Electrodynamics/Tutorials/3/4/1
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.
Introduction
Section 3.4 covers multipole expansions. What's this all about?
Well, remember how we used to check our answers by looking at distances far away from our charge distributions? If the charge distributions looked like a point charge with a charge of the total charge of the distribution, then we probably got our answer right.
However, that's not always going to be the case, especially when the net charge is zero.
Let's look at example 10.
Example 10
An electric dipole is two equal and opposite charges separated by a distance s. Find the approximate potential far away at a point P.
We can do superposition and write down the exact equation without thinking very hard:
- <math>V(P) = {1 \over 4 \pi \epsilon_0} \left( {q \over r_+} - {1 \over r_-} \right)\;</math>
What's the curly r? We can use the law of cosines:
- <math>r_+^2 = r^2 + \left( {s \over 2}\right)^2 - r s \cos \theta\;</math>
- <math>r_-^2 = r^2 + \left( {s \over 2}\right)^2 - r s \cos (\pi - \theta)\;</math>
The "other" angle pi minus theta is simply going to flip the sign of cosine.
- <math>r_-^2 = r^2 + \left( {s \over 2}\right)^2 + r s \cos (\theta)\;</math>
Easy peasy.
Now, if we massage this a bit, first by pulling out an r^2:
- <math>r_+^2 = r^2\left(1 + {s^2\over 4 r^2} - {s \over r} \cos \theta\right)\;</math>
Then we can see what happens when <math>r >> s\;</math>. The middle term falls into 0 (cross it off):
- <math>r_+^2 = r^2\left(1 - {s \over r} \cos \theta\right)\;</math>
Square root:
- <math>r_+ = r \left(1 - {s \over r} \cos \theta\right)^{1/2}\;</math>
Invert (we want 1 over r, after all)
- <math>{1 \over r_+} = {1 \over r} \left(1 - {s \over r} \cos \theta \right)^{-1/2}\;</math>
Griffiths mentions a binomial expansion. I'm going to help you out a bit...
- <math>(1 + x)^n = 1 + {n \over 1!}x^1 + {n(n-1) \over 2!}x^2 + {n(n-1)(n-2) \over 3!}x^3\;</math>
The condition is that x must be between 1 and negative 1. We certainly satisfy this condition. cosine can only be -1 to 1, and s is much smaller than r.
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Let's plug in what we have, n=-1/2, and x=-s/r cos theta, right? So the first few terms are:
- <math>1 + \left({-1 \over 2}\right) \left(-{s \over r} \cos \theta\right) + ...</math>
We'll ignore the higher terms; we're physicist and we're approximating, so we can ignore pretty much whatever we feel like ignoring.
Now we have:
- <math>{1 \over r_+} = {1 \over r}\left(1 + {s \over 2r} \cos \theta\right)\;</math>
For the negative one:
- <math>{1 \over r_-} = {1 \over r}\left(1 - {s \over 2r} \cos \theta\right)\;</math>
(You can check that yourself.)
Now let's subtract the two:
- <math>{1 \over r_+} - {1 \over r_+} = {1 \over r}\left(1 + {s \over 2r} \cos \theta\right) - {1 \over r}\left(1 - {s \over 2r} \cos \theta\right)\;</math>
The 1's cancel, and the cosine terms combine:
- <math>{1 \over r_+} - {1 \over r_+} = {s \over r^2} \cos \theta\;</math>
Plug it back in:
- <math>V(P) = {1 \over 4 \pi \epsilon_0} {q s \cos \theta \over r^2}\;</math>
Which is neat and tidy and happy.
This says that while a single point charge goes 1/r, a dipole's potential goes 1/r^2. For quadrapoles and octopoles, they drop off by 1/r^3 and 1/r^4.
Note that q is not the total charge---that's zero for dipoles and above. It is the charge of 1/2 of the dipole.
(Draw them).
