Purcell in section 2.17 discusses the electric field $E = <Ky, Kx, 0>$, which has field lines in the shape of a hyperbola, $\phi = -Kxy$, zero curl, and zero divergence. Purcell states that since curl is zero, this could be an electrostatic field.

Elsewhere, he writes that any electrostatic field can be created by a distribution of charge.

Now, since divergence of $E$ is zero, there can be no static charge anywhere. So how could any set of charges create this field?

I thought perhaps the charges must be at infinity. They would of course then need to be of infinite charge. I tried to model this as the limit as $\alpha \to \infty$ of charges at distance $\alpha$ of magnitude $\alpha^2$, but was unable to complete this.

Feynman seems to discuss similar configurations.

How can the field $\phi = -Kxy$ be electrostatic and yet have divergence zero everywhere? Where are the charges?


I understand that any harmonic function is zero div, zero curl, and yet not necessarily constant.

I'm asking something quite different: Where can charges be placed to create the electrostatic field described?

That is, my question is not "Mathematically, how can this field be?" but rather "Physics says that a charge will have a region of non-zero divergence around it. This field seems to have no such regions. Where are the charges that create it"? My question is based on Coulomb's Law and Gauss's Law - it is a physics question (not "Do non constant zero div zero curl fields exist?", which is a mathematical question, the answer being "Yes, harmonic functions").


The linked question does not address my question. The linked question shows how a field can have zero divergence everywhere and not be constant.

However, it does not show where the charge could be to produce such a field, if it is a purely electrostatic field. By Gauss's law, a region containing a charge must have divergence non-zero somewhere.

Note that the accepted answer by LPZ states "Yes, by Gauss’ law, it cannot be within the bulk of the domain, so you’re only left with the boundary or more generally outside the region where the field is given by this formula as in the second example" - in other words, agreeing that the field described by Purcell is not possible unless we say his description only covers part of the field.

To the closers: Do you maintain that Purcell's description can apply to the entire field? If so, please explain how, and why LPZ is wrong. If not: Then this question is unrelated to the one you linked as a duplicate, and this question needs to be reopened.

If you're unable to state whether you agree with LPZ that no electrostatic field can be described everywhere by Purcell's equations, or you feel that such a field can exist, than this question needs to be reopened.

  • $\begingroup$ There is no sharp border between math and physics. One needs the former to describe the latter and the latter to understand the meaning of Math. In your Update there is a wrong statement: "a charge will have a region of non-zero divergence around it." That is not what Gauss law says. $\endgroup$ Nov 28, 2023 at 4:00
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 If it makes you happy, "If region $R$ has a charge within it, then region $R$ will not have zero divergence everywhere." Regardless, without pontificating, are you able to answer the question? The existence of harmonic functions does not answer the question, nor does it imply that an electrostatic field which has zero divergence everywhere can contain charge. $\endgroup$ Nov 28, 2023 at 12:51

1 Answer 1


You can look at the required charge when your field is given by your formula in a finite domain $D$ and zero outside. Physically, this gives you a surface charge on the boundary $\partial D$ given by: $$ \sigma=\epsilon_0 E\cdot n $$ For example, if the domain is cylindrical of radius $R$ along the $z$ axis, in cylindrical coordinates: $$ \sigma=\epsilon_0 KR\sin(2\phi) $$ Or the charge element is: $$ dQ=\epsilon_0 KR^2\sin(2\phi)d\phi dz $$ As expected, it diverges as $R\to \infty$. You can do this for any domain $D$, though you’ll need to careful if there are som “angles”.

In practice, you could relax your requirement by asking what is the distribution of charges that reproduces your field as a linear approximation about the origin. Loosening the strict requirement, you can obtain many more solutions. By analogy with the Paul trap, you could generate your field by putting four charges on the corner of square with opposite corner charges having the same sign.

Btw, in practice, it is easier to impose voltage rather than charge. So, to get your potential, you’ll need electrodes that are shaped like hyperbolas.

Hope this helps.

  • $\begingroup$ So would you argue that there is no location of charges that can give Purcell's electrostatic field as he describes it ($phi = -Kxy$) everywhere, and that this field cannot exist without some type of boundary? $\endgroup$ Nov 27, 2023 at 18:14
  • $\begingroup$ Yes, by Gauss’ law, it cannot be within the bulk of the domain, so you’re only left with the boundary or more generally outside the region where the field is given by this formula as in the second example. $\endgroup$
    – LPZ
    Nov 27, 2023 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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