Skin effect explained in terms of the flow of charges? Can the skin effect for em waves incident on a conductor be explained by the movement of free charges within the conductor (given that they as $t\rightarrow \infty$, $\rho\rightarrow0$)? And if so how? Is there also a difference of what happens in good and bad conductors?
 A: Skin depth is precisely explained using that. The wave equation that is written in vacuum for deriving Maxwell's equation is somewhat modified. Basically you have
$$ \nabla \times B = \mu_0(J + \epsilon_0\dfrac{\partial E}{\partial t})$$
from Maxwell's equation. Whereas you have $J = 0$ in vacuum, hereby you have
$$ J = \sigma E$$
for a conductor, which implicitly states what you asked in the first part of your question about free charges being used for explanation of skin depth. Now this Maxwell's equation is combined with the other one
$$ \nabla \times E = -\dfrac{\partial B}{\partial t}$$
and what we get is an wave equation with an extra term, which has damped out solutions depending on distance penetrated. These damped out solutions explain the characteristic feature of skin depth. 
Now good and bad conductors are distinguished by the value of $\sigma$. I hope it sheds a bit more light on my answer to your twin question too in Why does the skin effect cause current to flow at surface of a conductor?
A: From Maxwell's equation
$$ \mathbf \nabla \times \mathbf B = \mu\left(\mathbf J + \epsilon\dfrac{\partial \mathbf E}{\partial t}\right)$$
Taking divergence of both sides gives
$$0=\mu \sigma \mathbf \nabla\cdot \mathbf E+\epsilon \dfrac{\partial}{\partial t}(\mathbf \nabla\cdot \mathbf E)$$
where we used $\mathbf J=\sigma \mathbf E\;.$ Now $\mathbf \nabla\cdot\mathbf E=\dfrac{\rho}{\epsilon}$.So equation for $\rho$ is 
$$\frac{\partial \rho}{\partial t}=-\frac{\mu\sigma}{\epsilon}\rho$$
This will give 
$$\rho=\rho_0\;\exp\left[{-\frac{\mu\sigma}{\epsilon}t}\right]\;.$$
Now, $\sigma$ depends on how good conductor it is.
