I'm currently taking a third-year electromagnetism course (we use Griffiths), and we have begun covering approximations of our potential function, $\text{V} =\int\frac{k \text{dQ}}{\textbf{||r||}}$, where dQ is the charge distribution that can depend on r. Mainly, we've used both the Taylor Series and the Multipole Expansion to help approximate our potential function at large distances.

However, I'm still very confused about the purpose of using these approximations. What do they tell us that our regular potential function doesn't? Is there something revealing if we use an approximation rather than the potential function? Is it "easier" to find the value of the potential using, for example, the Multipole Expansion, rather than just using our regular potential function? Why is the regular potential function not sufficient?

I haven't had a clear answer to this yet, and I've been told that it's because "we want something to tell us more than the fact that the potential is 0 when r is very large." However, I'm confused why that just isn't the answer? It makes sense that the potential is 0 at a large distance away from a charge distribution, why would it be anything different and how would the approximations give that? For example, in this passage in Griffiths:

enter image description here

How does an approximation (Taylor or Multipole) give us something "more informative?" As far as I'm aware, they will result in 0 anyways at a large distance. For example, Multipole expansion results in terms that depend on and go from $$\frac{1}{r}, \frac{1}{r^2}, \frac{1}{r^3}...$$

This, of course, will be 0 at large r, which is what our potential function also shows.

Moreover, for the Taylor Series, my professor did an example here:

enter image description here

enter image description here

For when z very large and also greater than R, I can see that the potential function goes to 0. Moreover, even after the Taylor Series expansion and the subsequent expression, it still goes to 0 at large z. However, I do admit that the Taylor Series expression does look like a point charge, which is interesting!

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    $\begingroup$ Have you tried the Wikipedia page on multipole expansion yet? It's not necessarily an answer but gives you more information about what it's all about. $\endgroup$
    – Triatticus
    Oct 31, 2023 at 0:01
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    $\begingroup$ They allow you to model potentials that you cannot even write down a function for in a simple manner. If you find a Taylor polynomial that is good enough, no need to perfectly model the whole potential! $\endgroup$ Oct 31, 2023 at 1:54
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    $\begingroup$ Closed form solutions are theoretical. Experimental solutions are often different; the multipole expansion is a bridge between them. It allows you to extract important physical information from an approximation of the field. $\endgroup$
    – auxsvr
    Oct 31, 2023 at 11:03
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    $\begingroup$ In addition to the comments above, the potential created by $N$ charges onto themselves can be computed in $\mathcal O(N)$ operations instead of $\mathcal O(N^2)$. That's thanks to the multipole expansion and a smart organization of the computation. This was a fundamental algorithmic discovery. See the FMM en.m.wikipedia.org/wiki/Fast_multipole_method $\endgroup$ Nov 1, 2023 at 1:59
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    $\begingroup$ To add to the understanding of informative in the book: sometimes it's not enough to know something is zero, but we'd like to know how fast or in which way it approaches zero $\endgroup$
    – Three Diag
    Nov 2, 2023 at 9:29

5 Answers 5


What does the electric potential of a water molecule look like?

enter image description here

Imagine a cartoon picture of a water molecule, in which the oxygen atom has charge $-2q$ and sits at the origin and the hydrogen atoms each have charge $q$ and sit at $x=\pm d\cos(\theta), y=0, z=-d\sin(\theta)$ with $d\approx 94$ pm and $\theta \approx 52.2^\circ$. What does the electric potential look like at distances very large compared to $d$?

Your question suggests that your response would be "the potential is basically zero." That is extremely unhelpful, however, if I asked my question with the intent of understanding how water molecules interact with one another. Those interactions may or may not be particularly strong, but if I want to understand the nature of them then I need something to work with.

