Polarization per unit volume

...or polarization "density"

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

Polarization "density"

Imagine that we...

Have an insulator.
Turn on an external electric field.
The field induces an "average polarization" throughout the insulator: That is to say, either an average charge separation, or a preferred direction in a collection of previously isotropic fixed dipoles.

You might be tempted to say...
"What difference does it make? that the material is polarized? 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), so no field!

That is... there is no field due to the polarized matter in the bulk of the body. And of course there are many more atoms in the "bulk" of the body than at the surface.
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 turns out to be the sum of the microscopic induced dipoles:
Consider a line of $n$ identical 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 (red) positive charge of one exactly overlaps with the (blue) 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 of this line is equal to the magnitude of the two charges times their separation: $$ \myv p_\text{total}=qn \myv d =n\myv p.$$
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.
$\myv P(\myv r)$ is a "point function" of position, $\myv r$, just like the charge density $\rho(\myv r)$ 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 in pieces. But we'll start by ignoring any applied electric 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 find the dipole moment at any location in an insulator if we know $\myv P(\myv r')$ from $\myv p=\myv P\,d\tau'$ and then add up ($\to \int$) the fields from all of those local dipoles to get the field: $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \int \frac{\myv \rr\cdot \myv P(\myv r')}{\rr^3} d \tau'.$$
"Bound" charge

Now we're going to work out that expression for the electric potential... $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \int \frac{\myv \rr\cdot \myv P(\myv r')}{\rr^3} d \tau'.$$
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) to start by remembering that... We can express $\myv\rr/\rr^3=(1/\rr^2)\uv \rr$ in terms of the gradient of a function, with respect to the source coordinate $\myv r'$, as: $$\myv \grad '(1/\rr) = \frac{\myv \rr}{\rr^3}.$$
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... $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \oint_{\cal S} \frac{\myv P\cdot \uv n}{\rr'} \,d a' - \frac{1}{4\pi \epsilon_0} \int(1/\rr)\myv \grad '\cdot \myv P d \tau'.$$
$\myv P$ has units of (charge*distance)/volume=charge/area, so surface charge density. So, the first term looks like the potential due to some surface charge density on a surface. $$\sigma_{b} \equiv \myv P\cdot\uv{n},$$ the bound surface charge density.
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.
Example 4.2

A few notes to go along with that problem...
Imagine that we have a sphere with a constant polarization density, in the $z$ direction, so $\myv P=(0,0,P)$, where $P$ is a constant.
The charge inside the sphere is given by $$\begineq \rho_b(\myv r) &= \myv \grad\cdot \myv P\\ &=\frac{\del}{\del x}0+\frac{\del}{\del y}0+\frac{\del}{\del z}P\\ &= 0 + 0 + 0 = 0. \endeq$$ And the charge on the outside of the sphere is $$\sigma_p=\myv P\cdot\uv n= P\uv z\cdot \uv r=P\cos\theta.$$ That is to say, this is the same situation as example 3.9 (towards the end), where the potential due to a surface charge $\sigma_0(\theta)=k\cos\theta$ on the surface of a sphere of radius $R$ was calculated.
The result of example 3.9, for the region outside of the sphere, was found to be (with $k\equiv P$): $$V(r,\theta)=\frac{P}{3\epsilon_0}\frac{R^3}{r^2}\cos\theta.$$ Since the dipole density is constant, the total dipole moment of the sphere (see Griffiths' 4.16) is the dipole density times the volume of the sphere, that is: $$\myv p = \myv P \frac 43 \pi R^3.$$ Solving this for the magnitude $P=|\myv P|$ and substituting into the potential: $$\begineq V(r,\theta)&=\frac{p}{\frac 43 \pi R^3}\frac{1}{3\epsilon_0}\frac{R^3}{r^2}\cos\theta\\ &=\frac{p\cos\theta}{4\pi \epsilon_0 r^2}= \frac{\myv p\cdot \uv r}{4\pi \epsilon_0 r^2} \endeq$$ Which we *recognize* as the field of a dipole, $\myv p$, at the origin.
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 $r$, $$\begineq 4\pi r^2 E_r &=\frac{1}{\epsilon_0} \frac{4}{3}\pi r^3 (-3k)\\ & & \Rightarrow E_r = \frac{kr}{\epsilon_0 }.\endeq$$
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).

Polarization per unit volume

Polarization "density"

A chicken and egg problem

Electric field due to polarized matter

"Bound" charge

Example 4.2

Problem 4.10

Problem 4.11

Image credits