Electric displacement

  • At this time of year winter is still hanging on...
  • And inspired by this morning's chapel...

Isn't it about time for a new field??





We're talking, of course, about a new vector field...

So, inside a dielectric, we have the "bound charge" $\rho_b = -\myv\grad \cdot \myv P$ that arises from any changes in polarization.

There may also be other charge that's been placed in the system. (That doesn't arise because of the dipoles.) E.g. Ion bombardment.

Since the charges from dipole effects was "bound" charge, call this other charge "free charge" (It is not "free" to move, by the way) $\rho_f$.

Both of these kinds of 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: $$\myv D \equiv \epsilon_0 \myv E + \myv P.$$

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

Two other ways of putting this ... $$\oint \myv D \cdot d \myv a = (Q_f)_{\text{enc}}.$$

And also, we could say that...

$\myv D$ 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 $r=b$: $$\sigma_b = \myv P \cdot \uv{n} =\{\begin{array}{lc} +\myv P\cdot\uv{r}=k/b & \text{ at } R=B\\ -\myv P\cdot \uv r=-k/a & \text{ at }\ r=a\end{array}\right. $$

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

 

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 $\myv E$ in terms of displacement and polarization, this means $$\myv D_{\parallel}^{\text{above}} - \myv D_{\parallel}^{\text{below}} = \myv P_{\parallel}^{\text{above}} - \myv P_{\parallel}^{\text{below}} .$$

image credits

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