(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

"Anelectric quadrupoleconsists of two oppositely charged dipoles in close proximity. (a) Calculate the field of the quadrupole shown in the diagram for points to the right of x = a, and (b) show that for x>>a the quadrupole field falls off as

[tex]\frac{1}{x^4}[/tex]"

---------(+q)-----(-2q)-----(+q)----------

the left charge is at position x = -a, the middle is at x = 0, and the right is at x = a.

2. Relevant equations

[tex]\vec{E}(P)=\sum{\frac{kq}{r^2} \hat{r}}[/tex]

3. The attempt at a solution

i found the charge to be this convoluted mess, but i don't see any ways of simplifying it.

[tex]\frac{kq}{(x-a)^2} \hat{i} - \frac{2kq}{x^2} \hat{i} + \frac{kq}{(x+a)^2} \hat{i}[/tex]

when i combined the fractions i got something even more horrifying.

[tex]kq \left[ \frac{x^2(x+a)^2 - 2(x+a)^2(x-a)^2+x^2(x-a)^2} {x^2(x-a)^2(x+a)^2} \right][/tex]

the way i understand it, when you show that something "falls off" you negate the a as x becomes very big, which makes sense... the a is very small and therefore pretty much negligible. however, when i did that and simplified, i got this:

[tex]\frac{0}{x^6}[/tex]

where did i go wrong?

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# Electric Field of a Quadrupole

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