# Understanding the Electric Displacement Field

1. Nov 19, 2013

### sigma_

I'm in an upper division undergrad E&M course using Griffths' Electrodynamics text, and I've been struggling to understand the intuition/motivation behind the displacement field, D.

1) D is defined as 0E + P), and is sometimes referred to as the flux density vector.

In LIH dielectrics, P is proportional to E by the relationship

2) P = ε0χeE

Making the substitution with the original definition:

3) D = (ε0E + ε0χeE) → D = εE (using the definition 1+χer)

To me, what these equations are saying is (please forgive my less than formal language, I'm making my best attempt)

1) The displacement field at some point within a dielectric (or anywhere, for that matter) is equal to the field within it, which is smaller in magnitude than the applied field, because it is weakened by the opposing field produced by the separation of charge at the outside edges of the dipole material (the "head" of one dipole canceling the oppositely charged "tail" of the neighboring dipole in the media) plus the Polarization (dipole moment per unit volume), which is in the opposite direction of the field generated by the dipole material which opposes the applied field (field due to free charge). By adding these two quantities, we essentially ignore the field of the dipole and retrieve something equal in magnitude to the field due to only the free charge.

This, to me makes sense, since the closed surface integral of D (dot) dA is equal to the free charge enclosed:

But what are 2) and 3) really saying?

If in 3) we say D = εE where E is the total field in the dielectric, multiplying by epsilon must serve to increase the magnitude of the RHS of the equation because D is also defined by the field of only the free charge, which must be greater than the field of the free charge minus the opposing field due to polarization.

but then how can 2) make sense? How can the polarization be larger in magnitude than the field in the dielectric?

I feel as though my understanding of something here must be completely backwards.

Last edited: Nov 19, 2013
2. Nov 19, 2013

### tiny-tim

hi sigma_!
E D and -P are the electric fields due to the total charge, the free charge, and the bound charge, respectively

since the bound electric field is (usually) in exactly the opposite direction to the free and total fields D and E (since it is caused by the molecular dipoles opposing D), it is conventionally written as minus P, so that all three letters represent fields in (usually) the same direction

i think it's that minus which is confusing you

3. Nov 19, 2013

### Jano L.

Yes, this is almost right (in SI the electric field enters the definition of $\mathbf D$ in the expression $\epsilon_0 \mathbf E$).

(bold mine)
While this may be correct inside dielectric in some cases (dielectric inside a thin capacitor), generally it is not true. It is true when the polarization is proportional to minus the electric field due to bound charges. This need not be so, because the polarization (by definition) satisfies different boundary condition from the electric field of bound charges; the condition is $\mathbf P = 0$ in vacuum, while the electric field due to the dielectric satisfies only $\lim_{|\mathbf r| \rightarrow \infty}\mathbf E_d(\mathbf r) \rightarrow 0$, and can be non-zero in vacuum.

For thin capacitor the electric field due to the dielectric $\mathbf E_d$ satisfies the same boundary condition as polarization, namely $\mathbf E_d = 0$ and hence in this case what you have inferred is correct.

But in general, $\mathbf E_d$ does not have to satisfy this condition. For example, imagine a long rod of length $L$ made of dielectric material, with permanent and uniform polarization $\mathbf P$ pointing along the long axis of the rod and with cross-section $\Delta S$. The bound charges appear only at the ends of the rod, and their electric field is that of a dipole of charge $P \Delta S$ and length $L$, non-uniform and non-zero almost everywhere in the vacuum and the rod. This is very different field from the polarization field, which is uniform in the rod and zero in the vacuum.

In electrostatics, the polarization $\mathbf P$ is best defined as expected average dipole moment of molecules that have at least some part in a small control volume divided by this volume (the volume should contain many , say >1000 molecules) or better, as expected average dipole moment of one molecule times number density of molecules.

4. Nov 19, 2013

### sigma_

Thank you! your post prompted me to do a little more algebra, keeping what you just said in mind.
Of course! P can absolutely be larger in magnitude than the total field in the dielectric!