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.

$$ \textbf{J} = \sigma \textbf{E} \\ \nabla \cdot \sigma \textbf{E} = - \frac{\partial\rho}{\partial t} \\ $$

  • $\begingroup$ The equation makes no sense, you are equating a vector and a scalar. It is also unclear what you are actually asking. $\endgroup$ Jan 9, 2021 at 17:40
  • $\begingroup$ I managed to write the wrong operator. Sorry for that. Post is edited now. $\endgroup$
    – Clone
    Jan 9, 2021 at 18:29

1 Answer 1


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 $\nabla \cdot \textbf{E} = \frac{\rho}{\varepsilon_0}$


$$\frac{d\rho}{dt}+ \frac{\sigma \rho}{\varepsilon_0}=0$$

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 $\textbf{E} = 0$

  • $\begingroup$ Thank you for the explanation! But how do I prove later mathematically that the field E is 0 inside the conductor? $\endgroup$
    – Clone
    Jan 9, 2021 at 18:28
  • 1
    $\begingroup$ Check out the answer now. $\endgroup$
    – Monopole
    Jan 9, 2021 at 19:45
  • $\begingroup$ Thanks for the edit! $\endgroup$
    – Clone
    Jan 9, 2021 at 20:48

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