# What do you mean when we say there cant be E.F 'inside' a conductor?

A metal is held by E.F...its an arrangement of electrons and kernels cause of which the metal sustains as a solid.

Since Columbian forces are holding the metal together, there has to be an E.F inside the metal, or the fields of electrons and the kernels as holding each other to form the solid...its cause of their mutual field, a metal exists.

Born2bwire
Gold Member
In a metal, or any conductor, there are no gaps between the valence energy levels, these are levels that bind an electron to an atom or to a covalent bond between atoms that connect the lattice, and the conduction energy levels, levels where the electron is free to move about the bulk material. We say there are no gaps but this is technically incorrect, the energy levels are discretized and Pauli's Exclusion principle prevents two electrons from occupying the same state (though we can have multiple states at the same energy level and such). However, the gaps are so small that the thermal energy from normal environments is larger than the gaps. So the thermal energy allows the electrons to go from the valence bands to the conduction bands without any external input and putter about as if the energy levels were continuous.

So in a metal, the electrons are free to move about, leaving behind positively charged ions (or we can talk about them as being holes but the rules for dealing with holes are a bit different than you would expect). In a normal environment, the Coulombic forces between these ions and the electrons sees to it that no local region obtains a net charge. Because if a small area gets a net negative charge, there is another area in the bulk with a positive charge and eventually the forces pull the electrons back making everything more or less neutral on an overall scale. However, if you apply an electric field, the electrons will move in the direction of the field due to the Lorentz force. However, the electric field from a negative charge points out in the radial direction. The field from a negative charge will oppose the applied electric field. Eventually, as the field pushes the electrons, they build up on the surfaces of the bulk because they cannot flow out of the material. They build up a net negative charge (and on the opposite sides a positive charge as those atoms give up their valence electrons to the applied field). The electric field between the electrons and ions opposes the applied field.

As long as there is a net electric field, the charges will move about in reaction but eventually they will create their own electric field that cancels out the applied field. Once the applied field is cancelled out, everything becomes static as any net movement of charge creates a net field that moves charges back until the field is gone and no forces act on the charges. So it is simply a matter of a system seeking a minimum energy level. The electrons are not impeded from moving up into the conduction bands and moving about the bulk and so a conductor will always redistribute the charges in a manner that cancels out an applied electric field.

Coulombic forces do not hold the metal together. The atoms of the metal bond with eachother via covalent bonds. In an effort to make a more stable outer shell (by either filling up the empty valence states with electrons or removing the extra valence electrons to empty the shell), the atoms create a hybrid orbital with neighboring atom(s). This hybrid orbital shares the needed electrons between the two of them. Coulombic forces are more at work with an ionic bond. This is where the atoms only need to strip or gain a few electrons, usually 1-3, to be more stable. Those with extra electrons only hold onto them weakly due to charge screening. This is where the inner electron clouds diminish the effective positive charge from the nucleus that is seen by the outer valence electrons. So in this case the extra electrons are stripped off and caught by other atoms who then become positive. Then, the ionic bond more closely resembles a coulombic attraction as the ions do not share their electrons like in covalent bonds.

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jtbell
Mentor
Since Columbian forces are holding the metal together, there has to be an E.F inside the metal,

On a microscopic level, there are certainly electric fields inside the metal. On a macroscopic level, these microscopic fields average out to zero in an electrostatic situation so that there is no net "long distance" force on conduction electrons, and no net motion of conduction electrons, i.e. no current.

This is thinking in terms of a classical model of conductivity, not a quantum-mechanical one, of course.

Born2bwire said:
In a normal environment, the Coulombic forces between these ions and the electrons sees to it that no local region obtains a net charge.

Equal charge distribution you mean.

Eventually, as the field pushes the electrons, they build up on the surfaces of the bulk because they cannot flow out of the material.

