# Dielectrics

We've examined electric fields in

- the vacuum with no charges (Laplace's equation holds)
- in conductors with lots of mobile charges ($\myv E=0$)

Insulators (dielectrics) are one kind of material that we have prominently ignored

### Dielectrics

Simplified picture of insulators (dielectrics):

- Plenty of charge inside
- But no free charges. Charges are bound (in molecules or on a crystal lattice) much more tightly than in a conductor.

Examples: Air, teflon, calcium carbonate

But a material is not alway a conductor or not always an insulator:

Put a high enough field on
insulating air, and electrons are forcibly stripped from atoms ("dielectric
breakdown")
and you get a conducting soup of ions.

In a semiconductor FET (Field Effect Transistor), an applied electric field can switch a region of the semiconducting material from conducting to non-conducting and back.

## Induced dipoles

When an electric field is applied to something like an atom which is electrically neutral, but is made up of positive and negative charges, the different charges move in response to the electric field, resulting in a small dipole moment.

The induced dipole is generally proportional to the applied field (for an isolated atom): $$\myv p = \alpha \myv E.$$

Values of $\alpha$ for a variety of atoms (units: $\frac {10^{-30}m^3}{4\pi \epsilon_0}$):

H 0.667 |
He 0.205 |
Li 24.3 |
Be 5.60 |
C 1.76 |
Ne 0.396 |
Na 24.1 |
Ar 1.64 |
K 43.4 |
Cs 59.6 |

#### Problem 4.4

*A point charge $q$ is situated a distance $r$ away from a neutral atom with polarizability $\alpha$. Find the force of attraction between them.*

We will:

- Find the field at the atom,
- Calculate the polarizaton of the atom,
- Use our previous calculations about the field of a dipole to...
- Figure out the field of the dipole back at the point charge,
- Use $F=qE$ to find the force on the point charge.

The electric field from the point charge is $E=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$.

The induced dipole is $\myv p = \alpha \myv E=\frac{1}{4\pi \epsilon_0}\frac{q \alpha}{r^2}$.

We calculated the electric field from a dipole. It was: $$\myv E_\text{dipole} = \frac{p}{4\pi \epsilon_0 r^3}(2\cos \theta\uv{r} + \sin \theta \uv{\theta})$$

Another possibility is that a molecule has a pre-existing dipole
moment, such as water. Then, when an electric field is applied, such molecules
will feel a torque turning them somewhat in the direction of
the field:

One might also imagine what happens in ionic solids...

Whatever the mechanism, the end result is that on the microscopic level, there are dipoles aligned with the field that weren't there before.

### Polarization

What if we polarize not one atom, but a big hunk of matter?

An external field induces dipole moments. There is some microscopic charge separation. I'd like to characterize this state of matter by something called...

the polarization vector:

$\myv P \equiv$ dipole moment per unit volume,

But you might be tempted to say...

"*What difference does it make?*
If I look at the microscopic level.... before, there were positive charges
and negatives charges, and afterwards, there is *still* a mix of positive and
negative charges that (as pictured) probably averages out to nothing (just
like before)!

But looking at the edges of a body full of dipoles, there is something significant going on:

Even if we only consider the charge at opposite surfaces, and the rest of the material as neutral, this leads to an effective dipole moment which is the sum of the microscopic induced dipoles.

Consider a line of $n$ dipoles, each made with charges $+q$ and $-q$ separated by a distance $d$: $\myv p_i = q\myv d$.

Now, they are all lined up such that the positive charge of one exactly overlaps with the negative charge of its neighbor. There is no net local charge between the edges, but there is still one left over charge at either end of the line, and so the dipole moment is... $$\myv p_\text{total}=qn \myv d = n\myv p.$$

So the net dipole moment is precisely the sum of the individual dipole moments, even if "inside" the line of dipoles, things sum to zero charge.

So, maybe it does make sense to talk about an "average" dipole moment,

$\myv P \equiv$ dipole moment per unit volume,

which is called the **polarization**.

The polarization is related to the local dipole moment in the same way as the charge density is related to the local charge. That is, the dipole moment $\myv p$ of a small volume $d \tau$ scales with the volume... $$\myv p = \myv P d \tau.$$

$P$ is a point function of position just like the charge density $\rho$ is a point function of position: Both of these are really local averages that don't actually take into account the atomic variation. They are something like the average over a volume which is large compared to atomic dimensions, but small compared to the macroscopic bodies we're considering.

