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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 \left(1 - {s \over r} \cos \theta\right)^{1/2}\;</math>
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| − | Invert (we want 1 over r, after all)
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| − | :<math>{1 \over r_+} = {1 \over r} \left(1 - {s \over r} \cos \theta \right)^{-1/2}\;</math>
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| − | Griffiths mentions a binomial expansion. I'm going to help you out a bit...
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| − | :<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>
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| − | 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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| − | (Be sure to like & share this video: I'm saving you a lot of time!)
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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:
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| − | :<math>1 + \left({-1 \over 2}\right) \left(-{s \over r} \cos \theta\right) + ...</math>
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| − | We'll ignore the higher terms; we're physicist and we're approximating, so we can ignore pretty much whatever we feel like ignoring.
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| − | Now we have:
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| − | :<math>{1 \over r_+} = {1 \over r}\left(1 + {s \over 2r} \cos \theta\right)\;</math>
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| − | For the negative one:
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| − | :<math>{1 \over r_-} = {1 \over r}\left(1 - {s \over 2r} \cos \theta\right)\;</math>
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| − | (You can check that yourself.)
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| − | Now let's subtract the two:
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| − | :<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>
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| − | The 1's cancel, and the cosine terms combine:
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| − | :<math>{1 \over r_+} - {1 \over r_+} = {s \over r^2} \cos \theta\;</math>
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| − | Plug it back in:
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| − | :<math>V(P) = {1 \over 4 \pi \epsilon_0} {q s \cos \theta \over r^2}\;</math>
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| − | Which is neat and tidy and happy.
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| − | 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.
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| − | 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.
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| − | (Draw them).
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| − | [[../Part 2]] | |
