Linear dielectrics (recursion approach)

I’ve been prepping for my fall course, Advanced Electricity and Magnetism for juniors and seniors, taught out of Griffiths great book. One of the topics that I want to deal with better this time around is electrostatics in materials (chapter 4, for those of you following along). I want to help my students better understand the relationships among:

  • The electric field, \vec{E}
  • The field of a dipole, \vec{p}
  • The polarization of a material, \vec{P}
  • The electric susceptibility, \chi
  • The electric permitivity, \varepsilon

What I’d like to focus on in this post, really for my own purposes in about two months or so, is how to get to \varepsilon from the susceptibility, \chi.

Ok, here’s the deal. When you expose a linear dielectric material, characterized by \chi, to an external electric field, \vec{E}_\text{ext}, the material becomes polarized, with each molecule shifting around its charge a little. This charge shift is characterized by the polarization of the material:

\vec{P}=\varepsilon_0\chi \vec{E}

It is the fact that the polarization is linearly related to the field that gives these types of material their name (linear dielectrics). Note that I didn’t put a subscript on the field in this case. In fact, that’s kind of the big deal here!

Once a material has become polarized, it’s like it has a bunch of little dipoles all throughout it. However, in the bulk, the plus charges that have moved a little away from their own minus charges tend to get close and cozy to minus charges near by. The net result is very little meso-scopic charge in the material. That argument breaks down at the boundary, of course, and it’s there that you get some leftover plus charge on one end and minus on the other.

That separation of charge produces its own electric field that fights against the original electric field. That’s where the absence of a subscript in the equation above becomes so interesting. The polarization of the material is a function of the total field inside the material, \vec{E}_\text{ext}+\vec{E}_\text{pol}.

So first there’s an external field that polarizes the material, then that polarization creates a field that makes the total field a little smaller, that then makes the polarization less so the field it makes doesn’t fight back so hard. It goes on and on and on. How can we get out of that recursive relationship?

Analytic approach

There is a rather straightforward analytical result that Griffiths describes in his book. The gist is this:


Here we note that the separated charges act just like a parallel plate capacitor (with \sigma=P):


Now we use the field due to the polarization:

\vec{E}=\vec{E}_\text{ext}-\frac{1}{\varepsilon_0}\varepsilon_0\chi \vec{E}

and solve:



Relatively straight forward and giving the correct result. I simply wondered if you could simulate how this end result happens dynamically.

Single steps

The first thing I tried was to turn on a field, determine the polarization, fix the field, and repeat. I expected this to converge nicely and it did:

low chi recursion

low chi recursion: chi = 0.4, final field = 1/1.4 ~ 0.71

I started with an external field of one and for each iteration I found the correct polarization field correction. After applying that, the total field changed which meant the polarization field changes. You can see that just 10 iterations was enough to get the correct result.

There’s a problem with this approach, however. Check out what happens if \chi gets bigger than one:

high chi rescursion

high chi recursion: chi = 1.1

Here there’s a runaway situation. So what happened? In the first iteration, the field “correction” from the polarization was stronger than the original external field. This meant that in the next iteration the field was pointing in the opposite direction. This meant that the next iteration had the external and polarization fields reinforcing each other. This will happen any time \chi gets bigger than one.

Dynamical modeling

So how is this problem fixed? For me, it comes down to recognizing that the polarization will never contribute a field that is in the same direction as the external field. In the last example, in the first step, the material will begin to be polarized. As it does, the total field starts to reduce, which dynamically reduces the polarization. In other words, waiting for the “step” to finish before calculating the new field was the big mistake.

To come up with another model I could have my students work on, I went back to the beginning. How can we model the microscopic polarization? One easy approach, again laid out by Griffiths, is to think of the charges in a molecule to be connected by a spring in a viscous medium. When a field is turned out, this spring stretches, creating a charge separation (or polarization) and the stretching ends when the electric force is balanced by the spring force.

At the macroscopic level, can you model the whole polarization in that damped/driven spring model? Sure, why not. What I do is numerically solve the following problem:

m\ddot{P}=-k P-\beta \dot{P}+E_\text{ext}

where I can choose m, k, and \beta to model the situation of interest. Specifically, at the steady state I want the polarization to be equal to \chi E_\text{ext} so k=1/\chi. m and \beta are adjustable to see how fast the system settles down to the steady state. Here’s what it looks like with m=1, \beta=0.1, and \chi=0.4 (note that what I plot is actually the total electric field or E_\text{ext}-E_P since the polarization field fights against the applied field):

low chi recursion spring

Damped spring model with chi=0.4 (expect Etotal=0.71)

The field oscillates a little but eventually settles down to just where we expect (1/(1+\chi).

So can this approach handle \chi>1? You bet! Here’s an example with m=1, \beta=0.1, and \chi=10:

high chi recursion spring

Total field with chi = 10 (expected field = 1/11 =0.09)

What’s cool about that particular one is that there are points when the total field is in the opposite direction of the applied field but it, too, eventually settles down to the expected result.


I know that this exercise has helped me wrap my brain around all of this but I’m not sure which parts to use when I teach it. The analytical approach above is quite straightforward and I could easily just do that. However, since my students are quite capable of the Mathematica it would take to pull off this modeling, I’m very tempted to make something like this modeling one of my standards for the course. At the very least I want them to see the run-away solutions if they do the naive approach of letting the field settle before re-calculating the polarization.

As usual I end with more questions. What do you think about these approaches?

About Andy Rundquist

Professor of physics at Hamline University in St. Paul, MN
This entry was posted in math, mathematica, physics and tagged . Bookmark the permalink.

5 Responses to Linear dielectrics (recursion approach)

  1. Ryan S. says:

    As a student about to take the course I’d say that having the modeling standards would be interesting to work with, but I’d also wonder if part of that same standard would involve the student discussing the understanding via the modeling vs the understanding via the analytical solution presented in the textbook. I’d think that the student’s understanding of the Physics behind a concept like linear dielectrics would factor in almost as strongly as would the modeling of the concept, if not more so.

  2. bwfrank says:

    I think you should write this as an AJP paper…

  3. Andy "SuperFly" Rundquist says:

    Here’s a student attempting the standard “I can discuss the foundations of, usefulness of, and ramifications of linear dielectrics.”:

    Here’s my review of it:

  4. Aashish Priye says:

    Thanks a lot. That was very helpful. I liked your idea of presenting this system through such mathematical models. This will definitely help with the students learning curve.

    • Andy "SuperFly" Rundquist says:

      you’re quite welcome. I’ve really come to appreciate having this blog to jot these kinds of thoughts down. In the old days I’d come up with an explanation like this, do it once in the classroom, and then forget it before the next year.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s