Steady-state current and equation of continuity I am learning EM and a bit confused when it comes to steady-state current and the equation of continuity.


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*Equation of continuity:
$$\nabla \cdot\textbf{J}=-\frac{\partial \rho}{\partial t}\rightarrow\sigma\nabla\cdot\textbf{E}=-\frac{\partial \rho}{\partial t}\rightarrow\sigma\frac{\rho}{\epsilon}=-\frac{\partial \rho}{\partial t}$$

*Steady-state curent:
$$\frac{\partial \rho}{\partial t}=0$$
So, for a steady-state current, which also satisfies the equation of continuity, we have
$$\sigma\frac{\rho}{\epsilon}=0\rightarrow\rho=0 \, .$$
However, $\rho=0$ implies no current since
$$\textbf{J}=\sigma\textbf{E}=-\rho\mu\textbf{E} \, ,$$
which contradicts the fact that there is a steady-state current.
I know that either my math or my understanding is wrong.
 A: Hint:
What you have done also shows that $\nabla \cdot E=0$ for steady state. In fact for conductors a very small $E$ gives a good amount of current. $E$ has to be very small since $\sigma$ is very large.
So that means that as $\nabla \cdot E=0$ that amount of flux entering any gaussian surface drawn within the conductor equals the amount of flux leaving that gaussian surface.
One thing more: the steady state equation does not apply always. Because when current sets up in a wire there is an electric field within the wire that drives the current through it and there are electrons which get collected at bends or turns to drive the current throughout the wire.
A: Clearly, you don't understand this equations in a certain sense. Equation of continuity is suitable for any situation, so whenever you can write down$$\nabla \cdot\vec{J}=-\frac{\partial \rho}{\partial t}$$
Together with  Ohm's law $\vec{J}=\sigma\vec{E}$, we have$$\nabla \cdot\vec{J}=\sigma\nabla \vec{E}=-\frac{\partial \rho}{\partial t}$$
So far, this goes well. For Steady-state curent, $-\frac{\partial \rho}{\partial t}\approx 0$, so
$$\nabla \cdot\vec{J}=\sigma\nabla \vec{E}=0$$
In steady-state curent, there doesn't exist static charges or no accumulated charges, so $\nabla\cdot\vec{E}=0$. So it is self-consistent.
