Dipole term in a quadrupole expansion

AI Thread Summary
The discussion focuses on calculating the potential from a system of three charges arranged along the z-axis and expanding it in the case where a is much smaller than r. The potential is derived using spherical polar coordinates, resulting in a formula that includes terms for the monopole and quadrupole, but notably lacks a dipole term. This absence of the dipole term is explained by the configuration of the charges, which effectively cancels out the dipole contributions due to symmetry. Participants confirm that this cancellation is independent of the actual magnitudes of the charges, as long as the two charges are equal and opposite. The analysis concludes that the calculations and reasoning regarding the absence of the dipole term are correct.
yoghurt54
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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 canceled 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.
 
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yoghurt54 said:
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.
 
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}

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

What do you get for your distribution?
 
Last edited:
Doc Al said:
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?
 
yoghurt54 said:
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!
 
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