Electric displacement



Isn't it about time we hop into a new field too??



We're talking, of course, about a new vector field...the Displacement field, $$\myv D(\myv r) \equiv \epsilon_0 \myv E(\myv r) + \myv P(\myv r),$$ to deal with our "chicken and egg" problem of the electric field $\to$ polarization $\to$ more changes to the electric field $\to$ more changes to the polarization $\to$...

*Free* charge

So, inside a dielectric, we have the "bound charge" $\rho_b = -\myv\grad \cdot \myv P$. This is the charge that "piles up" if the polarization is changing with position (instead of constant).

There may also be other charge that's been placed in the system. (That doesn't arise because of the dipoles.) E.g. Read a bit about Ion implantation:

Materials Engineering Group

With a source of ions, and control over an accelerating voltage, you can bombard a material with ions:

  • The ion current * time of bombardment / charge per ion = "dose": the total number of ions hitting the surface. Depending on the ion, a certain fraction will bounce off the surface, and the remaining atoms are "implanted": they come to rest somewhere below the surface of the bombarded material.
  • Controlling the accelerating voltage controls the kinetic energy of the bombarding ions. The higher the accelerating voltage, the deeper the average depth of the implanted ions.

Since the charges from dipole effects was "bound" charge, We'll call this other, implanted charge "free charge" $\rho_f$. (It is not "free" to move, by the way. We're *not* talking about conductors!)

Both free and bound charges are real charges, and so should be related to the divergence of the (total) electric field by: $$\epsilon_0 \myv \grad \cdot \myv E = \rho = \rho_b+\rho_f = - \myv \grad \cdot \myv P + \rho_f.$$

Re-arranging the two divergences into the divergence of one thing... $$\myv \grad \cdot (\epsilon_0 \myv E+\myv P) = \rho_f.$$

We'll define the quantity in parenthesis as a new vector field:

The electric displacement field: $$\myv D \equiv \epsilon_0 \myv E + \myv P.$$

[In Thermodynamics we do this: Enthalphy, $H\equiv U +PV$ and call it a "Legendre transform".]

Apparently: $$\myv \grad \cdot \myv D = \rho_f.$$

Any time we know the divergence of a vector field, we can use the divergence theorem to write a "Gauss' Law" for the field: $$\oint \myv D \cdot d \myv a = (Q_f)_{\text{enc}}.$$

Also, we can devise some rules for drawing field lines for the $\myv D$ field: $\myv D$ field lines terminate on "free" charges (not just any charge).

Problem 4.15

A thick spherical shell (inner radius $a$, outer radius $b$) is made of dielectric material with a "frozen in " polarization $$\myv P (\myv r) = \frac{k}{r}\uv{r}.$$ There is no free charge in this problem. Find the electric field using Gauss' law with the bound charges, and also using the "new Gauss' law" for the displacement.

Calculating the bound charges $$\begineq\rho_b&=-\myv \grad \cdot \myv P\\&=-\frac{1}{r^2}\frac{\del}{\del r}(r^2 k/r)\\ &= -\frac{k}{r^2}.\endeq$$

There's also a bound surface charge at $r=a$ and $r=b$: $$\sigma_b = \myv P \cdot \uv{n} =\begin{cases} +\myv P\cdot\uv{r}=k/b & \text{ at } r=b\\ -\myv P\cdot \uv r=-k/a & \text{ at }\ r=a \end{cases} $$

Now we can start to use Gauss' law. For $r < a$ there is no enclosed charge, and the electric field is zero.

For $a < r < b$ the charge enclosed by a gaussian sphere is both the surface charge at $r=a$ as well as some of the bound volume charge $\rho_b$: $$\begineq 4 \pi E_r r^2 &= \frac{1}{\epsilon_0}\left(-4\pi a^2 \sigma_b - \int_a^r dr' 4 \pi r'^2 (-\frac{k}{r'^2})\right)\\ &=\frac{4 \pi}{\epsilon_0}\left(-a^2 k/a - k(r-a)\right) = -\frac{4\pi}{\epsilon_0}kr .\endeq$$

Re-arranging for $E_r$: $$E_r=-\frac{kr}{\epsilon_0 r^2}=-\frac{k}{\epsilon_0 r}.$$

For $b \lt r$: The total enclosed charge in a sphere of radius $r$ is 0, so the electric field is zero.

Using the displacement instead

Since there are no free charges, $\oint_\cal{ S} \myv D \cdot d \myv a = (Q_f)_{\text{enc}} = 0$,

So everywhere $\myv D = 0 $... $$0= \epsilon_0 \myv E + \myv P => \myv E = -\frac{1}{\epsilon_0} \myv P.$$

For $r<a$ and $r>b$: $\myv E = 0$.

For $a<r<b$: $\myv E = -\frac{1}{\epsilon_0}\frac{k}{r}\uv{r}$.

Curl of $\myv D$

$$\myv \grad \times \myv D = \epsilon_0(\myv \grad \times \myv E) + \myv \grad \times \myv P=\myv \grad \times \myv P.$$

In the case of the "bar electret" of problem 11, can you see where $\myv \grad \times \myv P \neq 0$?

So, if the curl is not reliably zero, there's no potential: $\myv D \neq \myv \grad \Phi.$

We did a group sketching exercise, sketching first the the E-field of an electret, then figuring out D.

Field of a polarized sphere

[Example 4.2] - using Legendre polynomials

Do it! Result is used in problem 4.16a

Boundary conditions between two dielectrics

We derived a couple of boundary conditions for the electric field near an interface.

So if we apply our new Gauss' law for $\myv D$ related to $Q_\text{f enc}$ to this pillbox, we find: $$\Delta D_{\perp} = D_{\perp}^{\text{above}} - D_{\perp}^{\text{below}}=\sigma_f $$

Check: Not $\sigma_f/\epsilon_0$??



The contour integral of the electric field on the path pictured below is still zero if there are different dielectrics on either side of the boundary (since this depended on $\myv \grad \times \myv E = 0$ which is still true):

This implies that $$\Rightarrow \myv E_{\parallel}^{\text{above}} - \myv E_{\parallel}^{\text{below}} = 0.$$

Substituting in for $\myv E=\epsilon_0\myv D -\epsilon_0\myv P $ in terms of displacement and polarization, and re-arranging a bit, these boundary conditions imply that: $$\myv D_{\parallel}^{\text{above}} - \myv D_{\parallel}^{\text{below}} = \myv P_{\parallel}^{\text{above}} - \myv P_{\parallel}^{\text{below}} .$$

image credits

UN