Is there a finite-sized charge distribution whose electric field is *exactly* that of a point dipole?

This kinda gets lost in the telling when electric dipoles are introduced in textbooks, and it ends up causing a good deal of confusion (as in e.g. this recent example). Point electric dipoles, and the electric fields they produce, are most often introduced in textbooks as the limit of two charges $\pm q$ a distance $d$ apart, in the limit where $d\to0$ while sending $q\to\infty$ in such a way as to keep $p=qd$ constant. This presentation, however, can make several aspects very confusing: is this a physical field, or just a mathematical construct, or just an approximation? What's with that limit, and how does one implement that $q\to\infty$ bit? More importantly, why even do any of this?

As a result, point dipoles can often be quite elusive targets to understand, and this is often compounded by the fact that they are generally only used as limits or as approximations, and never with any real credibility just by themselves as such. With this in mind, then:

• Are there reasonable, physical charge distributions with a finite size and no singularities (beyond possibly point charges) which produce an electric field that is exactly equal to the field of a point dipole, without any monkey business with limits or approximations or anything?

Yes, this is perfectly possible

The way you do this is by choosing a charge distribution which is 'completely dipolar' in some suitable sense, and this will produce an electric field which is also a pure dipole. In more technical terms, all you need to do is use a charge distribution with a separable angular dependence (i.e. $\rho(\mathbf r) = f(r) g(\theta,\phi)$) and then set that angular dependence to a suitable spherical harmonic.

The simplest example of this is a surface charge distribution spread over a sphere, of radius $a$, say, which has a dipolar dependence, i.e. the simple-as-day surface charge density $$\sigma(\theta,\phi) = \sigma_0 \cos(\theta),$$ which looks like the image below, where red is positive and blue is negative, and the darker mesh lines are contours separated by constant (I'm also partial to dipolar gaussian charges, with the volume charge density $\rho(\mathbf r) = A\, z\, e^{-r^2/a^2}$, but with that one the charge is never quite fully zero when away from the system.)

Once you do decide to look at this charge distribution, there are two main nice things that happen:

• The electric field outside the sphere is exactly that of a pure point dipole, and
• the electric field inside the sphere is exactly uniform.

This can be shown via a few simple steps:

1. The point-dipole electric field $$\mathbf E_\mathrm{out}(\mathbf r) = {\frac {1}{4\pi \epsilon _{0}}}\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{r^{3}}$$ is a valid electric field (conservative and a solution of the Laplace equation) when away from the boundary.
2. The uniform electric field $\mathbf E_\mathrm{in} (\mathbf r) = \mathbf E_0$ is obviously a valid electric field inside the sphere.

We take these two for granted (though if required they can be checked by direct differentiation), and this means that to show that the fields as claimed above correspond to our surface density we only need to show that they obey the right matching at the boundary. Thus, we have

1. The two fields match in a conservative fashion at the boundary, i.e. their components normal to the surface are equal (so you cannot harvest free energy by doing small loops around the surface). To show this, we have $$\hat{\boldsymbol \theta}\cdot \mathbf E_\mathrm{in}(\mathbf r) = \hat{\boldsymbol \theta}\cdot \mathbf E_0 = \hat{\boldsymbol \theta}\cdot \hat{\mathbf z}E_0 = E_0 a \sin(\theta),$$ where the azimuthal component along $\hat{\boldsymbol \phi}$ is in both cases obviously zero, and \begin{align} \hat{\boldsymbol \theta}\cdot \mathbf E_\mathrm{out}(\mathbf r) &= \hat{\boldsymbol \theta}\cdot {\frac {1}{4\pi \epsilon _{0}}}\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{r^{3}} \\&= -\frac {1}{4\pi \epsilon_0}\frac{\hat{\boldsymbol \theta}\cdot\hat{\mathbf z}p }{a^{3}} \\&= -\frac {1}{4\pi \epsilon_0}\frac{p \sin(\theta) }{a^{3}} \end{align} Since the angular dependence matches, all we need to do is set $E_0 = -p/ 4\pi\epsilon_0a^4$.

