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.

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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\sqrt{1 - {s \over r} \cos \theta}\;</math>

Invert (we want 1 over r, after all)

<math>{1 \over r_+} = {1 \over r} {1 \over \sqrt{1 - {s \over r} \cos \theta}}\;</math>