Charge-Dipole Derivation - Assumption That x >> a

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In this derivation:

https://cpb-us-e1.wpmucdn.com/sites.../1599/files/2017/06/taylor_series-14rhgdo.pdf

they assume in equation (8) that x >> a in order to use the Taylor Expansion because a/x has difficult behavior. Why does that assumption work? Meaning, why can we assume the dipole is that much smaller? What happens if x is not much greater than a, then what would be the process to find a solution?
 

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  • #2
jtbell
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They're simply saying that if you're far away from the charges (x >> a), then you can use the Taylor expansion to arrive at a nice simple formula for the field that goes like 1/r3. It's an approximation, but it gets closer to the truth, percentage-wise, as you go further from the charges.

If you're not far away from the charges, then you have to use the exact but somewhat messy solution, equation (8).
 
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  • #3
kuruman
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The exact solution that works for any value of x is given by Equation (8) in the reference you quoted. For example, when you are at x = 0, the electric field is zero as should be obvious and as predicted by equation (8). One considers the behavior at x >> a because in that limit the exact expression becomes simplified. This limit has its use when, for example, you consider dipoles of atomic size. The fields they generate are usually observed at distance much larger than the size of an atom in which case the approximation gives a very good description of the field.
 
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  • #4
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Great! Thanks for those responses. There was never any time to ask questions like this in class. Always trying to cover too many chapters from the text every week, so they move too fast.
 

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