Equation of Continuity statements I dont understand There are two statements about Equation of Continuity in my professors notes that I don't understand.
$$ \nabla \cdot \textbf{J} = - \frac{\partial\rho}{\partial t} $$

*

*The Equation of Continuity can be used to show that, in some cases,
the field (or flux) lines of current density close in on themselves
(i.e., each flux line forms a closed loop).


*The phenomenon of the electric field E being zero in a conductor
can be explained by the use of the Equation of Continuity.
The first item means that the divergence of J is sometimes non zero basically, right? So its when charge density is not constant?
Not sure how one would go about for the 2nd point though. I tried using Ohms law in point form, that $\rho = 0$ inside a conductor and $ \nabla \cdot \textbf{E} = \frac{\rho}{\epsilon}$ but I came nowhere.
$$ \textbf{J} = \sigma \textbf{E} \\ \nabla \cdot \sigma \textbf{E} = - \frac{\partial\rho}{\partial t} \\ $$
 A: Imagine you bring a charge close to a conductor, there will be an electric field inside the conductor for a short period of time. But the time period where electric field present in conductor is infinitesimal. This time varies from material to material but it's around $\approx 10^{-16}s$ for metals
$$\textbf{J}=\sigma \textbf{E}$$
$$\nabla \cdot \textbf{J} = \nabla \cdot \sigma \textbf{E} =-\frac{\partial \rho}{\partial t}$$
and you know $\nabla \cdot \textbf{E} = \frac{\rho}{\varepsilon_0}$
Thus
$$\frac{d\rho}{dt}+ \frac{\sigma \rho}{\varepsilon_0}=0$$
and solution for this differential equation is
$$\rho = \rho_0 exp({-\frac{\sigma}{\varepsilon_0}t})$$
by following this derivation, you can calculate the time $t$ for charges to move to the surface and leave electric field inside conductor zero.
Edit regarding your comment
$$ \lim\limits_{t \to\infty} \rho = 0$$
so you will have $\nabla \cdot \textbf{E} = 0$, get use of divergence theorem
$$\int_{V}\nabla\cdot\textbf{E} dV = \oint_{S}\textbf{E} \textbf{dS} = 0$$
Hence $\textbf{E} = 0$
