I have been struggling on what to put for $Q(total)$ in the equations for the dipole and quadrupole moment for the potential:

$$ V_{quad}(\mathbf r) \approx \frac{2Qd^2}{4\pi\epsilon_0} \left( \frac{1}{r^3}\right) \frac{3\cos^2\theta - 1}{2} $$

What does it mean by $Q$?

  • $\begingroup$ Please specify what is d and $\theta$ is angle between which two vectors. $\endgroup$ – Kutsit Mar 25 at 7:37
  • $\begingroup$ Can You please give the reference where You have found this equation? $\endgroup$ – Artem Alexandrov Mar 25 at 7:38
  • $\begingroup$ The answers by LonelyProf and Cinaed Simson explain where your expression for the electric potential comes from. In these "physical" dipole and quadrupole cases, the total charge is zero (the monopole term takes care of the net-charge of the system). So, in these dipole and quadrupole expressions, the "Q" is the unit of charge used to build the physical multipoles from point-charges. In this linear quadrupole, three point charges +Q,-2Q, and +Q (whose net charge is zero) are used. $\endgroup$ – robphy Mar 28 at 14:13

This is the formula for the potential at point $\vec{r}$ due to a quadrupole formed from a charge $-2Q$ at the origin and two charges of $+Q$, one located at $(0,0,d)$ and the other at $(0,0,-d)$. The expression is valid for $d\ll r$, the magnitude of $\vec{r}$, and $\theta$ is the angle that the vector $\vec{r}$ makes with the $z$ axis.

The derivation of the formula is a fairly standard exercise, similar to the analogous derivation for a dipole formed from two equal and opposite charges separated by a small distance.

As @Emilio Pisanty pointed out, other distributions of charge can be devised so as to generate the same potential.

  • 1
    $\begingroup$ Indeed. Though it's worth remarking that this is not unique $-$ there are plenty of other charge distributions which have the same far-field asymptotics. This is (one of) the simplest interpretations for how that field can be produced, but you could happily change it (say, by changing $Q$ to $4Q$ and $d$ to $\tfrac12 d$) without affecting the potential. $\endgroup$ – Emilio Pisanty Mar 25 at 13:42
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    $\begingroup$ Of course, you are right, and hence we end up with a single parameter, the quadrupole moment, instead of having to discuss the charges and their separation individually. Probably I should have mentioned that. $\endgroup$ – user197851 Mar 25 at 13:45
  • $\begingroup$ Probably a slip-of-the-keyboard... but note: V is an electric potential, not electrical potential-energy. $\endgroup$ – robphy Mar 28 at 13:55
  • $\begingroup$ Thanks @robphy corrected now. $\endgroup$ – user197851 Mar 28 at 14:27

The $Q(total)$, would be the same $Q$ as in the monopole term.

For the dipole term, you took $Q$, copied it, charge conjugated the copy, and placed the charges a $d$ distance apart. The net charge of the dipole was $0$.

Let $(Q,-Q)$ denote the dipole.

For the quadrupole, you took the dipole, copied it, charged conjugated the copy - which gave you 2 dipoles, namely, $(Q,-Q)$ and $(-Q,Q)$.

Then the 2 dipoles were joined together at the $-Q$ charges on a straight line so the quadrupole looked liked this: $(Q,-2Q,Q)$. The length of quadrapole was $2d$. The net charge of the quadrupole was $0$.

In short, the dipole and quadrupole terms measure the charge distribution of the charge $Q$ in the event the charge distribution wasn't a pure monopole.


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