If the electric field on a closed region is uniform then by Gauss's law you have $0=\nabla\cdot\vec{E}=\rho/\epsilon_0$, so $\rho$ would be zero.

Is the converse necessarily true? Could we conclude that the electric field is uniform when $\rho=0$?

  • $\begingroup$ $(x, -y, 0)$ is an example of a non-uniform divergence-free field. $\endgroup$
    – Puk
    Commented Mar 27, 2023 at 5:57

2 Answers 2


Is the converse necessarily true? Could we conclude that the electric field is uniform when $\rho=0$?


For example consider the field around a point charge. The field is non-uniform because it falls as $1/r^2$ with distance from the charge but the charge density is zero everywhere except at the point.

The divergence is zero everywhere except at the point charge, but a field can have a zero divergence and still be non-uniform.

  • 1
    $\begingroup$ Which question are you answering "no" to? The title question "Is it possible to have a non-uniform electric field such that $\nabla\cdot E =0$" or a body question like "Could we conclude that the electric field is uniform when ρ=0?" $\endgroup$
    – hft
    Commented Mar 27, 2023 at 23:01

This is quite an interesting question, and maybe one can add to the answer and comment above by providing a more mathematical perspective.

These sorts of problems are best expressed using differential forms. Depending on your background you may not know so much about these, but I will assume some basic knowledge. They are a staple in physics at a higher level and I believe it is better to catch them earlier than later.

The main ingredient is Poincaré's lemma. This states that, given a $p$-form $\omega$ such that $d\omega = 0$, locally one has $\omega = d\alpha$ for some $(p-1)$-form $\alpha$. The electric field is really a one-form, whilst the magnetic field is really a 2-form. The gradient of a one-form $E$ is $d^\dagger E = \star d \star E$ where $\star$ is the Hodge star, so that $\star E$ is a $(D-1)$-form, $d \star E$ is a $D$-form and so $d^\dagger E$ is a scalar. Here $D$ is the dimension of the manifold in question.

Let us work with the flat space $\mathbb{R}^D$ where we can apply Poincaré's lemma globally (topologically non-trivial spaces are important too in physics, but bring added complications and richness). By the Poincaré lemma, the equation $\nabla \cdot E = 0$, better expressed as $d \star E = 0$, is then solved by $\star E = d \alpha$ for some $(D-2)$-form $\alpha$. Or $E = \star d \alpha$ (up to a sign depending on your conventions for $\star^2 = \pm 1$). Even better, dualize $\alpha$ to a 2-form $\omega = \star \alpha$, so that $E = \star d \star \omega \equiv d^\dagger \omega$ - we have really discovered a dualized version of Poincaré's lemma. This is the formal solution on a local patch. On a general manifold, one would have to sew the 2-forms $\omega_i$ for each patch $U_i$ together.

For example, in $D=3$, take $\omega = xy dx \wedge dy$. Then $\star \omega = xy dz$ and $d \star \omega = x dy \wedge dz + y dx \wedge dz$ and $E = d^\dagger \omega = x dx - y dy$, yielding the solution in Puk's comment above. This is easily checked to be divergence-less. (More generally, one can add a homogeneous term such as a multiple of $x^{a-1}y^{b-1}z^{c-1}(a yzdy dz - b xz dx dz + c xy dx dy)$ to get the same field $E$).

One might feel that that example is rather unphysical - the electric field grows unboundedly towards infinity. So you might wonder whether non-constant solutions exist which have boundary conditions such that they decay at infinity. But now we know how to do that: just multiply, for example, the previous solution $\omega$ by a $e^{-r^2}$ factor and take $d^\dagger$.

These kinds of localised field bumps can actually happen in physics - they are related to solitons.

But beware - there are more Maxwell equations. For example, assuming the absence of time-varying magnetic fields, we have $\nabla \times E = 0$, or in forms $d E = 0$. This is as well as the above. Thus, by Poincaré's lemma, $E = dV$ for some scalar $V$. Combining with the above, this is $d^\dagger \omega = dV$. This has solutions precisely when $d d^\dagger \omega = 0$, or equivalently when $d^\dagger d V = 0$. If the field is divergence-less, $d^\dagger E = 0$ and so $V$ must be a solution of $d^\dagger d V = 0$ (there is a similar a restriction on $\omega$). The latter trivially extends to the Laplace equation $\Delta V \equiv (d d^\dagger + d^\dagger d) V = 0$. For example, Puk's solution still works, being only linear. In particular, take $V = \frac12(x^2-y^2)$.

But more generally, any solution to Laplace's equation with the required boundary conditions will give you an example. The complete solution of Laplace's equation in dimension $D$ is quite a rich subject which I can't go into here, but let us say that there are many solutions.

If we allow our fields to be time varying, one discovers the electromagnetic wave. Here the electric field is non-uniform but still a solution of Maxwell's equations in a vacuum.


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