# Charge-Dipole Derivation - Assumption That x >> a

• I
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?

jtbell
Mentor
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).

kmcguir
kuruman