Yeah...exactly, so there are pressurised electrons at one end; since there's a force application on them, it manifests that there's a field which's causing the electrons to get squeezed, and cause the electrons are in the metal, the field is of course also in the metal (:tongue2:)

but eventually they will create their own electric field that cancels out the applied field

A field can be said as 0 if there will be no force application on a test charge, in this case, there is a force application, that's why the electrons shift to the other side of the conductor.

Sorry I'm bad at chemistry so............but as far are I remember, this was so............or was it the packing structure....anyway.....don't remember.

the atoms create a hybrid orbital with neighboring atom(s).

Then what's the difference between metallic an covalent bond?

Born2bwire
Gold Member
Equal charge distribution you mean.
More or less, the charges are always in a state of movement, due to, at the very least, thermal motion. So the charge distribution more less is equal but there is always momentary perturbations as thermal motion or other effects cause charge to bunch up in one area but then Coulombic forces corrects this. On a macroscopic level, the metal is neutral and has uniform charge distribution.

A field can be said as 0 if there will be no force application on a test charge, in this case, there is a force application, that's why the electrons shift to the other side of the conductor.
At steady state, the charge distribution creates its own field that cancels out the applied field. There is no forces at this point. However, the charge distribution is maintained by the fact that the applied field is still being applied. Just like in the equilibrium case where Coulombic forces ensure uniform charge distribution, whenever a net amount of charge deviates, then the secondary field no longers cancels the applied field creating a net field in the conductor. Once this net field occurs, this field applies a force which then redistributes the charges correctly. But again, the movement of charges is mainly due to thermal motion and on a macroscopic level this doesn't have much bearing.

Then what's the difference between metallic an covalent bond?
A covalent bond is a sharing of an electron or electrons between two atoms to form a molecule. When you have a lattice of molecules, the molecules are not attracted/repelled by each other by covalent bonds but rather Van Der Waal forces. But a metallic bond, if I recall correctly, is like a universal covalent bond. Where a covalent bond shares the electrons between two atoms, a metallic bond shares the valence electrons universally amongst all metal atoms. So the metal lattice is held together by a metallic bond and not Van Der Waal forces like with molecules. There can be lattices held by ionic bonds, like with sodium chloride, or covalent bonds, like silicon.

doesn't current flow in a conductor because there is an electric field????
please if any one can clear this doubt.

Born2bwire
Gold Member
doesn't current flow in a conductor because there is an electric field????
please if any one can clear this doubt.

Only for the short time that it takes for the charges to configure themselves in such a manner that cancels out the electric field. If you have an AC voltage applied, the currents only travel on the surface of a perfect conductor and on a realistic metal the currents are still for all purposes confined to the surface of the conductor.

Born2bwire said:
At steady state, the charge distribution creates its own field that cancels out the applied field.

It cannot be stated that if there's not motion, there's no field...suppose I'm holding a 1c charge in front of a 10000 c stationary charge, i.e the 1c charge is in a steady state, so it cannot be that the 1 c charge cancels the effect of the 10000 C charge and that if put a test charge in between them, it wont move.

There is no forces at this point.

You basically mean to say in the above 10000 C charge scenario, there's no force on the test charge.

We're gonna talk about the after earth scenario later...after resolving this i.e.

And aaaa...lets forget chemistry for a while, cause I think that does not happen...I still remember the reason why metals are mailable and ductile and that was cause they were held together by a sea of electrons and so the change in geometry did not matter.

Caesar_Rahil said:
doesn't current flow in a conductor because there is an electric field????
please if any one can clear this doubt.

Current flows in a conductor to stabilise the disbalanced field in the conductor, i.e as long as the current flows, there's not field.

I think I'm getting it.

It should be remembered that there's no boundary holding those electrons...the thing that's holding them are the positive ion's E.F...that means the stationary state of the electrons is cause the fields get canceled.

Now I think I'm about to generate more doubts with this (i.e dynamics in a metal under the influence of a field).

So what if the metal has a net charge?

If a test charge is placed inside a charged metal, it will experience no force cause all of the positive ions will pull/repel it..............This one too solved.