Moreover, now is a good time to calculate the dipole moment in terms of the surface charge density: the $x$ and $y$ components vanish by symmetry, and the axial component gives \begin{align} p_z &= \int z \sigma(\theta,\phi) \mathrm dA = \int_0^\pi \int_0^{2\pi} a\cos(\theta) \sigma_0\cos(\theta) a^2 \sin(\theta) \mathrm d\phi \mathrm d\theta \\&= 2\pi a^3\sigma_0\int_0^{\pi} \cos^2(\theta) \sin(\theta)\mathrm d\theta = 2\pi a^3\sigma_0\int_{-1}^{1} u^2 \mathrm du \\&= \frac{4\pi}{3} a^3\sigma_0 . \end{align}

2. The normal components of the fields differs by the surface charge density (or, in plainer terms, if you zoom in enough, the surface charge density works like an infinite plane of charge, pushing the electric field by $\sigma/2\epsilon_0$ away from it in either direction). To see this, we calculate $$\hat{\mathbf r}\cdot \mathbf E_\mathrm{in}(\mathbf r) = \hat{\mathbf r}\cdot \mathbf E_0 = \hat{\mathbf r}\cdot \hat{\mathbf z}E_0 = E_0 a \cos(\theta)$$ and \begin{align} \hat{\mathbf r}\cdot \mathbf E_\mathrm{out}(\mathbf r) &= \hat{\mathbf r}\cdot {\frac {1}{4\pi \epsilon _{0}}}\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{r^{3}} \\&= {\frac {1}{4\pi \epsilon _{0}}}\frac {2\mathbf {p} \cdot {\hat {\mathbf {r} }}}{a^{3}} \\&= {\frac {1}{2\pi \epsilon _{0}}}\frac {p\cos(\theta)}{a^{3}}, \end{align} and we confirm that $\hat{\mathbf r}\cdot \mathbf E_\mathrm{out}(\mathbf r)- \hat{\mathbf r}\cdot \mathbf E_\mathrm{in}(\mathbf r) =\frac{1}{\epsilon_0}\sigma(\theta,\phi)$, where we have \begin{align} \hat{\mathbf r}\cdot \mathbf E_\mathrm{out}(\mathbf r)-\hat{\mathbf r}\cdot \mathbf E_\mathrm{in}(\mathbf r) & = {\frac {1}{2\pi \epsilon _{0}}}\frac {p\cos(\theta)}{a^{3}}-E_0 a \cos(\theta) \\ & = \left(\frac {p}{2\pi \epsilon_0a^4}- E_0\right)a \cos(\theta) \\ & = \left(\frac {p}{2\pi \epsilon_0a^4}+\frac {p}{4\pi \epsilon_0a^4}\right)a \cos(\theta) \\ & = \frac {3p}{4\pi \epsilon_0a^3}\cos(\theta) \\ & = \frac{1}{\epsilon_0}\sigma(\theta,\phi). \end{align}

Thus, since the electric fields as given are full solutions everywhere for this charge distribution, they are the fields it generates.

The above text has a bit more detail than strictly necessary, in the interest of providing an essentially self-contained resource, but what did we learn? Well, it shows that whenever one is bothered by the presence of a point dipole field $$\mathbf E_\mathrm{out}(\mathbf r) = {\frac {1}{4\pi \epsilon _{0}}}\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{r^{3}},$$ one can always conceptualize it as arising from this charge density, and all the conceptual issues fall away, without the need of tricky limits or unclear approximations.

Moreover, the method extends directly to any arbitrary multipolarity: do you want to conceptualize quadrupoles or octupoles? No problem, just plop in an appropriate solid harmonic as your $\sigma(\theta,\phi)$, and you will automatically have a charge density that produces the relevant multipolar field. Moreover, this method extends all the way to hexadecapoles as beyond (as showcased in Hexadecapole potential using point particles?) where methods like point-charge-hypercubes give out.

The Mathematica code used to produce the image in this answer is available through Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/IsKL8.png"].

• One could argue that using a surface charge qualifies as "monkey business with limits or approximations", since $\rho(\vec{r})$ is basically infinite on the surface. It's still well-defined in a distributional sense, which is all we technically need to get solutions for the fields & potentials, but it requires a bit more effort to make rigorous. It would be nice to see a true volume charge density $\rho(\vec{r})$ that creates an external dipole field; I suspect that a series of concentric shells with this charge distribution would do the trick. – Michael Seifert Jul 6 '18 at 15:15
• @MichaelSeifert indeed it does - any $\rho(\vec r)$ whose angular dependence at fixed $r$ is strictly dipolar will produce a strictly dipolar potential, independently of the radial profile (and heck, you don't even need the dipoles of each layer to be aligned). But also, if you're going to quibble with the possibility of surface layers of charge but you're OK with a volumetric $\rho(\vec r)$, then I would ask why you're so confident that you can execute that $\rho(\vec r)$ with exact faithfulness, and (given that electrons are point charges) what it even means . – Emilio Pisanty Jul 6 '18 at 15:27