It’s said that any excess charge inside a conductor will redistribute at it’s surface. Now there are two cases to it:-

1) A neutral conductor placed in an electric field 2) A charged conductor kept in space.

Now what we do to explain this redistribution is, suppose there is a net electric field inside a conductor. Free electrons will move opposite to the external electric field lines and cause a ‘induced electric field’ that will eventually cancel out any net electric field inside the conductor.

But by what law do the charges redistribute at the surface?

To explain it, gauss Law is used.

$$ \Phi = \frac {\text {Charge Inside}}{\epsilon} $$

As we know there can’t be electric field inside a conductor, flux through a Gaussian surface inside the conductor should be zero.

From this line of reasoning,we say charge inside the Gaussian surface should be zero.

But my problem is, value of $\epsilon$ is inifinite for a conductor. So to make flux zero, charge inside doesn’t need to be zero...

Where is this doubt factually/theoretically incorrect?

  • $\begingroup$ If your Gaussian surface is entirely inside the conductor the charge enclosed will be zero. $\endgroup$
    – garyp
    Commented Jan 31, 2020 at 17:15
  • $\begingroup$ That's my doubt. Why would charge inside be zero? The Gauss Law arguement is used to explain this but in gauss law, epsion is infinite for a conductor so to make flux zero, charge inside could be any finite value. How can i understand that charge inside it will be zero? By what law? $\endgroup$ Commented Jan 31, 2020 at 17:40
  • 2
    $\begingroup$ Can you cite a reference which suggests that the constant in Gauss's law is different inside a conductor? $\endgroup$
    – R.W. Bird
    Commented Jan 31, 2020 at 18:08
  • $\begingroup$ @AshishRajShukla Take a look at Newton's shell theorem for spherically symmetric shells and apply the result to a charged shell. The charges redistribute on ths surface because the state with the lowest potential energy is the symmetric state, and all isolated systems tend towards this. $\endgroup$
    – Physics
    Commented Jan 31, 2020 at 19:03
  • 3
    $\begingroup$ $\epsilon$ is NOT infinite for conductors. It is the conductivity $\sigma$ that is infinite. $\endgroup$ Commented Jan 31, 2020 at 22:40

2 Answers 2


The macroscopic argument is that, if there is a field inside the conductor, the free charges in the conductor will move, sothe only situation compatible with steady state is one where there is no field inside and all the charges are redistributed on the physical boundary of the surface.

The more microscopic argument - which involves the conductivity - is provided by the continuity equation and goes something like this.

The current flowing out of a closed surface is $$ I_{\textrm{out}}=\oint_S \vec J\cdot d\vec S =-\frac{dQ_{\textrm{encl}}}{dt} $$ where $\vec J$ is the current density. Since $Q_{\textrm{encl}}$ is just $\int dv \rho_v$ we get \begin{align} \oint_S \vec J\cdot d\vec S=\int_V \vec \nabla \cdot \vec J dv= -\int_V \frac{\partial \rho_v}{\partial t} dv \end{align} and since the volume is arbitrary we get the continuity equation for currents: \begin{align} \vec\nabla \cdot \vec J=- \frac{\partial \rho_v}{\partial t} \, . \end{align} By the microscopic version of Ohm's law $\vec J=\sigma\vec E$ where $\sigma$ is the conductivity so that $$ \vec\nabla\cdot \vec J=\sigma \vec\nabla\cdot \vec E = \frac{\sigma \rho_v}{\epsilon}= \frac{\partial \rho_v}{\partial t} $$ where the microscopic form of Gauss' law $\vec\nabla\cdot \vec E=\frac{\rho_v}{\epsilon}$ has been used. Solving in $t$ using separation of variables yields $$ \rho_v(\vec r, t)=\rho_v(\vec r, 0) e^{-\sigma t/\epsilon}\, . \tag{1} $$ This equation states that the density of charge at location $\vec r$ in a conductor decreases exponentially in time from its initial value at this location. In particular:

  1. If $\sigma t/\epsilon$ is large, then $e^{-\sigma t/\epsilon}$ is small and the charge density at that point is very small,
  2. If $\sigma t/\epsilon$ is small, then $e^{-\sigma t/\epsilon}$ is near 1 and the charge density at that point does not change much.

For any material $\epsilon=\epsilon_r\epsilon_0$ and $\epsilon_r$ is typically of size $\sim 1$ to $100$, while $\epsilon_0\sim 10^{-11}$ so numerically:

  • For a good conductor like copper, $\sigma>10^{4}$ and $\sigma t/\epsilon \sim 10^{15} t$ is very large except for extremely short times. Thus, steady state where there is no charge density $\rho(\vec r, t)\approx 0$ inside, is reached very rapidly. In a perfect conductor where $\sigma\to \infty$ the charge inside reaches $0$ within an arbitrarily small time.
  • For a good insulator like quartz, $\sigma<10^{-15}$ so $\sigma/\epsilon$ is quite small and $\rho(\vec r, t)\to 0$ very slowly. In fact the assumptions behind this simple model must be revised when the relaxation times are that large.

The way I understand it (which may not be entirely correct, but I'm gonna take a shot at this), electrical permittivity measures the ease at which electric charge can move through a material, and thus how polarizable the material is. Therefore, the charge contained inside some area in the material actually depends on $\epsilon$. I would guess that, if you could write $q_{enc}$ as a function of $\epsilon$ (and of $E$), you would end up with a function that approached zero as $\epsilon$ approached infinity.


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