"Aha!" You say, "I can simply write down the answer:"

$$ V(x,y,z) = -\frac{2q}{4\pi \epsilon_0 \sqrt{x^2+y^2+z^2}}+ \frac{q}{4\pi\epsilon_0\sqrt{(x-d\cos(\theta))^2+y^2+(z+d\sin(\theta))^2}}+ \frac{q}{4\pi\epsilon_0\sqrt{(x+d\cos(\theta))^2+y^2+(z+d\sin(\theta))^2}}\tag{$\star$}$$

Okay, fair enough. So now, can you tell me how one water molecule interacts with another? Can you compute, for example, the torque that this water molecule applies to another which is located at some other position in a different orientation? In principle, sure. But it's going to be messy, and tedious, and not particularly illuminating.

Is there a better solution? From your time sitting in calculus classes, you may realize that you could approximate these results assuming that $d$ is small. This might provide a nearly equivalent answer which is far simpler to write down, and gives you vastly more insight into the behavior of water molecules.

In particular, notice that $$(x\pm d\cos(\theta)^2+y^2+(z+d\sin(\theta))^2 = x^2+y^2+z^2 \pm 2xd\cos(\theta)+2zd\sin(\theta) + d^2$$ $$\equiv r^2\left(1 \pm \frac{2xd}{r^2}\cos(\theta) + \frac{2zd}{r^2}\sin(\theta) + \frac{d^2}{r^2}\right)$$

Taking the square root and computing the reciprocal yields

$$\frac{1}{r}\left(1 \mp \frac{xd}{r^2}\cos(\theta) - \frac{zd}{r^2}\sin(\theta)\right)$$ where we have approximated $(1+ \ldots)^{-1/2}$ and neglected $d^2/r^2$, which we take to be far smaller than the terms we have kept. Applying this approximation to each term in the nightmare above yields the following:

$$V(x,y,z) \approx -\frac{2q}{4\pi\epsilon_0 r} + \frac{q}{4\pi\epsilon_0 r}\left(1- \frac{xd}{r^2}\cos(\theta) - \frac{zd}{r^2}\sin(\theta)\right)$$ $$+\frac{q}{4\pi\epsilon_0 r}\left(1 + \frac{xd}{r^2}\cos(\theta) - \frac{zd}{r^2}\sin(\theta)\right)$$

$$= \frac{(-2q + q + q)}{4\pi \epsilon_0 r} - \frac{2qzd\sin(\theta)}{4\pi \epsilon_0 r^3}$$

Hopefully you agree that this is a vast improvement in clarity. Things are improved further by noting that the first term cancels out because the total charge of the molecule is $-2q+q+q=0$, leaving

$$V(x,y,z)\approx \frac{\mathbf p \cdot \mathbf r}{4\pi \epsilon_0 r^3}\tag{$\star\star$}$$ where $\mathbf p \equiv -(2q)d\sin(\theta) \hat z$ is the molecule's dipole moment. I'm sure you will agree that $(\star\star)$ is vastly preferable to $(\star)$ both in conceptual clarity and computational efficiency - and for $d\ll r$, the difference between them is a small fraction of a percent.

What we've actually done is compute the first two terms of the multipole expansion, which formalizes what we did by hand and makes it very efficient. The monopole contribution vanishes because the net charge is zero, but the dipole moment is nonzero and so the second term in the expansion survives (as do others, but they are much smaller at large distances). We can now use it to answer my questions from before:

Can you compute, for example, the torque that this water molecule applies to another which is located at some other position in a different orientation?

Sure. The electric field due to the dipole is $$\mathbf E = -\nabla V = -\frac{\mathbf p}{4\pi\epsilon_0r^3} + \frac{3(\mathbf p \cdot \mathbf r)\mathbf r}{4\pi \epsilon_0 r^5}$$ (have fun calculating that for $(\star)$). An elementary result tells us that the torque on a second dipole $\mathbf p'$ is $$\boldsymbol{\tau} = \mathbf p' \times \mathbf E = -\frac{\mathbf p'\times \mathbf p}{4\pi \epsilon_0 r^3} + \frac{3(\mathbf p\cdot \mathbf r)(\mathbf p' \times \mathbf r)}{4\pi \epsilon_0 r^5}$$

can you tell me how one water molecule interacts with another?

