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Let's say that we have an uncharged spherical conductor. If we hold a charge +q near the conductor, then a charge distribution is induced. The positive charge q pulls negative charges over to the near side (on the surface of the conductor) and repel plus charges to the far side. This results in the electric field being cancelled within the conductor.

However, how can positive charges move to the far side? How can they move at all? I thought that positive metal ions don't move but only the electrons move (and since it's a conductor, I am assuming it consists of positively metal ions and free electrons).

If the positive metal ions would be able to move to the far side and the electrons move to the near side, then in the middle of the conductor you would have nothing/empty vacuum/no atoms, so how would that be possible?

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Regarding the same picture and charge distribution, I was also wondering if it is correct that a solid metal sphere can never have a volume charge, but only a surface charge, since the electric field must be zero inside a conductor always right?

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how can positive charges move to the far side? How can they move at all? I thought that positive metal ions don't move but only the electrons move (and since it's a conductor, I am assuming it consists of positively metal ions and free electrons).

The positively charged atomic nuclei do NOT move from one side of the sphere to the other. What happens is that some of the negatively charged electrons leave that side, leaving a net positive charge behind.

In electrolyte solutions and plasmas, positive ions do move, but in solids, the nuclei are essentially fixed in their position relative to their neighbors.

Regarding the same picture and charge distribution, I was also wondering if it is correct that a solid metal sphere can never have a volume charge, but only a surface charge, since the electric field must be zero inside a conductor always right?

That is essentially correct. A minor correction is that if there is a current in a conductor, there is an electric field $\vec{E}$ which is proportional to the current density $\vec{J}$ times the volume electrical resistivity of the conductor $\rho$.

$$\rho\vec{J}=\vec{E}$$

This is known as the "microscopic" version of Ohm's law.

Since the resistivity of most metals is quite small, the electric field will be small unless the current density is very large. The small electric field due a flow of current in a conductor is often ignored in practice.

However, even though there may be a small electric field within the body of a conductor in which there is an electric current, it is still true that charges do not accumulate in the interior of a conductor with uniform current density and uniform resistivity.

This can be shown by taking the divergence of the microscopic Ohm's law. If $\rho$ and $\vec{J}$ are uniform, then $\nabla \cdot \rho \vec{J} = 0$. Therefore $\nabla \cdot \vec{E} = 0$. Therefore by Gauss's Law, the charge density in that area of uniform current density $\vec{J}$ and uniform resistivity $\rho$ must also be $0$.

[Note that in this answer, I used $\rho$ as the symbol for resistivity. In the context of Gauss's Law, the symbol $\rho$ is often used for a different purpose, i.e. as the symbol for charge density. Please don't be confused by this.]

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  • $\begingroup$ Thanks for the clear answer! Regarding the movement of charges, is it then incorrect of Griffifths to say that the 'positive charges are repelled to the far side' since only the negative charges move? Also, will the positive nuclei be able to move by a large external electric field? And in your formula for the microscopic Ohms law, this E is the electric field inside the conductor? What is its direction? $\endgroup$
    – Stallmp
    Mar 15 at 7:32
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    $\begingroup$ Yes, it is not physically accurate to say that the positive charges move to the far side. It is the electrons that move in a solid. Nuclei do migrate very slowly in a hot solid, although still at a relatively tiny rate. If they move too fast, you no longer have a solid, but a liquid or semi-liquid like molten glass. If there were an electric field strong enough to cause nuclei to migrate significantly, the shape of the object would deform as well. *** The Electric field in an isotropic conductor (i.e. one with no preferred direction of conduction) is in the direction of the current flow. $\endgroup$ Mar 15 at 12:17
  • $\begingroup$ Thank you very much once again! So basically, there is only an electric field inside a conductor in the case of a current carrying conductor? $\endgroup$
    – Stallmp
    Mar 15 at 12:51
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    $\begingroup$ Yes, unless you want to get down to the atomic level. Since charge is not continuous, but composed of discrete electrons and protons, at the atomic level, there are electric fields. But these fields average out when we consider the scales of the conductors we buy at a hardware store. If you were doing research on graphene nano-tubes, you might have to take into consideration the electric fields in your graphene molecules. But then again, at that scale, there are currents that average out as well. Electrons are not still, even though at macroscopic scales, we might treat them as "static". $\endgroup$ Mar 15 at 13:02
  • $\begingroup$ Thank you very much! $\endgroup$
    – Stallmp
    Mar 15 at 14:26

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