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I've been looking on the net for electric point dipoles but couldn't find something useful (at least to me). The strange thing about an electric point dipole (the dipole moment pointing to the right on the x-axis, which could just as well point in any direction, but this the convention) is that in the dipole point the charge is zero, of which you expect no electrical field lines to emerge. Yet, it radiates electric field lines emerge outward from the point and entering the point symmetrical on the other side of the vertical axis. I know an electric dipole at a great distance can be approximated by a point dipole. But the "point" is, why constructing a point dipole? Does this make calculations easier, while keeping in mind a point dipole doesn't really exist?

Now is this a mathematical construct, with no connection to reality (at least, I can see none, except for making calculations easier so this point dipole has no realization in the real world)? If we bring two equal charges infinitely close together, the charges will add up to zero when being on top of each other. And above that their distance becomes zero so how can we speak of a moment (you can, of course, let the charges grow to infinity, but these are fixed)? So how can a neutral point object have field lines coming out and getting into it?

Is this picture the depiction of the passing of the real electric dipole moment to a (supposed) non-existent electric point dipole? I can't understand how an electrically neutral point creates electric field lines, wich is the case as the blue and red charge coincide. And what's the use (apart from maybe simplifying calculations, in which case it's certainly non-existent)?

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  • $\begingroup$ "You can let the charges grow to infinity but these are fixed". Not so - in the derivation of a dipole field the product of charge and distance is kept constant as the charges approach each other; they have to go to infinity as the distance goes to zero. $\endgroup$ – Floris May 31 '17 at 0:18
  • $\begingroup$ That's just another poorly written wikipedia article (or at least the animation). Wikipedia is not the place to learn physics. $\endgroup$ – David Hammen May 31 '17 at 0:30
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    $\begingroup$ Re Now is this a mathematical construct, with no connection to reality, Yes, it's a mathematical construct. So what? It has a definite connection with reality. "Of course it is happening inside your head, Harry, but why on earth should that mean that it is not real?" $\endgroup$ – David Hammen May 31 '17 at 0:34
  • $\begingroup$ "Wikipedia is not the place to learn physics." Uh oh... Quickly regrets virtually all of his physics knowledge... $\endgroup$ – DoublyNegative Sep 6 '18 at 18:50
  • $\begingroup$ @DavidHammen-Allright, mathematically it's all sound and clear, and of course, there is a connection to reality ("even though it's all happening in the mind Harry, just like the perception of the gods, doesn't mean they are not real, other evidence than scientific evidence is used in the last case to prove that they exist"). But then the question is: what is the connection? By the way, besides articles on the net, books, physics journals, your own thinking, etc., I think you can learn a lot about physics on Wikipedia. $\endgroup$ – descheleschilder Sep 15 '18 at 14:54
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There are no observed point-like electric dipoles (though at least one type of particle is predicted to have one: the $W$-boson). The smallest permanent dipoles I, personally, know have been seen are molecular in size (e.g. water, carbon monoxide, and phosphine). You can even build up ferroelectric materials that have a macroscopic permanent dipole moment like magnets have a magnetic one.

That said, the dipole fields are a useful component in calculations that even pop up in practical applications. When I say, "dipole fields" that isn't a typo, there are two types of dipole fields. The first is the type discussed in the question, the "outside" dipole field that has a potential that falls like $r^{-2}$ and electric field that falls like $r^{-3}$. That field is useful when calculating the fields outside of an imaginary sphere centered on the origin that contains all of the charges.

The "inside" dipole field is something you've already encountered in your studies. A constant electric field is, in fact, the field produced inside of a dipole. With the point charges in the original question, you just take the limit the other direction in distance, but also with increasing magnitude of charge, such that the electric field halfway between them is constant. "Inside" multipole fields are used when you're working inside of an imaginary sphere sitting at the origin that contains no charges.

The mixed case is difficult to describe with words, but straightforward algebraically. The basic of it is to separate the charges into the two above cases, then add up the results.

In terms of building a practical dipole, if you place a conducting sphere in a constant electric field the charges you induce on its surface produce a dipole field that is, in the ideal situation of uniform external field and perfectly spherical conductor, also perfect. In more detail, they produce an "outside" dipole field outside of the conductor, and the exact "inside" one needed to cancel the external field inside the conductor.

Notice how I said earlier that a constant electric field is a dipole field, and it induces a dipole in a spherical conductor? That's not an accident. The math behind dipoles, and other multipoles, has a deep connection to rotations (group theory), linear algebra, etc that are at the center of more advanced physics.

The other thing to keep in mind is that even though we haven't seen any physical point-like electric dipoles, point-like magnetic dipoles are everywhere: electrons, muons, all of the quarks, etc.

