I understand that electrons are very mobile in a conductor. But instead of all the charge accumulating on the surfaces of the conductor, as shown below in fig.1, couldn't the charges accumulate with a gradient across the entire conductor? That is to say, can't the positive and negative charges redistribute themselves throughout the conductor such that their density is greatest at their respective edges(surfaces) but lower as we move into the conductor? enter image description here enter image description here

Fig.2 describes what I mean. The various shades of the color describe the density of charge in the conductor. The darker the shade, the denser the accumulation of the positive charge and the lighter the shade, the denser the accumulation of negative charge.

Why can't this sort of charge distribution happen? In this case won't the electric field inside the conductor be uniform (and as a result not cancel out the external electric field)? If so, how? Or is there any other reason?

  • 2
    $\begingroup$ Charges inside a conductor always arrange themselves in such a way that net electric field inside it must be zero so there cannot be a charge inside this follows from gauss law $\endgroup$
    – Matt
    Apr 25 '17 at 17:31
  • $\begingroup$ No. What I was asking is that can't the charges redistribute themselves in the second manner and yet produce an electric field at each point within the conductor that is equal and opposite to the external electric field? $\endgroup$
    – Pranoy De
    Apr 26 '17 at 14:56

By Gauss' Law, if the charges were distributed as shown in your second figure, there would be a net electric field within the conductor. This would cause the charges to redistribute themselves until they reach the configuration shown in your first figure.


I had the same doubt once! And as everybody here have said. It's proof uses Gauss' Law. Trust me! It's a beautiful concept, exploiting the inverse square law dependence of electric field. Gauss's Law

$$\int E.dS = \frac {Q_{enclosed}}{\epsilon}$$ If a bounded region of space accommodates a finite amount of stationary electric charge (net charge), then the law guarantees that there should be some electric field poking out of the surface that bounds the charge

And if a field exists in a conductor, then the conductor is not happy, it's not in electrostatic equilibrium. The time order for accomplishing electrostatic equilibrium in electrical conductors is around a few femtoseconds 10^-15 !

But when there is a constant potential drive as in the case of a wire connected to a battery, I guess there might the be sort of a gradient of charge distribution as you say.

One needs to realise Gauss's Law, It's Proof, It's elegance.

And as why the inverse square law allows us to see field emanating from charges as something real spreading.


A system in non-equilibrium will always go (if possible) for a configuration of equilibrium.

Now, suppose you have a system such as that of Figure 2. In this case you will have an electric field inside the conductor, which will make the electrons move and rearrange in another configuration.

Since the final configuration will be of equilibrium, you cannot have any field inside the conductor (otherwise, charges will start moving) and the charges MUST be at the surface of the conductor (or they will produce an inside field).

  • $\begingroup$ Yes, but how do you say that the second charge distribution shall cause an unequal electric field inside the conductor(that will not cancel out the external electric field at every point within the conductor)? Can you prove that mathematically? $\endgroup$
    – Pranoy De
    Apr 26 '17 at 15:02
  • $\begingroup$ You can define a gaussian surface inside the conductor. Say you define it as a square that surrounds one type of blue in your figure. By Gauss' Law, since there a total amount of charge inside this surface, there will be an electric field. You know that the field is unequal because if you define another surface, which surrounds another type of blue, you will get a different charge, therefore, a different field. I'm assuming you've seen Gauss' Law. $\endgroup$
    – matrp
    Apr 26 '17 at 22:23
  • $\begingroup$ Okay, I think I understood your question wrongly. Forget my last comment if that's the case. I'll think about it. $\endgroup$
    – matrp
    Apr 26 '17 at 22:41
  • $\begingroup$ @Pranoy De: Re. your first comment to this answer, remember that the E field in Gauss' Law is the total E field at the Gaussian surface. So by Gauss' Law, the total E field in your second scenario will be non-zero. Gauss' Law is not just finding the E field due to the charge within the Gaussian surface (which as you ask could combine with the externally induced field to net a zero field) but the total field. $\endgroup$ Jul 1 '20 at 17:25

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