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I'm trying to understand how to think of and visualize a conductor where charge is added.

As I understand, conductor are defined as materials (where the nucleus is stationary) that permit electrons to flow freely from atom to atom. They have the properties that $\vec{E} = 0$ inside and $\rho = 0$ inside.

For the case where you add a positive charge, the electrons would move so as to cancel the effective electric field of the added charge, and electrostatic equilibrium would be obtained. Thus $\vec{E} = 0$ which implies by Gauss's Law ($\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0})$, so $\rho = 0$.

That's fine but, what happens if we remove an electron? How would the conductor react to this? Would it simply reconfigure to minimize repulsion forces and be left with a positive $\rho$ inside and a non-zero $\vec{E}$?

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  • $\begingroup$ Any charge you remove or add to the bulk, will alter the potential at that point. Due to the potential difference thus created w.r.t the surrounding regions, charges eill flow in/out of that point to equalise the potential. This will continue untill the excess charge reaches the boundary.(a little maths can show this). This will again make the intrrior devoid of excess charges and keep the electrostatic field zero. $\endgroup$ – Lelouch Jul 21 '16 at 17:49
  • $\begingroup$ @Lelouch Okay but isn't it the case that in a conductor only the electrons move from particle to particle since the nucleus of the conducting material is stationary, so which excess charge reaches the boundary in the case where you remove an electron from the material? $\endgroup$ – Alex Jul 21 '16 at 17:55
  • $\begingroup$ Do you know about en.wikipedia.org/wiki/Electric-field_screening? $\endgroup$ – CuriousOne Jul 21 '16 at 20:12
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When you put +ve charge on a conductor, you are really removing the same amount of -ve charge, because only the electrons move.

Gauss' Law assumes that charge is infinitely divisible, and can be spread uniformly throughout a volume or over a surface. This is a good approximation when the charge is on the order of $1 \mu C$, corresponding to about $10^{13}$ electrons. Then volume and surface charge densities $\rho$ and $\sigma$ have meaning.

However, when the charge is made up of a small number of electrons - and even more so when it is only one electron - these quantities do not have any meaning unless they are averaged over time. For the same reason, if the charge on the conductor is an excess or deficit of a small number of electrons, the electric field may not be zero at all points inside, because the excess charge is not spread uniformly over the surface.

Because of the random high-speed motion of the conduction electrons throughout the conductor, the electric field $E$ inside it may not be very well defined at all when the excess/deficit charge is so small, and will fluctuate with a standard deviation proportional to $\frac{1}{\sqrt{n}}$ where $n$ is the excess number of excess/deficit electrons.

Even when there are $10^{13}$ excess electrons, no particular group of $10^{13}$ electrons stays on the surface while the original conduction electrons stay inside. All of the electrons are identical and move around equally, and it is only on average that an excess of electrons will be found near the surface. As more electrons are added or removed, the region in which the excess or deficit of electrons is likely to be found - and in which the electric field is not close to zero - will get nearer and nearer to the surface.

If you add or remove charge, there is always a redistribution of the average charge, even when a single electron is added or removed, because the electrons are moving around randomly at high speed. They are not fixed to the surface.

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  • $\begingroup$ Thanks for your very informative response. Why does the random high speed motion of the conduction electrons result in larger fluctuations of $E$ for smaller numbers of excess/deficit electrons? Is it simply because as you said there is just not enough to spread to negate the effects of the added or removed charge? $\endgroup$ – Alex Jul 22 '16 at 11:25
  • $\begingroup$ @Alex : Sorry - my Answer is more speculative than is justified by my knowledge of electrons in metals. I will think some more about this issue and update it as soon as I can. $\endgroup$ – sammy gerbil Jul 23 '16 at 11:18

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