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| − | {{:Electrodynamics/Tutorials/Video_Introduction}}
| + | * [[/Part 1]] |
| − | | + | * [[/Part 2]] |
| − | == Introduction ==
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| − | Section 3.4 covers multipole expansions. What's this all about?
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| − | 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.
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| − | However, that's not always going to be the case, especially when the net charge is zero.
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| − | Let's look at example 10.
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| − | == Example 10 ==
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| − | An electric dipole is two equal and opposite charges separated by a distance s. Find the approximate potential far away at a point P.
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| − | We can do superposition and write down the exact equation without thinking very hard:
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| − | :<math>V(P) = {1 \over 4 \pi \epsilon_0} \left( {q \over r_+} - {1 \over r_-} \right)\;</math>
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| − | What's the curly r? We can use the law of cosines:
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| − | :<math>r_+^2 = r^2 + \left( {s \over 2}\right)^2 - r s \cos \theta\;</math>
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| − | :<math>r_-^2 = r^2 + \left( {s \over 2}\right)^2 - r s \cos (\pi - \theta)\;</math>
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| − | The "other" angle pi minus theta is simply going to flip the sign of cosine.
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| − | :<math>r_-^2 = r^2 + \left( {s \over 2}\right)^2 + r s \cos (\theta)\;</math>
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| − | Easy peasy.
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| − | Now, if we massage this a bit, first by pulling out an r^2:
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| − | :<math>r_+^2 = r^2\left(1 + {s^2\over 4 r^2} - {s \over r} \cos \theta\right)\;</math>
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| − | Then we can see what happens when <math>r >> s\;</math>. The middle term falls into 0 (cross it off):
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| − | :<math>r_+^2 = r^2\left(1 - {s \over r} \cos \theta\right)\;</math>
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| − | Square root:
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| − | :<math>r_+ = r\sqrt{1 - {s \over r} \cos \theta}\;</math>
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| − | Invert (we want 1 over r, after all)
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| − | :<math>{1 \over r_+} = {1 \over r} {1 \over \sqrt{1 - {s \over r} \cos \theta}}\;</math>
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