Charge density of an uncharged sphere This post follows directly from my question about EM waves in conductors EM-wave equation in conductors with source terms
A more specific question in light of this.
Given an uncharged conductor:
We know $2$ things that are true:
$$ \vec J = \sigma\vec E $$
$$ \int \rho dV = 0 $$
In a conductor,
Maxwells equation:
$$ \vec\nabla\vec E = \rho / \epsilon_0 $$
is now set to $ \vec \nabla\vec E = 0 $
this implies $ \rho = 0 $
But, if $ \vec J = \sigma \vec E $
and  $ \vec J = \rho \vec v $
which means we have  no Current density inside the conductor
However using these equations for a EM wave inside a conductor, the amplitude DECAYS exponentially DUE TO current density ( energy conservation)
so my final question is: if $ \rho = 0 $ how is there a current density to effect an EM wave inside the conductor?
 A: First of all an uncharged conductor implies $\rho=0$, which implies $\vec\nabla\vec E=\rho/\epsilon_0=0$. Not the other way around. Having a net-charge of $0$, or $\int_V\rho dV=0$ still allows charge separation due to electric and magnetic fields provocated in the conductor, which is nothing but current (charge flow).
Now there are still cases where the net charge $\int_V\rho dV$ is $0$ but $\rho\neq 0$. The thing is that in reality there is no such case. Or more specifically, the only cases where you have a volume without charges at all would be a vacuum and even so if there is a photon passing thrue it could split into electron and positron creating a charge distribution.
So specially when talking about a conductor, that consists of metal for example, there will be neutral atoms, which naturally have a charge distribution within them. When we write $\rho=0$, we mean there is no big amount of charged ions or electron accumulation in a specific place. There are only neutral atoms, of which the electrons (in the case of a conductor) can freely move and are thus able to create a current density.
