Quadrupole Moment and Potential

[Destroxia]

Homework Statement

Consider a physical dipole consisting of charge +q and −q separated by a distance a.
• At first consider the origin of coordinates to be located in the middle of the dipole and a positive charge to be located at a distance y = a/2 and negative charge at position y = −a/2. Find the first two nontrivial terms to the potential of this system at large distances.
• Second, place the charge −q at y = 0 and then place the other charge at y = a. Again, find the first two nontrivial terms to the potential of this system at large distances.
• Discuss the difference. In both cases use spherical coordinates.

Homework Equations

$V(r) = \frac 1 {4\pi\epsilon_0} \sum_{N=0}^{\infty} \frac 1 {r^{(n+1)}} \int (r')^n P_n(cos(\theta))\rho(r')d\tau'$

$V(r)_{mono} = \frac 1 {4\pi\epsilon_0} \frac Q r$

$V(r)_{dip} = \frac 1 {4\pi\epsilon_0} \frac {p \cdot \hat{r}} {r^2}$

$Q = total charge$

$p_dip = qd$

The Attempt at a Solution

This is just my attempt for the first bullet point:

The V(r)_{mono} is 0, because the total charge is $Q = +q - q = 0$.

The $p_dip = \frac {qa} 2 \hat y - \frac {qa} 2 (-\hat y) = qa \hat y$

The $V(r)_{dip} = \frac 1 {4\pi\epsilon_0} \frac {qasin(\theta)sin(\phi)} {r^2}$ because $\hat{y} \cdot \hat{r} = sin(\theta)sin(\phi)$ within sphereical coordinates.

So, this gives me the first non-trivial term. My question from this is, how do I get the term for the quadrupole potential, or even the quadrupole moment, for that matter? I've tried getting it using the multipole expansion formula I gave above, but it doesn't simplify down as nice, and I'm not sure what to do with some of the components.

$V(r) = \frac 1 {4\pi\epsilon_0} \sum_{N=0}^{\infty} \frac 1 {r^{(n+1)}} \int (r')^n P_n(cos(\theta))\rho(r')d\tau'$

$V(r) = \frac 1 {4\pi\epsilon_0} \frac 1 {r^3} \int (r')^2(\frac {3cos^2(\theta)-1} {2}) \rho(r')d\tau'$

I'm not at all sure what to do with this integral, or even how to do it, or it's bounds. I've found some other derivations online, but I don't know anything about tensors, or kronecker delta functions. Is there anyway to just get into a simple enough form I can deal with like $V(r)_{dip}$ ?

 

[BvU]

Hi,

You sure you are supposed to do it this way and not by developing $$
V(\vec r) = \frac Q {4\pi\epsilon_0} \left [ {1\over |\vec r - {\vec a\over 2}|} - {1\over |\vec r + {\vec a\over 2}|} \right ] \quad ?$$

PS avoid things like $qasin$ in $TeX$ by using \sin to get $qa\sin$

 

[Destroxia]

Hi,

You sure you are supposed to do it this way and not by developing $$
V(\vec r) = \frac Q {4\pi\epsilon_0} \left [ {1\over |\vec r - {\vec a\over 2}|} - {1\over |\vec r + {\vec a\over 2}|} \right ] \quad ?$$

PS avoid things like $qasin$ in $TeX$ by using \sin to get $qa\sin$

I'm not to sure what the actual way I'm supposed to do it, I just figured it was multiple expansion because it said it wants the first 2 "terms", and it was in the multipole expansion chapter of griffith's introduction to electrodynamics.

 

[BvU]

All I have available is a 1999 pdf where he actually works out the first case (example 3.10: A physical electric dipole consists of ...) in polar coordinates up to order $r^{-2}$, so only the first nontrivial term. And problem 3.29 asks to determine the quadrupole and octopole terms (so you're lucky your exercise only wants only the first two nonzero terms... )

Your multipole expansion yields the same expressions anyway, I expect: your charge distribution is indeed two delta functions and these are (for a physicist) defined with $\int f(x) \delta(x-x') dx = f(x')$