that you have charges creating $E$, and some of them are always tied to a partner opposite charge so it is convenient to look at them as dipoles generating polarization density in the material. And because that is also sometimes a bit complicated you disregard the polarization density and invent these bound charges instead.

After all of this, if you think these bound charges are a bit unpredictable, you just mash it all up in $D$ and hope to forget about anything that isn't only the free charges in your problem - sometimes it works and sometimes it doesn't.

$\endgroup$
1
  • $\begingroup$ Do you think the following statement would find consensus in the physical community: "Whenever we place a material (which has 0 charge density in the absence of external fields) in a nonzero external electric field, any charge density in the material that is induced by this electric field is considered 'bound charge'"? Because this would at least explain that the polarization density is finite inside the metal, which leads to the often stated "static permittivity of a metal is infinite". $\endgroup$
    – LLang
    Commented Nov 3, 2018 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.