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

"An electric quadrupole consists 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

\frac{1}{x^4}"

---------(+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

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

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.

\frac{kq}{(x-a)^2} \hat{i} - \frac{2kq}{x^2} \hat{i} + \frac{kq}{(x+a)^2} \hat{i}

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

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]

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:

\frac{0}{x^6}

where did i go wrong? And how do I start the second part of the question:?

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# Homework Help: Field generated by electric quadrupole

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