Maxwell's equations specify two vector and two scalar (differential) equations. That implies 8 components in the equations. But between vector fields $\vec{E}=(E_x,E_y,E_z)$ and $\vec{B}=(B_x,B_y,B_z)$, there are only 6 unknowns. So we have 8 equations for 6 unknowns. Why isn't this a problem?
As far as I know, the answer is basically because the equations aren't actually independent but I've never found a clear explanation. Perhaps the right direction is in this article on arXiv.
Apologies if this is a repost. I found some discussions on PhysicsForums but no similar question here.
Best Answer
It isn't a problem because two of the eight equations are constraints and they're not quite independent from the remaining six.
The constraint equations are the scalar ones, $$ {\rm div}\,\,\vec D = \rho, \qquad {\rm div}\,\,\vec B = 0$$ Imagine $\vec D=\epsilon_0\vec E$ and $\vec B=\mu_0\vec H$ everywhere for the sake of simplicity.
If these equations are satisfied in the initial state, they will immediately be satisfied at all times. That's because the time derivatives of these non-dynamical equations ("non-dynamical" means that they're not designed to determine time derivatives of fields themselves; they don't really contain any time derivatives) may be calculated from the remaining 6 equations. Just apply ${\rm div}$ on the remaining 6 component equations, $$ {\rm curl}\,\, \vec E+ \frac{\partial\vec B}{\partial t} = 0, \qquad {\rm curl}\,\, \vec H- \frac{\partial\vec D}{\partial t} = \vec j. $$ When you apply ${\rm div}$, the curl terms disappear because ${\rm div}\,\,{\rm curl} \,\,\vec V\equiv 0$ is an identity and you get $$\frac{\partial({\rm div}\,\,\vec B)}{\partial t} =0,\qquad \frac{\partial({\rm div}\,\,\vec D)}{\partial t} =-{\rm div}\,\,\vec j. $$ The first equation implies that ${\rm div}\,\,\vec B$ remains zero if it were zero in the initial state. The second equation may be rewritten using the continuity equation for $\vec j$, $$ \frac{\partial \rho}{\partial t}+{\rm div}\,\,\vec j = 0$$ (i.e. we are assuming this holds for the sources) to get $$ \frac{\partial ({\rm div}\,\,\vec D-\rho)}{\partial t} = 0 $$ so ${\rm div}\,\,\vec D-\rho$ also remains zero at all times if it is zero in the initial state.
Let me mention that among the 6+2 component Maxwell's equations, 4 of them, those involving $\vec E,\vec B$, may be solved by writing $\vec E,\vec B$ in terms of four components $\Phi,\vec A$. In this language, we are left with the remaining 4 Maxwell's equations only. However, only 3 of them are really independent at each time, as shown above. That's also OK because the four components of $\Phi,\vec A$ are not quite determined: one of these components (or one function) may be changed by the 1-parameter $U(1)$ gauge invariance.