# Do conductors have bound charges?

In Griffiths, the total charge density of a dielectric is: $$\rho$$ =$$\rho_b +\rho_f$$.

I have several questions:

1. Can the total charge density of a conductor be written as the equation above?

2. Do the conductors have the bound charges?

3. Do the conductors have displacement electric?

$$\underline{\textbf{My idea:}}$$

Since a conductor doesn't have a charge inside it, its charge will be "pushed" to the surface, it is not considered a bound charge, because the charges on the surface are not caused by polarization, then it can't be written as the equation above, and since it doesn't have a bound charge, hence it doesn't have displacement electric.

Please correct me if my reasoning and concept are incorrect.

• By "displacement electric", do you mean electric displacement field $\mathbf D$? Commented Jul 20 at 21:27
• @JánLalinský Yes Commented Jul 21 at 14:00

You are correct. In a typical conductor, like a metal, the charges are all considered free charges. That includes both the mobile electrons and the immobile lattice protons.

The reason is, as you say, the charge density is not due to polarization.

1. yes, but since $$\rho_b=0$$ it would be a little silly.
2. no
3. yes, e.g. in radio frequency antennas
• Ok, thanks a lot~ Commented Jul 18 at 15:56
• > In a typical conductor, like a metal, the charges are all considered free charges That isn't true. Protons and non-conducting electrons stuck in bonds in dielectrics are not called free charge, because they do not participate in DC current when it is present. So naturally the same definition is appropriate in metals. Conductivity is much smaller than if all electrons participated, only few of them do. So there is nothing free about the protons and stuck electrons. Only conduction electrons and electrons jumping between bonds in semiconductors to make holes move are free charges. Commented Jul 20 at 0:07
• @JánLalinský They are free charges in the sense of Maxwell’s macroscopic equations in matter. They do not arise as $\rho_b=-\nabla \cdot \vec P$. That makes them free charges
– Dale
Commented Jul 20 at 1:42
• Even bound charges in dielectric do not arise as $\rho_b = -\nabla\cdot \vec{P}$. That is just a formula for density of net bound charge in terms of average dipole moment per unit volume. When $\rho_b = 0$ in a dielectric, this does not mean the charges there are not bound. Similar in metals. They are bound, and their net charge density vanishes. Commented Jul 20 at 11:42
• @JánLalinský you seem to think that “bound charge” means “a charged particle that doesn’t move”. It does not. Bound charge is a term in classical EM in matter. It is specifically the term $\rho_b=-\nabla \cdot \vec P$. So when $\rho_b=0$ there is no bound charge. It doesn’t matter if there are charged particles that are not moving.
– Dale
Commented Jul 20 at 12:44

The language at the outset is different from what is used in a chemistry of materials perspective. It also leaves a sense of sloppiness, throwing around terminologies and using somewhat circular sounding arguments.

A fundamental charge is a physical entity such as an electron, proton, or ion. Bound electrons are in atomic or molecular orbitals. Recognizing that we not are discussing the plasma phase, we can agree that protons are bound charges (locked in the nucleus of an atom). Finally, ions are bound when they are in ionic bonds, such as in NaCl. In solids, free electrons are those that are excited from a bound state to a higher energy level and subsequently delocalized through the material. Free ions are those that able to move in the solid, e.g. through vacancy defects.

Polarization is the physical separation of two different charges under an applied electric field. Polarization creates or increases a dipole. Only bound charges can be polarized because they can be displaced relative to other bound charges. Free charges are not displaced relative to each other, they will freely move (accelerate) in the direction of the applied field. This creates current.

From this fundamental starting point, going through the classes of materials by chemistry, we find that, when a material has the ability to support the creation of free charges under an applied electric field due to a voltage per length, current flow will dominate polarization. This covers the behavior of metals, semiconductors, and ionic conductive ceramics. This is not to say that such materials cannot be polarized. Indeed, under an applied field even in metals, we must appreciate that bound electrons in atomic orbitals may be induced to be displaced from their otherwise symmetrical orbits around the nucleus. This appreciation helps us distinguish optical properties of metals versus semiconductors. The opposite case to having free charges is easier to appreciate. When a material does not support the ability to sustain free charges, the only response mode to an electric field is polarization. We find the three main polarization modes are electronic, ionic, and molecular. Plasmon polarization, as a fourth mode, is distinct to optical response.

With this view, the answers to your questions are as follows: 1 - yes, 2 - yes, 3 - yes (assuming by “displacement electric” you mean displacement electric field). And yes, conductors indeed do have bound charges.

In a metal the free charges counter external macroscopic fields. Therefore there is no polarisation of the bound, inner, electrons. For ultraviolet frequencies and beyond this is no longer true and bound charges will contribute to the response.