Electric Induction: Exploring Free Electrons & Internal Fields

[Omega0]

Hi,

I have been told that you take say a metal board and another board (both endless and paralell)
leads to the following effect:

The electrons are more or less freel in motion, they move to the surface until we reach an equilibrium. The electrons are assembled at the surface. Nice.

Next lesson: Internally there is no electrical field. It holds:

div(rho)=0 because rho is 0.

Okay, rho is zero when the electrons are missing at the other side of the board...
This counts only for the volume but not the surface!?

Do I get something trivial wrong or is there a deeper reason?

Thanks!

 

[WannabeNewton]

Given an arbitrary conductor in electrostatic equilibrium, $\rho = 0$ in the material of the conductor because $\vec{E} = 0$ in the material of the conductor and from Gauss's law we then have $\vec{\nabla}\cdot \vec{E} = 0 = 4\pi \rho$. But of course there is charge density on the surface of the conductor as you yourself noted; the surface charge density $\sigma$ is given by the jump condition in the electric field when passing from the interior of the conductor to the exterior: $\hat{n}\cdot (\vec{E}_2 - \vec{E}_1) = 4\pi \sigma$ or equivalently $\hat{n}\cdot (\vec{\nabla}\varphi_2 - \vec{\nabla}\varphi_1) = -4\pi \sigma$ where $\vec{E}_2$ is the exterior electric field, $\vec{E}_1$ is the interior electric field, and $\hat{n}$ is the unit normal to the surface of the conductor. This is assuming there are no dielectric materials present (the modification to the jump condition in the presence of dielectrics is very straightforward anyways). This is all in Gaussian units by the way.

 

[Omega0]

Thanks, but this is exactly what I read in every book. You say "surface charge density is given by the jump condition in the electric field when passing from the interior of the conductor to the exterior" but why?
Take the not influenced object as electrically neutral. div(rho)=0 naturally holds. Apply a potential, you learn
that the jump conditions hold. Jumps are something where you mathematically should be alarmed.
But okay, it's fine if you accept that the distribution of the electrons is suddenly on the surface.

My question is: Why?
What about the other side?

There is another reason which is not explained in detail. Could you give me a curve describing the electron density in this conductur?

 

[WannabeNewton]

Why what? Why all the charges are on the surface of the conductor after reaching electrostatic equilibrium?

 

[sardar]

Hello,

It sounds like you are discussing the concept of electric induction. This is a phenomenon where a changing magnetic field can induce an electric current in a conductor. In this case, the metal boards you mentioned act as conductors and the electrons within them are free to move around.

When we bring the two boards close to each other, the changing magnetic field causes the electrons in one board to move towards the surface facing the other board. This creates a separation of charges, with one side of the board having an excess of electrons and the other side having a deficit. This is known as an induced electric field, and it is responsible for the movement of electrons within the conductor.

As for your question about the internal electric field, it is true that there is no electric field within the conductor itself. This is because the charges within the conductor are in equilibrium, and any excess charges will move to the surface to neutralize the electric field. However, there may still be an electric field at the surface of the conductor, as you mentioned. This is because the surface is where the charges are able to move and redistribute themselves to maintain equilibrium.

I hope this helps clarify the concept of electric induction and the role of free electrons and internal fields. Keep exploring and asking questions!