Related: How is the curl of the electric field possible?

According to my information:

$$\nabla \times \frac{\hat{r}}{{|\mathbf{r}|}^2} = 0 $$

for all $\mathbf{r}$ ( including $\mathbf{r}=0$; compare with $\nabla \frac{\hat{r}}{{|\mathbf{r}|}^2} = 4 \pi \delta(|\mathbf{r}|)$ that is zero except when $\mathbf{r}=\mathbf{0}$ ).

As the electric field created by a single charge is proportional to $\frac{\hat{r}}{{|\mathbf{r}|}^2}$ and the field created by a discrete or continuous distributions of charges is a linear combination of the previous, it could (?) be inferred that, for all electric field:

$$\nabla \times \mathbf{E} = 0 $$

always, no matter if a) charges are in movement or b) charge is time dependent

That seems against to Faraday's law:

$$\nabla \times \mathbf{E} = - \frac{\partial}{\partial t} \mathbf{B} $$

please, what I'm missing ?

In other words, if we have a time dependent distribution of charges $\rho({\mathbf{r,t}})$:

$$ \mathbf{E(\mathbf{r,t})} = \int_{V'} \frac{1}{4 \pi {\epsilon}_0} \rho({ \mathbf{r',t}}) \frac{ \mathbf{r}-\mathbf{r'} }{|\mathbf{r}-\mathbf{r'}|^3} dV'$$

(replace integral by sum and $\rho$ by $q_i$ in case of punctual charges)

$$ \nabla \times \mathbf{E(\mathbf{r,t})} = \nabla \times \int_{V'} \frac{1}{4 \pi {\epsilon}_0} \rho({ \mathbf{r',t}}) \frac{ \mathbf{r}-\mathbf{r'} }{|\mathbf{r}-\mathbf{r'}|^3} dV' = \int_{V'} \frac{1}{4 \pi {\epsilon}_0} \rho({ \mathbf{r',t}}) \nabla \times \frac{ \mathbf{r}-\mathbf{r'} }{|\mathbf{r}-\mathbf{r'}|^3} dV' = \int_{V'} \frac{1}{4 \pi {\epsilon}_0} \rho({ \mathbf{r',t}}) ~ 0 ~ dV' = \int_{V'} 0~ dV' = 0 $$

that should (?) be true only in case of time independent distributions.

  • 1
    $\begingroup$ $E(x,t)$ is clearly a time varying field. Suppressing the coordinates, the correct equation for $E$ is $E=-\nabla \phi - \frac{1}{c}\frac{\partial A}{\partial t}.$ Hence $\nabla \times \big( E +\frac{1}{c}\frac{\partial A}{\partial t}\big )=0$ where $A$ is the vector potential. Or $\nabla \times E = -\frac{1}{c}\frac{\partial }{\partial t}\nabla \times A= -\frac{1}{c}\frac{\partial B }{\partial t}$ where $B=\nabla \times A$ which is Faraday's equation. $\endgroup$ Nov 19, 2019 at 3:34

3 Answers 3


When the curl is $0$ you are dealing with electrostatics, so of course $\frac{\partial \mathbf B}{\partial t}=0$. For a single, stationary point charge or a collection of such charges this is indeed the case.

Faraday's law always holds. When dealing with electrostatics it's still valid, but just a special case. The more general case is when you have time varying fields.

If you want to explicitly handle the electric field from a time varying charge distribution/current, use Jefimenko's Equations. You cannot just plug in $\rho(t)$ into Coulomb's law.

  • $\begingroup$ Thankd for your interest in the question. I've added to the question the phrase "no matter if charges are in movement or charge is time dependent" $\endgroup$ Nov 17, 2019 at 16:37
  • $\begingroup$ @pasabaporaqui I see. That doesn't change my answer though. As soon as charges are moving then you have time varying fields. $\endgroup$ Nov 17, 2019 at 16:38
  • $\begingroup$ That is the question, even if charges are time dependent or moving, the curl of inverse square distance is zero, thus, it seems the curl of the field must be also 0, against Faradays. In other words, you say that "when the curl is 0 you are dealing with electrostatics", but curl of inverse of distance equal zero seems (?) applicable to static and dynamic systems. See newest addition in the question. $\endgroup$ Nov 17, 2019 at 17:23
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    $\begingroup$ @pasabaporaqui This might help. You can't just plug in $\rho(t)$ into Coulomb's law. $\endgroup$ Nov 17, 2019 at 18:00

Your relation

$$ \mathbf{E(\mathbf{r,t})} = \int_{V'} \frac{1}{4 \pi {\epsilon}_0} \rho({ \mathbf{r',t}}) \frac{ \mathbf{r}-\mathbf{r'} }{|\mathbf{r}-\mathbf{r'}|^3} dV'$$

is only valid for static charges.

If the charges are moving, that is if $\rho$ actually depends on $t$ there are extra terms in $\mathbf{E}$, and these extra terms do contribute to the curl of $\mathbf{E}$.

  • $\begingroup$ Thanks. And the correct expression is ... ? $\endgroup$ Nov 17, 2019 at 18:14
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    $\begingroup$ @pasabaporaqui: Check out Jefimenko's equations. $\endgroup$ Nov 17, 2019 at 18:14
  • $\begingroup$ @MichaelSeifert: yes, this has been the hint given in previous answer. $\endgroup$ Nov 17, 2019 at 18:15
  • $\begingroup$ @pasabaporaqui Aaron Stevens gave you the reference, and Michael Seifert tried to do so, but there seems to be a bug in his answer. Try this : en.wikipedia.org/wiki/Jefimenko%27s_equations $\endgroup$
    – Alfred
    Nov 17, 2019 at 18:17
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    $\begingroup$ @pasabaporaqui Done $\endgroup$ Nov 17, 2019 at 18:49

I think the essential point, which is encoded in Jeffimenko’s equations, is that the information about the location of the charge propagates at finite speed, i.e. the speed of light. So if a charge is moving, the field at any given point is not the Coulomb field of the instantaneous location of the charge, but instead depends on where it was a time t=r/c ago, where r is the distance to the charge evaluated at that previous time. Since the charge is moving, the field at a given point will actually depend on a range of past locations of the charge, and this can give a field with nonzero curl.


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