Now what if a charged metal is put in a field?...the metal will experience a force and start to move, since the stationary components of the metal are the positive ions, the force acts on the positive ions cause of the field.

Since the nucleus are inside the metal, the field is inside the metal. :tongue2:

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jtbell
Mentor
dE_logics said:
i.e as long as the current flows, there's not field.

No, the current flows precisely because there is an electric field inside the conductor. The electric field exerts the force on the conduction electrons which causes them to move.

The statement that there is no (macroscopic average) electric field inside a conductor applies only under electrostatic conditions: when the conduction electrons are stationary (except for random thermal motion). A current-carrying wire is not an electrostatic situation. The conduction electrons move because of an electric field inside the wire, which is associated with the potential difference between the ends of the wire.

The statement that there is no (macroscopic average) electric field inside a conductor applies only under electrostatic conditions: when the conduction electrons are stationary (except for random thermal motion). A current-carrying wire is not an electrostatic situation.

that solves my doubt.
thank you very much.
also can it be proven that electrostatic field very near the surface of a charged conductor is always perpendicular to it. I tried using Gauss' Law but of no use

It's good that you are thinking this through dE logics and it seems that the feedback you are getting is giving you a greater understanding.I,however, am having a little trouble following your reasoning because of your references to the word "kernel".What exactly does this word mean in the context you are using it?Excuse my ignorance but I just googled and I am still none the wiser.

jtbell said:
The statement that there is no (macroscopic average) electric field inside a conductor applies only under electrostatic conditions: when the conduction electrons are stationary (except for random thermal motion). A current-carrying wire is not an electrostatic situation. The conduction electrons move because of an electric field inside the wire, which is associated with the potential difference between the ends of the wire.

Ok...I get it, thanks for telling!

Kernel -

I don't know...but by kernels I mean the positive ions of the metal...my book referred it as 'kernel'...I too searched for it after your notification...thanks for telling.

Ok...in a metal influenced by a field, yes, there will be no field at the places where the electrons reside (i.e at certain places cause of the disturbance caused by the field) but, at other places there should be a field since the electrons moved to the other side, it means there exists a field inside a metal at other places and if a negative test charge is put at that place, it will move to one side, as with the electrons.

jtbell
Mentor
by kernels I mean the positive ions of the metal...my book referred it as 'kernel'

Is your book in German, by any chance? German "Kern" = English "nucleus".

Is your book in German, by any chance? German "Kern" = English "nucleus".

...no actually it was my XII grade book.

I have the word in my notes I made at that time -

"Sea electron theory-
According to this theory-
1.a metal has a no. Of free electrons, as metals loose there valance electrons as they have very less affinity to them(the ionisation energy is very less)
2.These electrons move freely in the metals. The remaining nucleus and the shell below it gains a +ve charge and are named kernel.
3.The kernels now occupy fixed positions called lattice sites, and surrounding them is a gas of electrons, the electrons can be also refered as sea, so this theory is also called electron sea theory. The electrons are said to be de localised.
4.Now an electron is attracted by many kernels each of being of different charge, this electrostatic force binds the atoms of the metal together, this force is called cohesive force, the bond is called metallic bond."

Old document...used MS office at that time. Stupid choice.

A few ambiguities still remaining......

Born2bwire
Gold Member
Which ones?

dE said:
Now what if a charged metal is put in a field?...the metal will experience a force and start to move, since the stationary components of the metal are the positive ions, the force acts on the positive ions cause of the field.

Since the nucleus are inside the metal, the field is inside the metal.

Then -

dE said:
Ok...in a metal influenced by a field, yes, there will be no field at the places where the electrons reside (i.e at certain places cause of the disturbance caused by the field) but, at other places there should be a field since the electrons moved to the other side, it means there exists a field inside a metal at other places and if a negative test charge is put at that place, it will move to one side, as with the electrons.

jtbell
Mentor
2.These electrons move freely in the metals. The remaining nucleus and the shell below it gains a +ve charge and are named kernel.

