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.
Best Answer
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.