# Distribution of charge over a conductor?

Excess charge in a conductor is supposed to distribute itself evenly over the surface of the conductor. If the conductor has a charge of $2e$ how would the charge distribute itself? There's no way you can distribute $2$ electrons over the conductor so that the charge is evenly distributed, unless you can have less than $1e$ of a charge.

I have absolutely no knowledge of quantum mechanics, so please keep answers intuitive rather than technical.

## 1 Answer

It's tempting to think of an electron as a little point, but in most systems electrons are delocalised, that is they do not have a well defined position. As a general rule, if an electron is tightly bound to something then it more localised than if it's nearly free. So for example the electron in a hydrogen atom is bound to the proton so it's localised to a sphere with a radius of a tenths of nanometres (though it's still impossible to say exactly where within that sphere the electron is. By contrast electrons in the conduction bands of metals behave as if they are almost free, and indeed you can calculate their properties assuming they are free and get reasonably good answers.

The point of all this rambling is that if you add two electron to a conductor those two electrons don't have a precisely defined position but are (at least in principle) spread out over the conductor. So the charge density on the surface can be even because the two electrons are delocalised over the whole surface of the conductor.

I should add that this is a somewhat idealised scenario and would apply only to a perfectly spherical superconductor that is totally isolated. In real conductors imperfections would change the delocalisation of the electrons. Likewise bringing up test charges or anything similar to the conductor will interact with the electrons and partially localise them.