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

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

  2. 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.

  • $\begingroup$ Where did you get $\sigma \frac{\rho}{\epsilon} = -\frac{\partial \rho}{\partial t} from?$ $\endgroup$ – Yashas Mar 19 '17 at 5:41
  • $\begingroup$ @YashasSamaga: the divergence of static electric field is $\frac{\rho}{\epsilon}$ $\endgroup$ – user147507 Mar 19 '17 at 6:03
  • $\begingroup$ Two questions: why does steady state current imply $\frac {\partial \rho}{\partial t}=0$? and where does $\sigma=-\rho \mu$ come from? $\endgroup$ – GeeJay Mar 19 '17 at 6:17
  • $\begingroup$ @JayJay: (1) Because for steady-state current, charge density does not vary with time; (2) Conductivity equals charge density times electron (charge) mobility. Thank you. $\endgroup$ – user147507 Mar 19 '17 at 6:32
  • $\begingroup$ The only places I think error might have crept in is that a) ohm's law doesn't hold here because it's an empirical law or b) $\sigma=ne\mu =Ne\mu /V =Q\mu /V \neq \rho \mu $ because Q/V is just the aggregate charge density and not a function of spatial coordinates as $\rho=dq/dV $ is. $\endgroup$ – GeeJay Mar 19 '17 at 7:28

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.

  • $\begingroup$ But $\nabla \cdot\vec{E}=\frac{\rho}{\epsilon}=0\rightarrow\rho=0$, which makes $\textbf{J}=\sigma\textbf{E}=-\rho\mu\textbf{E} =0$. How come can the current density = 0 while a steady-state current exists? Thanks. $\endgroup$ – user147507 Mar 19 '17 at 15:36
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    $\begingroup$ Hey, the $\rho$ in $\nabla \cdot\vec{E}=\frac{\rho}{\epsilon}$ represents the whole charges in a closed Gaussian surface, so including positive charges and electrons. In a medium, no matter Conductor or Insulator, these charges always have a homogeneous distribution, so there is no net charges in a closed Gaussian surface. For the difinition of current density $\vec{J}=\rho \vec{v}$, the $\rho$ is the Mobilizable charges or say Kinetic electrons, so it exists in a current-carrying wire forever. $\endgroup$ – Feynman Mar 20 '17 at 0:21


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.

  • $\begingroup$ I still don't get it. Please explain to me. Thanks. $\endgroup$ – user147507 Mar 19 '17 at 15:39
  • $\begingroup$ @Marmousi It's simple. Try to see it this way. In a conducting wire density of electrons is very large so sigma is very large hence by ohm's law (that you have written) we get E tending to 0 which is right . Now take any gaussian surface within the conductor , now since in steady state there is no accumulation of charge the amount that flows in flows at out too. At any time the flux entering = flux leaving . No net flux. And in your 1st equation where you write E=rho/epsilon you are not taking its del. It will be del. (sigma.rho/epsilon) which is not what you have written . That's wrong...... $\endgroup$ – Shashaank Mar 19 '17 at 15:47
  • $\begingroup$ When does Marmousi say $E = \rho / \epsilon$? $\endgroup$ – GeeJay Mar 19 '17 at 16:11
  • $\begingroup$ @Shashaank: I understood everything you said but it doesn't seem to address the source of my confusion: the contradiction in which the current density = 0. How could J = 0 while there is a current? Besides, I moved sigma out of the $\nabla$ because I let it be constant. Thanks. $\endgroup$ – user147507 Mar 19 '17 at 16:12
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    $\begingroup$ @Marmousi OK. The Wikipedia article says that sigma =n*e*(mu) where n is only the electron density (for current due to electrons) .While here rho is the total charge density which is 0. While according to Wikipedia it's n (electron density) which here is not 0 . Hence there is current.en.m.wikipedia.org/wiki/… $\endgroup$ – Shashaank Mar 19 '17 at 16:43

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