Electrodynamics/Tutorials/3/4/2

Video Intro

Hi, this is Jonathan Gardner.

We're covering [section reference] of Griffiths Introduction to Electrodynamics.

I'm going to move fast, but you can always rewind.

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As always, questions in a video response or comments.

Let's get started.

Intro

We've derived the multipole expansion of the potential. As long a we're looking at points far away from the charges, then this expansion can give us some additional insight into what's really going on.

The Monopole Term

The monopole term is:

<math>V_{\text{monopole}}(P) = {1 \over 4 \pi \epsilon_0} {Q \over r}\;</math>

Note that Q is the total, net charge of the distribution. r is not the distance from the object to the point, but the distance from the origin you've chose to the point.

This means that this terms behaves as if you have all the charge on the origin. This works because as you get farther and farther from the object, the "center of charge" of the object will move closer and closer to the origin, comparatively speaking.

The Dipole Term

The dipole term becomes important when the net charge is zero.

<math>V_\text{dipole}(P) = {1 \over 4 \pi \epsilon_0} {1 \over r^2} \int r' \cos \theta\ \rho \ d\tau\;</math>

r', as you might recall, is the vector pointing to the tiny volume we're integrating. We can rewrite r' cos theta in terms of r hat, the direction of the point P from the origin.

<math>r' \cos \theta = \hat r \cdot \vec{r'}\;</math>

r hat won't change over the integration volume, so we can draw that out, and we're left with:

<math>V_\text{dipole}(P) = {1 \over 4 \pi \epsilon_0} {1 \over r^2} \hat r \cdot \int \vec{r'}\ \rho \ d\tau\;</math>

Now we have a constant part that depends only on P, and a part that depends only on the configuration of charges relative to the origin. Note that the integral is adding up vectors, the sum of which is dotted with the r hat vector pointing in the direction of P.

The integral is named the dipole moment, p. It's a vector.

<math>\vec{p} = \int \vec{r'} \rho\ d\tau\;</math>

And once we have that, the dipole potential is just:

<math>V_\text{dipole}(P) = {\vec{p} \cdot \hat r \over 4 \pi \epsilon_0 r^2}\;</math>

By now, you should know how to translate integrals like the dipole moment to point, line, and surface charges.

<math>\vec{p} = \sum q_i \vec{r_i'}\;</math>

<math>\vec{p} = \int \vec{r_i'} \lambda\ dl\;</math>

<math>\vec{p} = \int \vec{r_i'} \sigma\ da\;</math>

Physical Dipole

Let's plug in the example we did earlier. We had two point charges, opposite each other, separated by s.

<math>\vec{p} = q\vec{r_+'} - q\vec{r_-'} = q(\vec{r_+'} - \vec{r_-'}) = q \vec{s}\;</math>

As you move further and further away, this dipole term will dominate the others.

Also, if you shrunk s and stay put, then the dipole term will also dominate.

If you wanted a "pure" dipole, whose potential is exactly 1/r^2, you'd need to let s go almost all the way to zero. But the dipole moment would shrink, so you'd have to compensate by increasing the charge q. That is, real dipoles with an actual separation between the charges don't behave like pure dipoles where the charges are infinite by the separation is zero. But it's good enough to assume it's a pure dipole if you've got a very, very small separation and strong charge.

Dipole Moments Are Vectors

One final word on dipoles: They are vectors. They follow superposition, so if you have two dipoles close to each other, just add the dipole moments to get the dipole moments of both of them.

Let's take a square of four charges, like charges in opposite corners, and opposite charges in adjacent corners.

What's the dipole moment? No matter how you calculate it, it's going to be zero. You can add the dipole of the top to the dipole of the bottom, or the dipole of the sides, or the dipole of opposite corners (0 and 0 = 0), it doesn't matter.

This is, of course, a quadrupole, which has its own potential different from dipoles.

Thanks for your time!

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