Supposed charge distribution on a metallic sphere

Question

Usually when we talk about charge distribution on a sphere we say that excess charge resides on the surface of metallic sphere. As is said here:

• The net electric charge of a conductor resides entirely on its surface.

But when thinking about the distribution of positive charge on such a sphere I find that the net charge is distributed evenly over the entire sphere (even inside of it).

• Reason: Since matter is made up of atoms and in solid metal only negative charge is mobile therefore positive charge on a sphere can be attributed to the absence of charge on it. Now when we place positive charge on a sphere (i.e., to take some negative charge from it) we find that an electric field is generated in it which causes the inner electrons to move out and distribute evenly all over the sphere causing a uniform charge distribution throughout the sphere.

This is quite contrary to what I read in books. So:

• What is wrong with the logic that I give?

Answer 1

Your logic is not correct because you assume these electrons will be taken from the bulk of the material whereas in reality the surface atoms are sufficient to provide for the necessary electrons. For example, in the case of a neutral sphere and an external charge, the total induced charge on the sphere is zero, i.e. the electrons that depart the positive surface charge density areas match identically with the electrons that arrive to the negative surface charge density areas, all without actually needing to get electrons from the inside of the material.

But how many electrons does this one sheet of outermost atoms provide? Let's assume we have a 10 g Fe cube. $\rho_{_\text{Fe}}=7.874\times10^{6}\ \text{g}\cdot\text{m}^{-3}$, $\mu_{_\text{Fe}}=55.745\ \text{g}\cdot\text{mol}^{-1}$, so one side of this cube is $1.08\times10^{-2}$ meters and it has $1.08\times10^{23}$ atoms in it. Valence of $\text{Fe}$ is 2, so it has, in total, $2.16\times10^{23}$ electrons.

Now let's put this cube in an electric field of magnitude $E=1\times10^6 \text{V/m}$. The surface charge density this field induces will be $\sigma=E\cdot\epsilon_0=1\times10^6\times8.8542×10^{-12}=8.8542×10^{-6}\ \text{C}\cdot\text{m}^{-2}$. One side of this cube is $1.173\times10^{-4‬}\ \text{m}^2$, so the total charge induced in one side of it will be $a^2\,\sigma=1.04\times10^{-9}\ \text{C}$. Charge of a single electron is $1.60217662×10^{-19}$, so the number of electrons on this side is $6.492\times10^{9}$.

So, out of a sea of $2.16\times10^{23}$ electrons, only $6.492\times10^{9}$ goes to produce this "net electric charge of a conductor". Notice that I used 1 million Volt/meters as my electric field (a rather strong field), and still only used 1 out of every $10^{14}$ electrons in the material.

Can I get this many electrons from the face of the cube that lies opposite to the negative charged face (so we will say it is positively charged)?

Yes. The depth necessary to provide as many electrons as we need is $a\cdot N_{used}/N_{total}\approx 10^{-16}\ \text{m}$. The lattice parameter (spacing between sheets of atoms) of Fe is $\approx 10^{-10}\ \text{m}$. So all the necessary electrons can be gathered by the one outmost sheet of atoms (literally the surface sheet), and we still use only 1 in 1 million electrons from there.

In the end, if the conductor is near ideal, the electrons that depart from "positive charge density" surface areas will go to the "negative charge density areas". As it is shown above, for achievable electric field values, this process can be achieved by changes happening only in the surface regions (even only on the surface sheet of the materials), and the relative change in electron count is around 1 in a million in more extreme cases.