Solving Maxwell's equation:How there is no free charge but still free current? In my textbook, I can't understand how the author solved Maxwell's equations of electromagnetism in a conducting medium. The author assumed that there are no free charges inside the medium such that Maxwell's 1st eqn. becomes,$$\vec{\nabla}\cdot\vec{D}=0$$ but in the problem 4th eqn. still remains $$\vec{\nabla}\times\vec{H}=\vec{J}+\frac{\partial\vec{D}}{\partial t}$$Now as per my understanding only free charges give rise to free current $\vec{J}$ so if there is no free charge shouldn't $\vec{J}$ be also zero?
 A: Free current is indeed produced by free charges, but zero charge density does not correspond to an absence of free charges.
Whenever you have identical positive and negative free charge densities, the free charge density is zero. Positive and negative free charges can move independently of each other in such a way that the total charge density is zero everywhere (think about a loop of wire, with electrons traveling around the loop with constant drift velocity). When they do this, they create a free current, even when the free charge density is zero.
A: I have thought about your question for some time, and I'm fairly confident that I've found a solution to the conundrum. Beware that, like you, I'm a "new student", so I caution you to take the following with a grain of salt. I hope that a more expert user can (in)validate the arguments below eventually.
Let's treat the problem from a physical standpoint first. I think you are more interested in the mathematics, but perhaps a little physical intuition may help set the stage for the latter part. 
It's pretty clear why you can indeed have a current density without charge density: if the former is a macroscopic density, then the value of $\rho$ at a point will be the total charge cointained in a small volume around that point, a volume which is not greater than the resolution of the measuring devices at play, but which can still contain a great deal many particles (you can call this volume "infinitesimal"). Specifically, such total charge may be $0$ (e.g., there is an equal number of electrons and protons in our infinitesimal volume), but particles may still enter and exit the infinitesimal volume, thus producing a current, without changing the total charge, as long as the number of, for example, electrons going in and out is the same. 
In a wire this may well be the case, as there is usually the same number of electrons and protons in it, and the electrons which move to produce the current usually don't "pile up" anywhere.
But let's move on to the mathematics. What is the definition of current density? To define it, let's introduce four auxiliary functions:


*

*The negative charge density $\rho^-$. This is a function of space and time which yields the total amount of negative charge contained in an infinitesimal volume at a given instant: $\rho^-=\frac{dQ_{neg}}{dV}$. By integrating this function over some volume, the total negative charge in that volume is obtained

*The positive charge density $\rho^+$. Same as above, mutatis mutandis.

*The negative velocity field $\bf v^-$: again a function of space and time yielding the $velocity$ of the negative charge carries at a given point and instant of time. Perhaps you can think of ${\bf v}(x,y,z,t)$ as the average velocity at instant $t$ of the negatively charged particles contained in the infinitesimal volume sorrounding the point $(x,y,z)$.

*The positive velocity field $\bf v^+$. Same as above, again with the due modifications.


It should be clear that the total charge density $\rho$ is given by $\rho^-+\rho^+$. Now, define the current density as ${\bf J}=\rho^-{\bf v^-}+\rho^+{\bf v^+}$. The current through any surface is given, as usual, by the flux of $\bf J$ through that surface.
Here you can see that $\rho=\rho^-+\rho^+$ may well be $0$, but $\bf J$ need not be $0$ as well. Hence you can have no net charge anywhere, but still have a current (density). Going back to the neutral wire case, it is evident that, the protons being still and the electrons moving, the positive velocity field is null, and only the "negative" terms account for the current.
