Is there a fourth component to the electric field and magnetic field?
Actually, the electric and magnetic fields from one combined tensor called the electromagnetic field tensor. This is a rank-2 tensor and takes the form* $$ F^{\mu\nu}=\left(\begin{array}{cccc} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{array}\right) $$ It has the following properties:
- It is anti-symmetric (so $F^{12}=-F^{21}$)
- It is traceless
- It has 16 elements, but only 6 distinct values
- When multiplied by its dual tensor ($G^{\mu\nu}$) it gives a Lorentz invariant value of $4\mathbf{B}\cdot\mathbf{E}$
- The inner product, $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2)$, is also a Lorentz invariant
You can also derive Maxwell's equations through the tensor by applying $\partial_\mu$ to it. Gauss's law and Ampere's law come from $$ \partial_\mu F^{\mu\nu}=4\pi J^\nu $$ where $J^\mu=\left(\rho,\,\mathbf{j}\right)$ is the four-current. The magnetic Gauss' law and Faraday's law come from applying the Bianchi identity to get $$ \partial_\gamma F_{\mu\nu} + \partial_\mu F_{\nu\gamma} + \partial_\nu F_{\gamma\mu}=0 $$ Or more concisely, $$\partial_{[\mu}F_{\nu\gamma]}=0$$
I am an astrophysicist, so I use cgs units; in SI, all electric fields have a factor of $1/c$.
This is more an extended comment to address the comments to Kyle's answer
For example if there was a time component to the electric and magnetic field
In a relativistic context, the electric and magnetic field components are not components of separate, related vector fields but, rather, are components of a 2nd rank tensor field; the electric and magnetic fields are part of one geometric object, not two.
Indeed, a clue to this is found in the fact that the magnetic field is, in 3D, a pseudo-vector field rather than a vector field.
So, in fact, the question "what is the time component of the electric and magnetic field" actually presumes a falsehood; it presumes that the electric and magnetic fields are separate but related four-vectors.
But, they're not.
Since a rank 2 tensor has two indices, we can properly speak of the time-time component, the time-space components, and the space-space components of the tensor but not the time or the space component(s).
To answer this question, you need to have a full geometric understanding of the Maxwell equations and what they represent.
Maxwell's equations are a garden-variety system of PDEs. In STA notation, it's simply
$$\nabla F = -J$$
We take it for granted that $F$ is a bivector, and thus has 6 components, and that $J$ is a vector, and thus has 4 components. But this equation describes up to eight separate equations. Why is that?
For an arbitrary bivector field $K$, the derivative $\nabla K$ can have both vector and trivector terms. That Maxwell's equations have only a vector source term is actually quite significant: this is part of the physical content of Maxwell's equations. We're saying the EM field is determined only by a vector current.
What would happen if there were a trivector current source term? It would be "magnetic" charge (magnetic monopoles) and an associated current. So right away we can appreciate what that source term would denote.
But wait, there's more! Let's say we had both electric and magnetic currents then. What kinds of fields could produce them?
As you've been trying to get at, these are the other two components of a field that could go into this differential equation. They are a scalar field and a pseudoscalar field. I'm not familiar with how these fields would manifest themselves, or what they would do.
So why don't we find out?
Let $\lambda$ be the scalar field. How would this affect Maxwell's equations with just a current source term?
Let $F = e_0 E + B$, where I've implicitly denoted that the magnetic field is a bivector. You can identify it as a vector instead and then consider $\epsilon_3 B$, but the net effect is pretty minimal.
Maxwell's equations then break down as
$$\nabla \cdot F = \partial_t E - e_0 \nabla_3 \cdot E + \nabla_3 \cdot B = -\rho e_0 - j$$
and
$$\nabla \wedge F = -e_0 \nabla_3 \wedge E - e_0 \partial_t B + \nabla_3 \wedge B = 0$$
Adding a scalar field $\lambda$ would affect only the vector part, with its gradient:
$$-J = \nabla \cdot F - e_0 \partial_t \lambda + \nabla_3 \lambda$$
So all in all, this would probably appear as some kind of extra current not associated with the motions of electric charges--or perhaps it would be indistinguishable from electric currents in some way, save that it permeates all space as a continuous function. It would look like there are some currents everywhere, in some sense. You can see why we don't even consider the existence of such a field. Unless it's very small, we would've detected it some time ago, since it interacts with the electric current source term.
An analysis of the magnetic pseudoscalar field would probably end up the same way.
So, is the Faraday tensor actually missing two extra components, a scalar field and a pseudoscalar field? I'd say no, but if you discover otherwise, you'll probably win a Nobel Prize. Good luck with that. As I said in the other question, do not be fooled into thinking that, just because $Fx$ has eight components, there are missing components of the Faraday field. There very likely aren't any such missing components. You can see this by considering what those components would do in the vanilla Maxwell equations, as I have done here.
Edit: some corrections on the relationship between this scalar field and gauge fixing.
This scalar field would remove the freedom to change $A$ through gauge transformations, as $\lambda$ would specify the divergence of $A$. Recall that gauge fixing relies upon the ability to perform the transformation,
$$A \mapsto A + \nabla \chi$$
For some scalar field $\chi$. This can be done because $\nabla \wedge (A + \nabla \chi) = \nabla \wedge A = F$, so the EM field is unchanged.
But if $\nabla \cdot A = \lambda$, then adding the gradient of a scalar field would change the value of $\lambda$ measurably, in all but the simplest cases:
$$\nabla \cdot (A + \nabla \chi) = \lambda + \nabla^2 \chi$$
Now, you would be restricted to gauge functions $\chi$ that are strictly harmonic. Harmonic fields are usually those that arise from some choice of boundary conditions--i.e. this would correspond to some choice of boundary condition, and the field contribution from currents would be unchanged. Of course, it strains imagination to consider how one would reasonably do this for gauge fixing. And if you found a transformation that preserves $\lambda$, it would not leave $F$ invariant in general.
So the supposed existence of this function $\lambda$ would have profound consequences toward gauge fixing. It does not absolutely forbid it as I originally thought, but it puts serious constraints on fixing that we probably would've run into by now.