What would the electric field and potential of three point charges (for example -q, -q and +2q, equally spaced) look like in the far-field regime as opposed to the near-field regime? We are only being taught about the near-field regime and I'm curious to know what happens when you 'zoom out'!


This is known as a multipole expansion. Basically, when you are interested in the electric potential created by an compact assembly of charges, and you're looking at distances much larger than the spatial extension of the collection of charges, you can express the electric potential as a power series of $m_n/r^n$ (times some function of $\theta$ and $\phi$), with each $n$ corresponding to a successive "pole". In many cases, it is enough to only keep the term corresponding to the lowest non-zero $m_n$, as this will give you the asymptotic behaviour for $n\to \infty$.

$n=1$ is the "monopole" term. It is proportional to the total electric charge. Intuitively, if you have an excess of positive (resp. negative) charge $\delta q$ over the whole distribution, you can expect that the distribution of charges will mostly be seen at long distance as a point charge $\delta q$, because the field created by the other charges will approximately cancel out. Thus $V(r\to \infty) \simeq V_{\mathrm{monopole}}(r) = \frac{\delta q}{4 \pi \varepsilon_0 r}$.

The $n=2$ term, which decays faster than the monopole ($1/r^2$ instead of $1/r$) is called the "dipole" term. This term and all higher order terms come from the fact that even if you have an equal number of positive and negative charges, but they are situated at different position, their field will only approximately cancel out at long distance. The dipole term appears when the barycenter of all negative charges is separated from the barycenter of positive charges. Assuming a total negative charge $-Q$ equal to the number of positive charges $Q$, and noting $\overrightarrow{d}$ the distance between the negative barycenter and the positive barycenter, the long-distance field created by this "dipole" is $V(r \to \infty) \simeq V_{\mathrm{dipole}}(r) = \frac{Q \overrightarrow{d}\cdot \overrightarrow{u_r}}{4 \pi \varepsilon_0 r^2}$, where $\overrightarrow{u_r}$ is a unit vector from the origin to position $\overrightarrow{r}$. Note that $V_{\mathrm{quadrupole}} = 0$ in the plane for which $\overrightarrow{d}\cdot \overrightarrow{u_r} = 0$, because in that case you are at the same distance from (the barycenter of) negative charges than (the barycenter of) positive charges, so their field roughly cancel out.

If both monopole and dipole terms are $0$, the next leading term would be quadrupole. This becomes harder to describe non-mathematically, but this will happen in most cases when the net charge is zero and the barycenter of positive charges is at the same place as the barycenter of negative charges (except if you have some higher order symmetry). For instance, one of the simplest case of the quadrupole term leading would be $2$ positive charges and $2$ negative charges all arranged on the corners of a square, alternating positive and negative charge. In this case the long distance field would decay as $1/r^3$.

In your example, (one charge $-q$ at position $-a$, another charge $-q$ at position $0$ and $2q$ at position $a$), the leading term would be the dipole term, because the net charge is zero but the barycenters of positive and negative charges are at different positions.

You can then easily deduce the electric field from the electric potential by taking the gradient.

For reference, this is the electric field created by two-point charges, which would correspond to the dipole term at long distances (at distances shorter than the separation between charges, the exact field depends on the exact distribution of charges).

Electric field created by a dipole

  • $\begingroup$ Good answer. As an interesting extension, note that if the charges were ordered $-q$, $+2q$, and $-q$ along the axis, then the dipole moment would vanish and the quadrupole term would dominate. $\endgroup$ – Michael Seifert Nov 8 '20 at 18:39
  • $\begingroup$ Thank you very much for the detailed explanation! If I wanted to draw a plot of the electric field lines and/or contours for surfaces of constant electrostatic potential, how would this look? $\endgroup$ – aoifeo Nov 9 '20 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.