Why is electric field inside a conductor zero?

[Doc Al]

Imagine just 4 electrons in a circular disk. They'll form a square. Each will be in equilibrium. But in the vicinity of each electron the e-field will be non-zero. Shall I draw a diagram and calculate the e-field somewhere in the middle between electrons, on the surface?

What does this have to do with the field inside a conductor? In a conductor, you can redistribute as many electrons as needed to cancel any external field. (Of course, if the external field is so incredibly humongous that all available electrons within the conductor are still not enough to cancel the field--then all bets are off. )

I don't seen any point of diagramming the field of 4 electrons. I don't imagine anyone here is saying the the field from those electrons is zero everywhere.

Also: The only reason those electrons might be in equilibrium is if something is holding them in place (and thus exerting a force on them). Are they attached to the disk? Or are you picking 4 electrons on the edge of the disk? Why?

 

[Ulysees]

As shown below, E-field can be non-zero even though all charges are in equilibrium. You could do it with 4 electrons, or with 4000000000 electrons. The idea is the same, between electrons the field is non-zero.

Now the Shell Theorem of gravity likewise for electrostatics predicts that with enough charges evenly distributed on the surface the total field inside adds up to zero. But it does not add up to zero on the surface, between charges. The point is, equilibrium of charges does not imply zero electric field everywhere, only where each charge is.

The field inside can be calculated numerically for any conductor based on the relation between surface curvature and charge density. For most charged conductors, the sum will NOT be zero. Take a cube for example. All charge goes to the corners of the cube. This is predicted by the relation between curvature and charge density. Shall I draw a cube and the related 6 E-field components?

Are (the 4 electrons) attached to the disk? Or are you picking 4 electrons on the edge of the disk? Why?

Of course they are not attached, they stay on the surface due to the (electromagnetic) surface forces (the light blue arrows). This disk only has N=4 electrons. A real disk only has a finite number N of electrons. Both have fields between charges.

 

[Doc Al]

Again: What does this have to do with the field inside a conductor?

 

[Ulysees]

That's not the only issue. Someone made an incorrect statement, and I am politely correcting.

The point is, equilibrium of charges does not imply zero electric field everywhere, only where each charge is.

 

[Ulysees]

And on the burning issue of the field inside an arbitrary conductor, the answer was given too:

The field inside can be calculated numerically for any conductor based on the relation between surface curvature and charge density. For most charged conductors, the sum will NOT be zero. Take a cube for example. All charge goes to the corners of the cube. This is predicted by the relation between curvature and charge density. Shall I draw a cube and the related 6 E-field components?

 

[Doc Al]

If you think you have shown that electrostatic equilibrium does not imply zero electric field everywhere within a conductor, you would be wrong.

 

[Ulysees]

Isn't there space between electrons?

 

[Ulysees]

Hey don't forget the classical conductor is only a mathematical model, ultimately a real conductor is nothing but particles in free space.

 

[Doc Al]

As shown below, E-field can be non-zero even though all charges are in equilibrium. You could do it with 4 electrons, or with 4000000000 electrons. The idea is the same, between electrons the field is non-zero.

When we speak of conductors and electrostatic equilibrium, we are talking about all the electrons--not just those on the surface.

Now the Shell Theorem of gravity likewise for electrostatics predicts that with enough charges evenly distributed on the surface the total field inside adds up to zero. But it does not add up to zero on the surface, between charges. The point is, equilibrium of charges does not imply zero electric field everywhere, only where each charge is.

The point of the shell theorem is that the field from a uniform shell of charge is everywhere zero inside the shell.

The field inside can be calculated numerically for any conductor based on the relation between surface curvature and charge density. For most charged conductors, the sum will NOT be zero. Take a cube for example. All charge goes to the corners of the cube. This is predicted by the relation between curvature and charge density. Shall I draw a cube and the related 6 E-field components?

You seem to think that if you put a charge on a conducting cube, that all the charge would end up at the corners. Not so! (If you think I'm wrong, provide a standard reference.)

Again, the surface charges will rearrange themselves so as to make the field equal to zero everywhere inside the conductor. If this were not so, there would be current flowing inside the conductor! (Which is certainly not electrostatic equilibrium.)

 

[Ulysees]

I realize that in the classical model the conductor is populated with "continuous charge everywhere". Ie there is not such thing as "between charges". There is no such thing as a particle. In that model, all space is occupied by continuous charge. Therefore all space has to have a zero E-field, I grant you that.

Now in a real conductor, it's all tiny particles with a lot of space between them. So between particles, there is enough space where the E-field can be non-zero and is non-zero.

Now you'll be happy to see me admit that the classical model of the conductor is more practically useful. And that is why zero E-field inside is assumed in most cases. But if we give plenty of positive charge to the metal (ie remove most of the valence electrons), then the E-field inside will not be zero because there are not any valence electrons to move under the influence of the field at most places inside.

When we speak of conductors and electrostatic equilibrium, we are talking about all the electrons--not just those on the surface.

Those on the surface attract our attention because all others are canceled out by the nuclei they are attracted to. So extra electrons we add, go to the surface.

The point is, equilibrium of charges does not imply zero electric field everywhere, only where each charge is.

The point of the shell theorem is that the field ...

Where it says "The point is", I do not refer to the point of the Shell Theorem, I refer to the point that I am trying to make that between particles the E-field can be and is non-zero (and only in the spherical shell is it zero everywhere inside).

You seem to think that if you put a charge on a conducting cube, that all the charge would end up at the corners.

Alright, some will not if the amount of charge is small. But if you put so much positive charge that the lattice almost runs out of valence electrons, the charge will all go to the corners. The non-zero E-field inside will then be balanced by inter-atomic and intra-atomic electromagnetic forces.

 

[zarbanx]

okkk ulysses listen wht doc al is sayingi found a solid proof of wht i am listing below:-
1-field is ALWAYS zero inside a conductor(which includes a conducting shell) even when there is an external field and even when there is a charge inside.
2-the potential at all points is same whether there is an external electric field or non uniform distribution of charge due to a charge kept in the cavity inside the shell.

the thing abt four charges at corners is not wht i wanted to discuss in this thread.It is a conductor we r talking abt and if there is a field inside the conductor then electrons(which r present in excess in the conductor) will move to nullify the field.

 

[Ulysees]

In the cases that concern you (under external field is one case, perhaps a charged conductor should be the other case), what do you believe the field is between two atoms of copper?

EDIT: Is there movable charge between all possible pairs of copper atoms, in an object made of copper?

 

[dioib]

I found this thread looking for a pop science explanation of the matter. Well it's two years now, but I am going to share my point for what it is worth:
[STRIKE]Ulysees[/STRIKE] had a point and his/her argument is valid as long as you're off the continuum limit. Consider a 1-d conductor, circle in shape, with 4 electrons on (hence nonzero net charge), free to move around by construction (being a conductor). This is a classic conductor with just not enough charge to lend itself to Shell Theorem or other differential argument. Symmetry enforces the electrons to form an square (inside the conductor). Yet, being a conductor, neither the surface (circle) is an equipotential one¹ nor the electric field is zero inside.

But you can play all those Shell games in the continuum limit where differentials make sense.

1- At the location of each electron the common argument that renders the surface equipotential works fine, but as you move on the circle, away from location of the four electrons, the argument fails, because there is no electron to rely on.