Curl of electric field 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.
 A: 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.
A: 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 $\mathbf{E}$.
A: 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. 
