Electric field inside a closed conductor.

[Shreyansh]

I recently studied Gauss' theorum according to which there is no electric field inside a closed conductor. But an isolated chage emits field lines in all possible directions. So why is it that there are no electric lines from the surface of the closed conductor inside it?

 

[Orodruin]

Assuming that the conductor does carry a net charge, the charge will arrange itself at the surface of the conductor. This is due to the fact that the divergence of the electric field is proportional to the charge density and the electrical current is proportional to the electric field. Therefore, the charge density inside the conductor decreases exponentially with time (in a perfect conductor, this happens instantaneously).

 

[Shreyansh]

Assuming that the conductor does carry a net charge, the charge will arrange itself at the surface of the conductor. This is due to the fact that the divergence of the electric field is proportional to the charge density and the electrical current is proportional to the electric field. Therefore, the charge density inside the conductor decreases exponentially with time (in a perfect conductor, this happens instantaneously).

That is alright that the charge gets arranged on the surface of the conductor. But what I wnated to ask is why the charge on the surface emits electric field only outside and not inside like I have shown. For a symmetric body the opposite lines would cancel each other but then what for an irregular shape?

 

[Orodruin]

But what I wnated to ask is why the charge on the surface emits electric field only outside and not inside like I have shown.

Every small surface charge does give a contribution to the electric field inside the conductor as well. However, it is going to be arranged on the surface in such a way that the contributions from all of the surface charges cancel out, producing a net electric field of zero inside the conductor.

 

[Delta2]

Because in a more intuitive explanation, the electric field inside the conductor is canceled out one way or the other. For example in a solid metal sphere, the electric field in the center of the sphere, from a charge dq at point A of the surface of the sphere, is canceled out by the electric field of charge dq at the diametrical point A'. With similar reasoning( which might be abit more complicated i ve to say ) you can prove that the electric field in any point inside the sphere is also zero.

EDIT oops, i am a slow typist lol...

 

[Delta2]

For conductors of irregular shape if we suppose that the electric field is not zero inside, then from Ohm's law there will be current density $\vec{J}=\sigma\vec{E}\neq 0$ but also $\vec{J}=\rho\vec{v}$ where $\rho$ the charge density and $\vec{v}$ the velocity of the charge. So it has to be afterall $\rho\neq 0$ but we know with reasoning of post #2 that the charge density is very soon zero everywhere inside the conductor.

 

[Orodruin]

but also $\vec{J}=\rho\vec{v}$ where $\rho$ the charge density and $\vec{v}$ the velocity of the charge

This is not true, you can have a current density without a net charge. The only thing necessary is that negative and positive charges move differently. The general idea is instead to insert $\vec J = \sigma \vec E$ into the continuity equation, which tells you that
$$\frac{\partial \rho}{\partial t} = - \nabla\cdot \vec J$$
since there is no net charge production. Inserting Ohm's law into this gives us
$$\frac{\partial \rho}{\partial t} = - \sigma \nabla\cdot \vec E = -\sigma \frac\rho{\varepsilon_0}$$
where we have assumed $\sigma$ to be constant and Gauss's law $\varepsilon_0 \nabla\cdot \vec E = \rho$. If you want a static situation where the charge density does not change with time, then $\rho = 0$ (if not it decays exponentially with the conductivity divided by the permittivity as the decay constant).

 

[Delta2]

This is not true, you can have a current density without a net charge. The only thing necessary is that negative and positive charges move differently

I ll have to agree with you in the general case, however in most conductors the current carriers are the free electrons, we don't have current carriers of positive charge that is free protons or free nucleus or free positrons (!).

 

[Orodruin]

I ll have to agree with you in the general case, however in most conductors the current carriers are the free electrons, we don't have current carriers of positive charge that is free protons or free nucleus or free positrons (!).

Yes, but $\rho$ is not the charge density of free electrons, it is the charge density. Therefore $\vec J$ is still not equal to $\rho \vec v$. The charge density that has to equal to zero is the total charge density in the conductor.