The 6 components of the electromagnetic field are in fact components of an antisymmetric Minkowski tensor. It's easier to remember the relations in terms of the four-potential:
$$F_{\mu \nu} = \partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
The time-space components are (latin indices run from ##1## to ##3##, and I use the (+---) convention of the metric)
$$F_{0n}=\partial_0 A_n-\partial_n A_0=-\frac{1}{c} \partial_t A^n-\partial_n A^0=E^n=-E_n,$$
and the space-space components
$$F_{mn}=\partial_m A_n-\partial_n A_m=-\partial_m A^n + \partial_n A^m=-\epsilon^{lmn} B^l=-B^{mn}=-B_{mn}.$$
This is in a fixed inertial reference frame.
For the following we also need the Hodge dual
$$(\dagger F)^{\mu \nu}=\frac{1}{2} \epsilon^{\mu \nu \rho \sigma} F^{\rho \sigma}.$$
After some algebra one gets
$$(\dagger F)^{0 n}-B^n.$$
You can rewrite this in a coordinate-free way. Take a reference frame with the time-like unit vector called ##u^{\mu}##. For the original frame, used above we have ##(u^{\mu})=(1,0,0,0)##. So you can define four-vector electric-field components by
$$E^{\mu}=F^{\mu \nu} u_{\nu}, \quad B^{\mu}=(\dagger F)^{\mu \nu} u_{\nu}.$$
In the reference frame, where ##u^{\mu}=(1,0,0,0)## a single monopole sitting at rest in the origin of the spatial coordinate system, there is no time component. You have
$$\vec{E}=0, \quad \vec{B}=\frac{g}{4 \pi r^3} \vec{x}$$
where ##g## is the magnetic charge of the point monopole. It's completely analogous to electrostatics.