# Homework Help: Linear Quadrapole Problem

1. Oct 20, 2007

### Old Guy

1. The problem statement, all variables and given/known data
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:
$$\frac{3qd^2}{4\pi\epsilon_{0}r^4}$$

2. Relevant equations

3. 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?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 21, 2007

### Staff: Mentor

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.

3. Oct 21, 2007

### 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.

4. Oct 21, 2007

### Old Guy

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

5. Oct 21, 2007

### Meir Achuz

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

6. Oct 21, 2007

### 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

Last edited by a moderator: Oct 22, 2007
7. Oct 21, 2007

### 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?

8. Oct 22, 2007

### Staff: Mentor

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)

Last edited: Oct 22, 2007
9. Oct 22, 2007

### clem

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