Electric displacement field in a conductor

The photo given below is from the solution manual to Griffiths Electrodynamics, it claims that the displacement field inside a conductor a zero. Now I do not see why it is obvious , $$D = εE + P$$, now I know that the net Electric Field inside a conductor is zero so the first term definitely goes to zero but what about the Polarization term , how do I make sense of that?

• try to have a look at the examples at the end of this answer physics.stackexchange.com/q/798254 Feb 1 at 0:02
• If there is a no E field at a point, is it in your view that a tiny dipole moment could form at the point (polarization)? If so what causes that tiny dipole moment when the static (long lasting) result is of no E field? The positive and negative charged attract, only the presence of an E field at a point can create a tiny dipole. Free charge also needs force to move, but unlike a dipole moment, after it moved, it can rest in its new position with zero force on it. A dipole moment needs lasting force (E field) to maintain itself. Feb 6 at 14:12

I believe the questions stems from viewing $$\vec{E}$$ and $$\vec{P}$$ as independent.

But $$\vec{P} = \epsilon_0 \chi \vec{E}$$

Thus: $$\vec{D} = \epsilon_0\vec{E}+\epsilon_0 \chi \vec{E}=\epsilon_0 (1+\chi) \vec{E}$$

Thus if $$\vec{E}$$ is zero then $$\vec{P}$$ is also zero.

E in all cases is the final and total, all cause field, at the point. It doesn't refer to just an external field, but E is the final, when all said and done, field at the point in question. As that's the case, when there is no field E at the point, then there's also no polarization at the same point (the lasting polarization after everything moved in place is a function of the final field at a point).

I've noticed from the comment below that you described the situation in metal as polarization. But note that polarization refers to charge bound in place forming a dipole under an external field. Whereas movement of free charge from within the metal to its boundry, isn't considered polarizaion, but simply movement of free charge.

In fact, if I'm not mistaken, a perfect conductor never polarizes - the concentration of free charge at the boundry may result in a perpendicular and "out" pointing field away from the boundry (away from the metal and towards empty space). But that's free charge rather than a polarized bound charge, and what's more, the actual resulting field is just outside the material in space, and not just inside it. With the field being just outside the boundry, there is nothing that can polarize - it's just empty space.

• This answer is only true for linear dialectics. Feb 2 at 17:48
• @JerroldFranklin but also linear conductors, isn't it the case? Anything linear from a perfect dielectric to a perfect conductor? Feb 2 at 18:21
• Anyhow, thanks for your commnet. Feb 2 at 19:04

I think that in this case it can be quite useful to revisit the microscopic interpretation of polarization in classical electromagnetism.

In a dielectric material with no external electric field we have molecules that have overall no electric charge, and we may distinguish between two cases:

1. The charges within a single molecule are uniformly distributed an the single molecule has dipole moment.
2. The charges within a single molecule are not uniformly distributed and the molecule has an intrinsic dipole moment (take water as the classical example). However the average dipole moment is zero since the singular dipoles are disordered.

Now we apply an external electric field and we see that such field causes polarization within the material. In case $$1$$ we may think that the electric field tends to 'stretch' the molecules and separates the positive and negative charges causing the presence of an average dipole moment that we measure using $$\textbf{P}$$. In case $$2$$ the electric field causes the molecule dipoles to order among each other resulting again in an average dipole moment $$\textbf{P}$$.

What happens in a perfect conductor?

If we have a perfect conductor then applying an external electric field does not cause the molecules to 'stretch' into dipoles but rather they split into electrons and ions and then travel to the opposite sides of the conductor until the reach the boundaries, thus all the charges are on the external surface and you have no dipole moment in the bulk. That is way you can safely assume $$\textbf{P}=0$$ in the bulk of a conductor.

Please keep in mind that this is a just a simple interpretation of the actual phenomena that happen at the atomic scale, yet it works well in classical EM.

The polarisation field, at least in linear materials, is proportional to the electric field. The electric field is zero in an ideal conductor and thus so is the polarisation field. Note, there is no distinction between an "applied" electric field and the net electric field in this relationship, it is simply the electric field where you are measuring the polarisation.

To put it another way, the induced dipole moment per unit volume (the definition of the polarisation) is zero inside a (perfect) conductor because there is no electric field inside a conductor and so there are no induced dipoles in the material.

My original answer was wrong, so I am replacing it with this answer (in Gaussian units), which I believe is correct. In a dielectric with a permanent polarization, $$\bf P$$, any electric field will move free charges until the electric field produced by the free charges and by the polarization vanishes. Therefore, the electric field, $$\bf E$$, will be zero in a conductor, even one with a permanent polarization. In what is called a ferroelectric material, the electric field may modify $$\bf P$$, but will not completely extinguish it. This means that $$\bf D=E+4\pi P=4\pi P$$ will not vanish in a ferroelectric conductor. If the divergence of $$\bf P$$ vanishes, there will be no free charge inside the conductor, and all charge will be on the surface. However, if $$\bf P$$ has a divergence, there will be a charge density inside the conductor, given by $$\rho=\nabla\cdot P$$. Griffiths must have assumed that $$\bf P=0.$$

• a homogeneous $\mathbf P = const$ is enough for $\nabla \cdot \mathbf P =0$ to hold inside Feb 4 at 23:56
• Yes. I said "Griffiths must have assumed that P=0." because the quote from Griffiths said "D=0". Feb 6 at 2:50

This is a result of the electrostatic balance of a conductor.

First, it is obvious that $$\mathbf{E}$$ inside the conductor is zero, because any tiny electric field will cause the motion of the free electrons, until $$\mathbf{E} = 0$$ is satisfied everywhere in the conductor. This is also the mechanism of electrostatic screening.

Then,$$\mathbf{P}$$ is defined as the density of the electric dipoles induced by external electric field. However, because the electric field itself is screened by the free electrons, no dipoles can be induced. So $$\mathbf{P}$$ is also zero.