Field generated by electric quadrupole

[kjintonic]

Homework Statement

"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.

Homework Equations

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

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:?

 

[tiny-tim]

Homework Statement

"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.

Homework Equations

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

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

Hi kjintonic!

At the top, the x^4s cancel, and you get a^2x^2s etc,

and the bottom is x^6 + …

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:
$$\frac{0}{x^6}$$

No, you don't let a –> 0, you just let a stay fixed, and let x —> ∞.

So a^2x^2 + … / x^6 + … = a^2/x^4 + …

(oh … and you have to type [noparse]$$before and$$ after any LaTeX! [/noparse] )

 

[sru]

potential due to the quadrupole is;
V=qd²[3cos²θ-1]/4╥εx³

Now,
Radial component of electric field, E_x=-∂V/∂x=-∂/∂x[qd²(3cos²θ-1)/4╥εx³]

and

Transverse component of field, E_θ= - (1/x)[∂V/∂θ]
=-(1/r){∂/∂θ[qd²(3cos²θ-1)/4╥εx³]}

therefore,resultant field is, E=√(E_x²+E_θ²) which will be proportional to 1/x^4.

 

[kjintonic]

Thanks a lot for the help. I simplified that combined faction and got \:
(6a^2x^2-a^2)/(x^6-2a^2x^4+a^4x^2).

I can't see any more way to simplify it :(

 

[tiny-tim]

Thanks a lot for the help. I simplified that combined faction and got \:
(6a^2x^2-a^2)/(x^6-2a^2x^4+a^4x^2).

I can't see any more way to simplify it :(

(what about[noparse] $$and$$?[/noparse])

Why do you want to?

That obviously falls off as 1/x⁴.

(If you're not convinced, just divide top and bottom by x⁶)

 

[kjintonic]

o cool :D Thanks a lot. One last question. This does prove that the quad falls off as 1/x^4 but how do i get field to the right of x=a?