
#1
Aug2407, 12:47 PM

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 [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}{(xa)^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(xa)^2+x^2(xa)^2} {x^2(xa)^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? 



#2
Aug2407, 12:56 PM

P: 11

Use the binomial approximation, and learn to love it:
(1+b)^2 ~= 1  2b + 3b^2 for b<<1 Start with your first expression...forget combining fractions. Factor an x^2 out of all the denominators to get terms like the one I just wrote. Apply the approximation and collect like terms. You'll notice that most terms will cancel and what you will be left with is the desired x^6 dependence. 



#3
Jul1509, 06:27 PM

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: [tex]\frac{3Q}{4\pi\epsilon_0 r^4}[/tex] where [tex]Q=2aq^2[/tex] Any help please? 



#4
May1111, 11:35 AM

P: 1

Electric Field of a Quadrupole
I think you can get out of it with a mix of the 2 posts above...
Of course (at least i hope so, since it's timestamp is 2007) this is no longer useful to OP, but I'm giving some elements for the community (this very thread pops out in 1st page of a famous web search engine for 'electric quadrupole field') The x^6 term IS zero. This is clear from using a=0 in the equation. What you have to do in order to solve the problem is to factorize by Q = 2qa² (*not* 2aq²) and let a/x > 0 afterwards only (letting Q be fixed, as it is the electric quadrupole moment ; alternatively you might say, letting 'a' stay a fixed parameter and increasing x). You'll end up with a ~Q/x^4 term as above post suggests. 



#5
Mar712, 01:18 AM

P: 1

If you just do simple algebra, and use [itex](x+a)(xa)=x^2a^2[/itex] on the original poster's "more horrifying expression," a lot of terms cancel and you are left with:
[itex]E=2kqa^2\frac{3x^2a^2}{x^2(x^2a^2)^2}[/itex] This is still exact, no approximations. As [itex]x\rightarrow\infty[/itex]: [itex]E=2kqa^2\frac{3}{x^4}[/itex] which is the expected result. (Dimensional analysis shows Exitwound's expression for [itex]Q[/itex] should be [itex]Q=2qa^2[/itex], not [itex]Q=2q^2a[/itex]). 


Register to reply 
Related Discussions  
Electric quadrupole  Advanced Physics Homework  0  
Electric Quadrupole  Introductory Physics Homework  0  
Electric Quadrupole  Introductory Physics Homework  0 