The structure of the dipole field is universal, so studying the ideal electric dipoles helps you understand the very real magnetic ones. Compare the electric field of an electric dipole $$\mathbf{E}(\mathbf{r}) = \left(\frac{1}{4\pi\epsilon_0}\right)\frac{3(\mathbf{p}\cdot\hat{r})\hat{r} - \mathbf{p}}{r^3}$$ with the magnetic field produced by a magnetic dipole $$\mathbf{B}(\mathbf{r}) = \left(\frac{\mu_0}{4\pi}\right)\frac{3(\mathbf{m}\cdot\hat{r})\hat{r} - \mathbf{m}}{r^3}.$$

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    $\begingroup$ Well, both nucleons and electrons are predicted in the Standard Model to have nonzero electric dipole moments (example, and links therein), though admittedly they're some five to twelve orders of magnitude smaller than our current detection capabilities, so there's that. $\endgroup$ – Emilio Pisanty Dec 30 '17 at 1:03
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Are point dipoles a thing, i.e. a physical object that you can find in the world? Well, not really, but does it matter? In particular, let me flip the question around for a bit: are point charges physically realizable? Well, so far as we can tell from the experimental data, yes, electrons and quarks are pointlike particles, but then again that's not something you really need to know in order to use point charges in electrostatics. In fact, even if they weren't pointlike (like, say, atomic nuclei), we still would (do) approximate them as point charges without losing any sleep over it.

Now, let's unpack one of your more interesting statements:

So how can a neutral point object have field lines coming out and getting into it?

Here you're taking the concept of electric field lines rather too literally: they are a pictographic way of understanding the field, but nothing more. We say they "start" and "end" at charges because it models streamlines of a fluid with nonzero flow out of or into some sample volume, but that's it. And, as far as the streamlines go, here's the important bit: no matter how small a volume you choose around the origin, the electric field of a point dipole will always have zero flux (or, more heuristically, you'll have the same number of lines coming in and coming out). Putting the boundary at the origin doesn't work: you need the field to be nonsingular for those manipulations to be valid. (Also, putting a gaussian surface right through a point charge is an excellent way to break Gauss's law.)

Now, is the point-dipole electric field a valid solution of the electrostatic equations? Why, yes it is, and it even has a charge density to go with it: $$\rho(\mathbf r) = \mathbf p \cdot\nabla\delta(\mathbf r) = p_z \delta(x)\delta(y)\delta'\!(z).$$ Here the $\delta'\!(z)$ is the derivative of the delta function, which can indeed be defined (in something called the distributional sense). It's a bit singular, but frankly, so is the delta-function charge density of a point particle; you're already breaking normal mathematics to do those, and adding point dipoles doesn't change anything.

I suspect that that sounds like cheating you (though I assure you that it isn't), but here is the thing: we don't study point dipoles because they may or may not exist in the real world. Instead, we study them because the fields they produce are an excellent approximation to real-world fields: sometimes as part of a systematic expansion, but quite often they are such good approximations (i.e. miles better than all the other approximations you're doing in that calculation) that we take them as exact. For more details, see this previous answer of mine, but the short of it is that the dipole field is the leading or subleading term in a systematic expansion that is really helpful in conceptualizing, understanding, and calculating, the electric fields of arbitrary charge distributions.

Similarly, it's important to emphasize that the point-dipole fields are perfectly legitimate electrostatic fields, and that there are perfectly reasonable finite-size charge distributions (which I describe in this Q&A) that exactly reproduce the fields of a point dipole. So, even if you don't buy the limiting procedure, the fields themselves are unquestionably physical.

And, finally, what's with that limiting procedure? To be frank, I think it's described sub-optimally in many textbooks, but that's neither here nor there. How do working physicists think of and use point dipoles in their everyday physics? Well, most matter is electrically neutral, so when seen from far away (or when interacting with homogeneous fields) its Coulomb term will be zero, but that doesn't mean that it doesn't produce interesting residual fields from imperfect cancellations of its components: the dipole field captures these residuals, and it puts them in a nice, clean approximation which, as it turns out, is virtually exact. And, in physics, virtually exact is about as close as you get to "physical existence".

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The definition of the dipole moment of charge distribution $\rho({\bf x})$ is given by ${\bf p} = \int_V {\bf x}' \rho({\bf x}') d^3 x'$. Start from the definition and all should be clear. An arbitrary distribution of charge will have other electric moments as well. Which one dominates the electric field calculation at large distance depends upon the distribution of charge. In many cases its the electric dipole moment that dominates.

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  • $\begingroup$ Here's the thing, though: what's $\rho(\mathbf x')$ for a point dipole? $\endgroup$ – Emilio Pisanty May 31 '17 at 1:50
  • $\begingroup$ It's two point charges. The dipole has a "point" like position not extent. $\endgroup$ – D_J_S May 31 '17 at 17:36
  • $\begingroup$ What you're describing is known as a finite dipole, and it is distinct from a point dipole, the source that produces the field $\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{r^{3}}$, which can be seen as the limit of a finite dipole $p$ of fixed magnitude as the charge separation tends to zero - and which is the true topic of the question, as expressed in the body of the text. $\endgroup$ – Emilio Pisanty May 31 '17 at 17:45

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