Hmm, OK. Maybe it's British terminology. U.S. textbooks (at least the ones I've used) also never use "+ve" and "-ve", but instead spell the words out: "positive" and "negative."

Born2bwire
Gold Member
The electric field of a charge is not a local effect, it goes on to infinity. So the field is not cancelled at the charge's location, it is cancelled throughout the entire volume of the conductor. The metal, as you put it, is a sea of electrons. All of the electrons do not move in response to the electric field. If there is a net field anywhere inside the conductor, then some form of charges will move in response to it until they setup their own secondary field that cancels the originally applied field. The whole process provides feedback for itself.

jtbell said:
Hmm, OK. Maybe it's British terminology. U.S. textbooks (at least the ones I've used) also never use "+ve" and "-ve", but instead spell the words out: "positive" and "negative."

No those were my notes, I used those +ve and -ve :rofl:

So the field is not cancelled at the charge's location, it is cancelled throughout the entire volume of the conductor.

It can be that at a place a field does not exist, and at other places it does...take 2 equal charges, put a test particle in the middle, there's no field there, but move the charge towards one of those charges, and you'll see an acceleration.

Same is happening here...I mean that's why the electrons moved to the other side...also you cannot say that cause of those electrons the field does not exist at that place...cause for the field to not exist despite the fact that there're charges surrounding you, all the charges should be arranged in a specific way...................so if you do not experience a field at a place with many charges surrounding you, it means that you will be able to feel the field at other places cause the effect of some charges will decrease, while for the others it will increase (or remain constant) as you move to your new location, so the particle has to experience a field at other places in such a scenario.

Same is the case here...if you do not experience a field at the places the electrons reside, it means you will be able to feel the field at other places, cause at those places, the effect of the electrons will reduce and the effect of the interfering field/positive ions (of the metal) will increase.

All of the electrons do not move in response to the electric field.

Can you pls explain how?...anyway, I did ask this once here...didn't get any answers.

What I think, this should not happen...moreover even if this happened, it will not help unless the distribution of these residual electrons is uneven.

If it IS uneven...then also the last problem still exists.......a charged metal in a field.
If there is a net field anywhere inside the conductor, then some form of charges will move in response to it until they setup their own secondary field that cancels the originally applied field.

There is no charge left to move!...all charges have been influenced by the interfering field (this has to do with the previous lines...residual electrons)

Born2bwire
Gold Member
The metal is not charged. It is, more or less, neutral. The electric field is only going to shift charges around the metal but unless you actually do some nifty stuff with grounding or friction you are not going to change the net charge on the metal.

All of the electrons will feel the electric field and move to some degree yes, but what I mean is that do not envision that a field is going to strip every atom in the metal of all of it's valence electrons and send them all to the far boundary of the metal. Look, if wikipedia is to be believed, copper is a face-centered lattice with a lattice constant of 0.3610 nm. Let's take a 1 m square cube of this crap. That means that 1 m cube has 12.126e28 atoms. Each atom has 1 valence electron, so we have a charge density of 3.406e9 C/m^3. Wikipedia actually says that it is 13.6e9 C/m^3 but hey, one order of magnitude accuracy ain't bad for back of the envelope. So let's say we have a 1 m thick infinite slab of copper and we place all the electrons on the back and assume we have the corresponding ions on the front, then we have something like a parallel plate capacitor with charge of 27.2e9 C/m^2. This would give us an electric field of 3.072e21 V/m. So when you can generate something on the order of 1 zettavolts per meter, then we can start to worry about running out of electrons.

If we assume a perfect conductor, which you can see above is a very very good approximation for most metals, then we assume there is infinite amount of charge available to move around in the conductor.

