# Electric Field of a Quadrupole

by xaer04
 P: 38 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?
 P: 292 This is a very old thread, but it's the exact same problem I'm working on. How does one apply the binomial approximation in this case? And is there a typo in the approximation that ChaoticOrder wrote down? I am at the very same equation mentioned above but do not know how to simplify it into something usable. The answer should be: $$\frac{3Q}{4\pi\epsilon_0 r^4}$$ where $$Q=2aq^2$$ Any help please?
 P: 1 If you just do simple algebra, and use $(x+a)(x-a)=x^2-a^2$ on the original poster's "more horrifying expression," a lot of terms cancel and you are left with: $E=2kqa^2\frac{3x^2-a^2}{x^2(x^2-a^2)^2}$ This is still exact, no approximations. As $x\rightarrow\infty$: $E=2kqa^2\frac{3}{x^4}$ which is the expected result. (Dimensional analysis shows Exitwound's expression for $Q$ should be $Q=2qa^2$, not $Q=2q^2a$).