Each water molecule creates an approximate dipole field, and other molecules feel a torque which encourages them to align with that dipole field. Two isolated water molecules will therefore seek to align with one another (this is part of hydrogen bonding), and networks of water molecules will assume a configuration which most effectively aligns them with one another (which results in the crystalline structure of ice).

  • $\begingroup$ Thank you for the answer! Just a few clarifications: - We don't assume that r is large enough such that it goes to infinity (as I thought before), but such that it is much larger than a specific variable? In this case, it is d. - Moreover, doesn't the dipole term depend on 1/r^2? Here it depends on 1/r^3. What changed? $\endgroup$ Oct 31, 2023 at 19:40
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    $\begingroup$ @opaque_dragon When we say a dimension-ful quantity (such as a length or a time) is large (or small), we always mean with respect to some other quantity of the same dimension (if there is no other such quantity in the problem, then it is scale invariant and the size of our quantity does not change the problem). In this case, “far away” means that $r$ is large compared to the size of the charge distribution, which means large compared to $d$. $\endgroup$
    – J. Murray
    Oct 31, 2023 at 21:52
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    $\begingroup$ @opaque_dragon As for your second question, note that there are factors of $\mathbf r$ in the numerator. If you divide them out (so you’re left with unit vectors in the numerator) then the dipole potential goes like $1/r^2$. $\endgroup$
    – J. Murray
    Oct 31, 2023 at 21:55
  • $\begingroup$ Nice Answer, and cool example! $\endgroup$
    – MrDBrane
    Nov 22, 2023 at 7:18

Most math problems lack exact symbolic solutions. But sometimes, when you break a problem down into an infinite series of problems, the individual problems are more tractable. This is especially useful when it turns out that only a few of the solutions contribute significantly to the result.


It is like approximating a function with Fourier components. You can approximate any sufficiently well behaved periodic function with its Fourier components. The more components you use, the more accurate the approximation.

It is the same with a multipole expansion. You can approximate the potential of any reasonable charge distribution.

The function you are approximating may not be be tractable. But Fourier series and multipole components are.

In the case of the example disk, at large distances $V \propto 1/r$. This is the behavior of a point charge. The lowest order multipole component tells you what equivalent point charge is most like this distribution. It tells you that the distribution does not consist of equal amount of + and - charge. In that case, the lowest order component would be a dipole or higher and the decay would be $V \propto 1/r^2$ or higher.

Higher order components tell you more about the shape of the charge distribution.


An electrostatic potential depends on the charge distribution. The charge distribution depends on both the total amount of charge present and the way in which the charges are arranged. When the total charge is $0$, then the electrostatic potential depends only on how the charges are arranged.

The multipole expansion formalizes this observation. It tells us that an arbitrary electrostatic potential can be broken into a power series of $r^{-1}$ with coefficients called poles. The first pole corresponds to total charge. The second to dipole moment. The third to quadrupole moment, and so on. All terms other than the first pole tell us about the arrangement of the charge distribution.

We say that the lowest order term in the power series dominates the behavior of the electrostatic potential (since subsequent terms vanish much faster). Hence, if there is a non-vanishing first pole, then the electrostatic potential is largely determined by the total charge. However, when the total charge vanishes (as is the subject of this section of Griffiths), then higher order poles may dominate.

By "looking at the charge distribution from far away" (large $r$), you are washing away all of the details given by looking at higher order poles in the multipole expansion. Hence, when you do not want to "look at the charge distribution from far away", the multipole expansion offers a clean way to write the appropriate electrostatic potential.

This leads to the electrostatic potential of an electric dipole, or even an electric quadrupole (as is briefly discussed in some problems in Griffiths). The multipole expansion is also done with classical gravitational potential.


I did a study using multipole expansion of kaolin platelets. The multipole method was useful in modeling the alternating plates of silica and alumina i.e. negative and positive plates bound by electrostatic force. The multipole expansion allowed the mathematical construction of these platelets and showed they minimize the torque between two platelets when the platelets were touching edge to face, which was seen in micrographs. So multipole expansions are useful for representing the fields and dynamics of complex systems.


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