You need to realize that this is really just an optimization scenario. The charges will move in such a way that they will setup an opposing electric field. There is no restriction to moving the charges because any interference from phonon interactions is negligible and the conduction band in a metal is, for all intents and purposes, continuous. We do not worry about "running out of charges" because even a modest metal like copper has enough charge to support mindblowing levels of electric fields should it be completely polarized. Under these circumstances, there is nothing preventing the system from arranging itself in the configuration of minimum energy, which is when the total electric field is zero inside the conductor. This is no different then the copper atoms settling into their face centered lattice at room-temperature in order to minimize the bulk's energy levels. Or how a handful of marbles will seek the minimum configuration when dumped into a bowl.

rcgldr
Homework Helper
Could a field be strong enough to ionize a conductor (repel or attact a significant number of the free electrons away from the conductor)?

Could a field be strong enough to ionize a conductor (repel or attact a significant number of the free electrons away from the conductor)?

I think so but a google moment is advised.I believe the phenomenom is called field emission.

Each atom has 1 valence electron, so we have a charge density of 3.406e9 C/m^3. Wikipedia actually says that it is 13.6e9 C/m^3 but hey, one order of magnitude accuracy ain't bad for back of the envelope.

That difference is quiet a lot...I mean, what we're doing here is adding a 0 to 3.406e9 C/m3

Did you mean 13.406e9 C/m3?

Anyway...it does not matter here...just for example's sake.

then we have something like a parallel plate capacitor with charge of 27.2e9 C/m^2.

Howcome?...I mean how can it make a parallel plate capacitor? the charge sustains on the metal itself cause a field, while in a capacitor, its cause of the high breakdown voltage of the dielectric material.

This situation is different.

Could a field be strong enough to ionize a conductor (repel or attact a significant number of the free electrons away from the conductor)?

So when you can generate something on the order of 1 zettavolts per meter, then we can start to worry about running out of electrons.

Under an influence of a field, the electron distribution will be -

According to you -

[PLAIN]http://img147.imageshack.us/img147/1690/quantumelectrondistribu.png [Broken]

According to me -

[PLAIN]http://img115.imageshack.us/img115/1690/quantumelectrondistribu.png [Broken]

According to you my principle is wrong cause if this was so, the charge density made would have been extremely high.

But using the same principle (that at one place here are tons of electrons, while at the other there are none) we can also model a lower charge density...when only a few positive ions will be exposed, i.e at the place where the electrons reside, the density of electrons will not be that high as compared to the natural charge density of the metal.

For example -
In this case, using the same principle, the charge gained is low -

[PLAIN]http://img115.imageshack.us/img115/1690/quantumelectrondistribu.png [Broken]

While with this is high -

[PLAIN]http://img152.imageshack.us/img152/1690/quantumelectrondistribu.png [Broken]

Infact, how can some electrons not move (when the field falls), while the others start moving...the field doesn't know how to discriminate among electrons, its gonna influence all of the electrons, so all the electrons will move together, leaving a few places in the metal completely electron-less...while at other places, even electron density.

We can say that E.F does not exist here -

http://img126.imageshack.us/img126/1690/quantumelectrondistribu.png [Broken]

Cause the electrons are not moving.

But at this place -

http://img2.imageshack.us/img2/1690/quantumelectrondistribu.png [Broken]

(i.e places where the electrons are not avilable)

The E.F will exist...now howcome?

First of all, the electrons moved cause a field does exist there, if there's no field the electrons will return back (I said this before)..............now it can happen that the cloud the electrons towards the right neutralise the field in the red sections highlighted above, as a result, no net filed in that place...BUT this can only be stated for a for a few places, and that place is the area where tons of electrons just accumulated.

Cause if we move away from area, the influence of the interfering field will increase,and that of the electrons will decrease.

We can't say that the positive ions will cancel out the interfering field's effect cause if we introduce a test charge, it wont actually matter cause it will be attracted by all the positive ions, canceling everything out...so the only thing left to influence the test charge is the interfering field.