#### A chicken and egg problem

- The applied electric field causes a polarization,
- dipoles give rise to a field,
- that field changes the electric field at the positions of polarizable matter...

We'll go after this by ignoring the applied field for the time being, and just calculating the...

### Electric field due to polarized matter

The far-field of a dipole $\myv p$ is $$V(\myv r) = \frac{1}{4\pi \epsilon_0}\frac{\myv \rr\cdot \myv p}{\rr^3}.$$

With our definition of the dipole moment of a small volume in terms of the local polarization, we can write the field of a medium characterized by $\myv P(\myv r')$ as $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \int \frac{\myv \rr\cdot \myv P(\myv r')}{\rr^3} d \tau'.$$

Now, let's wave the wand of mathematical magicness a bit, or else appeal to Saint Barbara (Chango)--patron saint of mathematicians (who is closely associated with lightning). If we consider the gradient with respect to the source coordinate $\myv r'$: $$\myv \grad '(1/\rr) = \frac{\myv \rr}{\rr^3}.$$

Using this to re-write the potential above: $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \int \myv P\cdot \myv \grad '(1/\rr) d \tau'.$$

One of the product rules for the divergence of a scalar times a vector is $\myv \grad \cdot (f \myv A) = f(\myv \grad \cdot \myv A)+\myv A \cdot (\myv \grad f)$. Using this in our potential: $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \left[\int \myv \grad '\cdot (\myv P/\rr) d \tau' - \int(1/\rr)\myv \grad '\cdot \myv P d \tau'\right].$$

Using the divergence theorem allows us to rewrite that first integral as a surface integral of a flux... $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \oint_{\cal S} \frac{\myv P\cdot d \myv a'}{\rr'} - \frac{1}{4\pi \epsilon_0} \int(1/\rr)\myv \grad '\cdot \myv P d \tau'.$$

The first term looks like the potential due to a surface charge. The units are right, so let's call this:
$$\sigma_{b} \equiv \myv P\uv{n},$$
the **bound surface charge**.

Indeed this is just the charge "sticking out" at the end of the line of charge we considered earlier.

The second term looks like the potential of a charge density: $$\rho_b \equiv -\myv \grad \cdot \myv P.$$

So we might formally re-write the potential in terms of these charge-like quantities as $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \oint_{\cal S} \frac{\sigma_b }{\rr'}da' + \frac{1}{4\pi \epsilon_0} \int \frac{\rho_b}{\rr} d \tau'.$$

OK! So now we can calculate the potential of some hunk of polarized matter by just adding up the potentials of "unbalanced charges"--the charge sticking out the edges of a polarized body, and the charge that piles up inside in regions where the polarization is changing--rather than by integrating over the polarization field.

#### Problem 4.10

*A sphere of radius $R$ carries a polarization $\myv P(\myv r) = k \myv r$.
Calculate the bound charges $\rho_b$ and $\sigma_b$, and find the field in-
and outside the sphere.*

**Inside** the sphere ($r < R$):
$$\rho_b = -\myv \grad\cdot \myv P = -\frac{1}{r^2}\frac{\del}{\del r}(r^2kr)=-(1)(r^2)3kr^2
= -3k.$$

The charge density inside the sphere is constant.

Now, draw that ol' sphere and apply Gauss' law: $$\oint_{\cal S} E_r da = \frac{1}{\epsilon_0}Q_\text{enc}$$

When the sphere has a radius $r$, $$\begineq 4\pi r^2 E_r &=&\frac{1}{\epsilon_0} \frac{4}{3}\pi r^3 (-3k)\\ & & \Rightarrow E_r = \frac{k}{\epsilon_0 r}.\endeq$$

**Outside** the sphere:

On the surface itself, there's a charge: $$\sigma_b = \myv P\uv{n} = P_r = kr|^{r=R} = kR.$$

So, the total charge enclosed in a sphere with radius $> R$ is: $$4\pi R^2 (kR) -(3k) \frac{4}{3}\pi R^3 = 0.$$

This means that $E_r=0$ outside the sphere.

#### Problem 4.11

*A short cylinder of radius $a$ and length $L$ carries a "frozen-in" uniform
polarization $\myv P$, parallel to its axis. (This is analogous to a bar magnet: it's called a 'bar electret'). Find the bound charge and sketch
the field for (i) $L \gg a$, and (ii) $L \ll a.$*

### Image credits

Ian Boggs, Werl Triptychon by Robert Campin (1438).