# Dipole term in a quadrupole expansion

## Homework Statement

There are three charges arranged on the z-axis. Charge $$+Q_2$$ at the origin, $$-Q_1$$ at $$(0,0,a)$$
and $$-Q_1$$ at $$(0,0,-a)$$.

Using spherical polar coordinates (i.e the angle $$\vartheta$$ is between $$r$$ and the positive z-axis), find the potential at a point with a distance $$r$$ from the origin, and in the case $$a<<r$$, expand the potential up to terms including $$(a/r)^2$$. Identify terms due to a charge, a dipole and a quadrupole.

## Homework Equations

Well, I found that before the expansion, we find that the potential V is:

$$V = 1 / 4 \pi \epsilon ( Q_2 / r -Q_1 (1 / \sqrt{r^2 + a^2 - 2*a*r*cos\vartheta} + 1 / \sqrt{r^2 + a^2 + 2*a*r*cos\vartheta}) )$$

The denominators of the $$Q_1$$ charges are derived from the cosine rule, and the fact that for the bottom charge, the angle made with the z-axis is $$\pi - \vartheta$$ which makes the cosine of that angle the negative of the cosine of theta.

## The Attempt at a Solution

Right, after taking out a factor of $$r$$ and expanding the square root denominators to the $$((a/r)^2 - 2(a/r)cos \vartheta)^2$$ term and ignoring terms greater that the degree 2 we get this:

$$V = 1 / (4 \pi \epsilon r) (Q_2 - Q_1(2 + (a/r)^2(3cos^2 \vartheta- 1))$$

I have a term for the charge and a term for the quadrupole, but no term for the dipole, as those terms cancelled when summing up terms in the expansion.

Have I done this right? Should there be no dipole term? I've been stuck on this for a couple of months.

Doc Al
Mentor
Have I done this right? Should there be no dipole term? I've been stuck on this for a couple of months.
I haven't checked your math, but it makes sense that there would be no dipole term. Think of the charge configuration as two opposing dipoles--they cancel.

kuruman
Homework Helper
Gold Member
The standard definitions for the monopole, dipole, etc. terms when you have discrete distributions of charges are

Monopole
$$Q=\sum q_{i}$$

Dipole
$$\vec{p}=\sum q_i \vec{r_i}$$

$$Q_{ij}=\sum q_iq_j (3x_ix_j-r^{2}_{i}\delta_{ij})$$

What do you get for your distribution?

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I haven't checked your math, but it makes sense that there would be no dipole term. Think of the charge configuration as two opposing dipoles--they cancel.

The two opposing dipoles - are they the pairings (above) $$-Q_1,Q_2$$ and $$Q_2, -Q_1$$ (below) ?

Is this independent of what the actual magnitudes of the charges are, so long as the two like ones either side of the central one have the same magnitude?

Doc Al
Mentor
The two opposing dipoles - are they the pairings (above) $$-Q_1,Q_2$$ and $$Q_2, -Q_1$$ (below) ?

Is this independent of what the actual magnitudes of the charges are, so long as the two like ones either side of the central one have the same magnitude?
Yes and yes.

Alrightey then, thank you very much!