Only under these circumstances -

http://img80.imageshack.us/img80/1690/quantumelectrondistribu.png [Broken]

i.e uneven distribution of electrons, it is possible to state that there's no field, cause the electrons will cancel things out.

But to prove the above model, we gotta see why this model -

[PLAIN]http://img152.imageshack.us/img152/1690/quantumelectrondistribu.png [Broken]

Is wrong, or how this model -

[PLAIN]http://img147.imageshack.us/img147/1690/quantumelectrondistribu.png [Broken]

is correct...then we gotta think.

Could a field be strong enough to ionize a conductor (repel or attact a significant number of the free electrons away from the conductor)?

You mean initiating breakdown?

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It took around 6hrs to upload those images.

It took around 6hrs to upload those images.

That doesn't mean no one's gonna respond!

Born2bwire
Gold Member
Look, I don't know how the charges distribute themselves exactly. I don't need to know or care. There would be a number of factors in the actual distribution like temperature, coulombic forces, the geometry of the bulk, the applied field, the model of the electron's moving through the lattice, etc.

If we have a static field that is applied perpendicular to a large slab of conductor, then an easy way to cancel it out would be to strip all the electrons of the atoms on one boundary of the slab and move them to the opposite boundary. Then you would have a net sheet of positive charge on one end and sheet of negative charge on the other just like a parallel plate capacitor. But this is not a stable configuration since the extra electrons on the one side would exert a larger coulombic force on the valence electrons in the neighboring atoms and the same on the other side due to the positive ions. Realistically, I would imagine the charge distribution to be something like a linear gradient. Given the very low level of the applied field, the gradient will be very very very small. If we have a perfect conductor, then it would just be ions on the boundary on one side and the necessary number of electrons on the other side.

Let's play our game again. If we move only one electron from each atom on the boundary of an infinite 1m thick slab of copper, then the charge density will be about 1.2292 C/m^2 along the boundaries of the slab. This means that the effective electric field would then be 1.388e11 V/m. So we can see that for copper, we would only need to move a very small amount of charge from the boundary to counter an applied field. The gradient of the charge will be very very small since the perturbation of moving the electrons from one end to the next will be very small. If we applied a field of 1 V/m, then we would need a charge density of epsilon, 8.8541878176e-12 C/m^2. So we woud have to move a very very small number of electrons to achieve this change in density along the boundaries. This will cause the electrons on the lattice to shift a little, but by a very small amount.

So, in an ideal conductor, we assume that all the electrons will move towards a boundary and all the ions to the opposite boundary. Classic example would be the charge distribution of a spherical shell conductor encasing a point charge. Since we only need to move a very small number of electrons to counter any realistic field in a good conductor, then this is still a very good approximation. Heck, if you have a sheet of copper that is 0.1 mm thick our little back of the envelope calculation will still give us an electric field of 100 megavolts per meter.

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QUOTE]Look, I don't know how the charges distribute themselves exactly. I don't need to know or care. There would be a number of factors in the actual distribution like temperature, coulombic forces, the geometry of the bulk, the applied field, the model of the electron's moving through the lattice, etc.[/QUOTE]

There should be just one answer, despite the geometry, temperature...........or anything else.

Its independent of virtually every factor.

Name one, which it should depend on...as for what you said -

Temperature - It will only increase the number of free electrons.

Coulombic forces - That's the subject here

geometry of the bulk - You mean E.F falling at a point rather than the whole conductor...really think about it, it doesn't matter...................the question that arises is the same.
But this is not a stable configuration since the extra electrons on the one side would exert a larger coulombic force on the valence electrons in the neighboring atoms and the same on the other side due to the positive ions.

I also gave the solution to this -

For example -
In this case, using the same principle, the charge gained is low -

http://img115.imageshack.us/img115/1690/quantumelectrondistribu.png [Broken]

While with this is high -

http://img152.imageshack.us/img152/1690/quantumelectrondistribu.png [Broken]

Such sort of distribution will occur until things stabilise -
The repulsive force among the electrons on one side + the attractive force on the electrons cause of the exposed kernel = Repulsion of the electrons cause of the field.

