Electric Field Approximation for Linear Quadrapole Problem

[Old Guy]

Homework Statement

Given the following charges, all on the x-axis:
-q at x=-d
+2q at x=0
-q at x=d

Show that the electric field at a point (x,y) = (0,r) (ie, on the y-axis a distance r from the origin) is approximately:
$$\frac{3qd^2}{4\pi\epsilon_{0}r^4}$$

Homework Equations

The Attempt at a Solution

We were asked for an approximation, but it seems to me that an exact solution is relatively straightforward. The direction of the field from the positive charge would be along the y-axis. The field resulting from the negative charges would be from y=r toward those charges, but the x-components of those fields would cancel. Consequently, shouldn't the field simply be the field produced at r by the positive charge, minus the y-components of the fields produced by the negative charges? Am I missing something?

 

[Doc Al]

You're right, the exact solution is straightforward--but it's messy. Start by writing down the exact answer and showing that it reduces to that simple approximation when R >> d.

 

[Old Guy]

Right, and for that I got $$\frac{2q}{4\pi\epsilon_{0}[\frac{1}{r^2}-\frac{r}{(r^2+d^2)^(\frac{3}{2})]$$

I understand the concept of taking r>>d, but don't see any way with the above equation that gets either d^2 or a 3 in the numerator, or r^4 in the denominator.

And thanks for your help!

 

[Old Guy]

$$\frac{2q}{4\pi\epsilon_{0}[\frac{1}{r^2}-\frac{r}{[(r^2+d^2)^(\frac{3}{2})]]$$

 

[Meir Achuz]

Expand the denominators of the exact expression in the geometric series
1/(1-x)=1+x+x^2+...

 

[Old Guy]

Well, I'm having trouble getting the equation to show. I'll try breaking it up:

For the field from the positive charges, I have $$\frac{2q}{4\pi\epsilon_0(r^2)}$$
For the negative charges, $$\frac{-2qr}{4\pi\epsilon_0[(r^2+d^2)]^(\frac{3}{2})}$$
Hopefully these will show

 

[Old Guy]

Meir's Suggestion

Thank you - I had thought of this, but don't see how to get the expression in the proper form. My terms are {1/r^2 - r/[(r^2+d^2)^(3/2)]}. I can get this into a form of
1-r^3/[(r^2+d^2)^(3/2)], but this is a form of 1-x, not 1/(1-x). Am I missing something obvious?

 

[Doc Al]

For the negative charges, $$\frac{-2qr}{4\pi\epsilon_0[(r^2+d^2)]^(\frac{3}{2})}$$

Rewrite $$(r^2+d^2)^{3/2}$$ as $$r^3(1+(d/r)^2)^{3/2}$$

Now you can take advantage of the approximation:
$$\frac{1}{(1 + x)^a} \approx 1 - ax$$
(when x << 1)

 

[Meir Achuz]

It's easier to get the potential and then E=-dV/dx.