And yes, what I mean by 'no electrons' is that there are no valence electrons of course, if you stripped all electrons from the atom, it'll make plasma...I guess.

Realistically, I would imagine the charge distribution to be something like a linear gradient

Yeah...if this actually happens, then there would have been no field inside the conductor in this case................but apparently, there seems no reason why will this happen.

I also modeled this scenario with the uneven distribution....but there really seems no way why will that happen.

If we have a perfect conductor, then it would just be ions on the boundary on one side and the necessary number of electrons on the other side.

It appears we have a common point.

I assumed a superconductor in all the above cases..........but even if its not a superconductor...how will it matter?

Ok...lets start a new thread about this...how exactly will be the electron distribution under application of the field since we're confusing the main question to what actually is being discussed.

So by now............this is the dependency tree of the whole set of questions -

1) Electrons distribution under the influence of the field (the new question) (https://www.physicsforums.com/showthread.php?p=2193813#post2193813).
1.1)How can we state that that there's no E.F inside the conductor?(This thread itself)
1.1.1)Why potential difference 'inside' the capacitor is different than what's out side it?(https://www.physicsforums.com/showthread.php?p=2179710#post2179710)

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Born2bwire
Gold Member
Temperature affects the distribution because it is a measure of the mean kinetic energy of the particles. This is a conductor, all valence electrons are free because the band structure is continuous in light of the thermal energies so it will not affect the number of free electrons (within reason). A high temperature causes variations in the lattice of the bulk and in the movement of the electrons. This will cause things to move around, requiring equilibrium to be microscopically re-established continuously. The coulombic forces will dictate the minimum energy state of the system. The electrons are free particles in a conductor but the ions are still fixed. There is a difference if we have a lattice that has atoms that give up 1, 2, or 3 electrons because then the ions can be stronger for a given lattice. In addition, the lattice and spacings dictate charge density and so on. The geometry affects the boundary conditions that the system can orient itself. It can also cause artificially high fields due to what is called field enhancement. Areas around sharp points can have very high charge densities that create large fields. The model of the lattice and such will dictate how the ions move (if possible) and how we can strip the atoms. Finally, if we have an electric field that is varying in space then it will affect the charge distributions. And because the extra shift to make everything happy again is probably very very small, the area between the surfaces will more or less remain neutral.

There is no one way that you can describe the charge distribution but we can make generalizations and solve for the distribution for a perfect conductor for certain geometries. In the end, for a perfect conductor we assume that we will induce charges on the surface of the conductor in accordance with the applicable electrostatic laws. This makes it simple for symmetric closed surfaces like a spherical shell enclosing a point source but it would be difficult without a computer to solve for a rectangular shell enclosing a point source.

Again, this is a simple energy minimization problem. The applied electric fields will move the charges freely around the conductor. They will naturally find the state that will cancel out any fields on the interior because they always move in a manner that counteracts the applied field. They will only reach equilibrium once there is no longer any field inside the conductor.

The problem with your diagrams is that they are too extreme. I tried to show you that even realistic conductors have more than enough electrons just on the surface to counteract any realistic applied fields. This is why we make the assumption that they are perfect conductors with an infinite number of electrons. The difference this assumption makes is minimal. So we do not have a situation where you will strip every single electron and thrust it to one side of the conductor. In reality, you will strip a few electrons per unit area on the surface and move them from one side to the next. This may cause an imbalance because if we perfectly cancel out just the applied field, we still have a charge distribution that is not what we would like at equilibrium without any applied fields. So the bulk needs to do a shift, but this shift, due to the very few charges moved per volume, will be very slight and it will probably be a very very very small gradient. It wouldn't surprise me if the shift in the charge densities would be overpowered by the changes in time due to thermal variations.

A perfect conductor is not the same as a superconductor, that point needs to be made certain.

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