Why don't electrons leave a negatively charged metal in air?

[Ulysees]

I'm familiar with lightning rods taking advantage of the mutual repulsion of charges to shoot off a corona discharge off the sharp end and start a thunder, but why doesn't corona discharge happen to all charged metals? What makes air such a good insulator, when it's just gases, relatively few molecules moving all over the place bouncing on each other, how can this be a good insulator?

 

[marlon]

To answer you question : think of the essential difference between a conductor and an insulator. What makes a metal a good conductor ? This property is not inherent to air !
wink wink, enjoy

marlon

 

[Ulysees]

> What makes a metal a good conductor ?

Dunno. If you were an electron in free space, would there be anything to stop you from moving? Space should be a great conductor, better than an electron-crowded metal. E/M waves move faster too.

You see what I'm saying? Air is closer to free space than it is to metal, is it not?

 

[peter0302]

I think you're confusing electric current in a conductor with the movement of a free electron through space or through air. They are not the same.

 

[Ulysees]

What about electrons jumping from one metal to another separated by space, like in an old amplifier lamp. It doesn't matter what we call current, I just want to know what's stopping electrons from leaving a metal based on first principles, quantum principles, wavefunctions and all that. That's why I'm asking you guys in this section.

From what I know there is nothing stopping an electron from jumping from a negatively charged metal to another in free space, and very little in air.

 

[Ulysees]

In other words, how do we account for the insulating properties of air from first principles.

 

[peter0302]

Air consists mainly of N2 and O2, both of which are highly stable molecules with the valence electron shell completely filled (triple bond for N2, double bond for O2). It takes a lot of energy to put an electron into the next shell, far more than would be given up when an electron leaves a negatively charged metal.

Metals by contrast do not have their valence shells filled. In fact, electrons in a metal can move fairly freely like water in an ocean because of this, jumping from atom to atom. For metals, even a small electric potential imbalance will cause current flow because the electrons, being totally free to move, will gravitate (no pun intented) toward the positive end of the imbalance. But, for air, because it requires so much energy for an electron to jump to the next shell level, this requires a huge potential difference, along the scale of lightning.

In a vacuum, electrons will only flow toward a positive charge, which is then simply governed by F = kq1q2/r^2.

 

[Ulysees]

Yes but electrons don't have to jump from air molecule to air molecule, they can just fly between molecules, right?

 

[peter0302]

No, with a lightning bolt, there is a current flowing through the air, making the air itself a conductor - i.e., jumping from atom to atom. It's because the air is ionized that a conductive path is formed. No one is certain, however, _why_ an ionized path is created in the air.

Again, remember that electrons "want" to be bound to atoms. They ideally want to be in a full valence shell, but they'd much rather attach themselves to a higher orbital spot in an atom than be free. Free electrons are usually only seen in labs or electronics like vacuum tubes or CRT televisions.

 

[Mentz114]

Yes but electrons don't have to jump from air molecule to air molecule, they can just fly between molecules, right?

Cathode rays are electrons traveling in space. But you have to make a pretty hard vacuum for them to travel a foot between emitter and target. Old vacuum tubes use a stream of electrons - note the name. The point is that at atmospheric pressure, electrons can't fly between the gas molecules very easily.

 

[Ben Niehoff]

This is a good question; one way to answer it is to use the Method of Images.

Consider an infinite conducting, charged plane (which is what any conductor looks like up close), and a point charge of the same sign, near the plane. The point charge induces an additional, opposite charge on the conducting plane, which can be modeled by considering a fictional "mirror-image" of the charge, sitting the same distance away, but on the other side of the conductor.

The repulsive force of the infinite plane is constant, but the attraction between the point charge and its image goes as 1/r^2. Therefore, no matter how strong the repulsion is between the charge and the plane, there is some distance r within which the attraction between the charge and its image is stronger. Call this distance D.

Therefore, to fully remove an electron from a conducting surface, it must be removed at least a distance D, or it will actually be attracted back to the surface! To remove the charge a distance D against this attractive force requires some energy; calculation will show that this energy is equal to the "work function" of the material in question (the same work function used to calculate the photoelectric effect, where electrons are knocked out of a metal by energetic photons).

 

[marlon]

> What makes a metal a good conductor ?

Metals are good conductors because they exhibit the property of conduction band electrons. Such electrons are not bound to one atom but can move from one atom to another in the socalled conduction band. Such bands exist thanks to :

1) putting many atoms together (you need a many particle system, single atoms do not have electronic bands)

2) the specific crystal structure, ie its symmetry, of that material, which makes up the electronic energy structure

marlon

 

[marlon]

Yes but electrons don't have to jump from air molecule to air molecule, they can just fly between molecules, right?

They can but they can also interact with air-molecules. Besides, electrons loose kinetic energy when passing through air ! This happens much less in a metal ! Why ? because electrons passing through a metal are essentially helping each other to pass through the conductor, a bit like "domino's". When you place a potential across a wire, the electric field at the end of the wire changes, this causes the electrons at the end of the wire to move. Once these electrons have moved, they change the electric field experienced by the neighbouring electrons, which as a result feel a net force causing them to move, thus again perturbing the field for their neighbouring electrons. This is repeated down the whole length of the wire, each electron perturbing the field of its neighbours causing it to move,

marlon

 

[Gokul43201]

Metals are good conductors because they exhibit the property of valence electrons. Such electrons are not bound to one atom but can move from one atom to another in the socalled valence band. Such bands exist thanks to :

1) putting many atoms together (you need a many particle system, single atoms do not have electronic bands)

2) the specific crystal structure, ie its symmetry, of that material, which makes up the electronic energy structure

marlon

I think you meant to say "conduction" instead of "valence."

 

[Jakell]

When you have a lighting rod, the tip is usually very pointy. At these sharp edges, you can build up an extremely large electric field compared to that along the length of the rod. The stronger electric field allows the air to become ionized much easier, dissipating much of the charge on the rod. If your rod is too pointy, though, the efield drops off too rapidly, and you do not get as much ionization.

Any charged piece of metal could do this, it would just have to be fairly pointy.

 

[marlon]

I think you meant to say "conduction" instead of "valence."

Opps, indeed, my mistake. Post corrected.

Thanks

marlon

 

[jostpuur]

The air is not insulator for the same reason as the solid insulators are, for sure! Solid insulators are insulators because the available momentum eigenstates form a filled ball, and the expectation value of the momentum of all particles remains zero. Such effect does not occur in gas, because there is no periodic potential on the background.

I don't think that considering the bound states of electrons on the gas molecules was helpful either. If the electrons cannot get onto the bound states of molecules, shouldn't the free electrons then just behave as a gas as if the electrons were like other molecules?

This is a good question; one way to answer it is to use the Method of Images.

Consider an infinite conducting, charged plane (which is what any conductor looks like up close), and a point charge of the same sign, near the plane. The point charge induces an additional, opposite charge on the conducting plane, which can be modeled by considering a fictional "mirror-image" of the charge, sitting the same distance away, but on the other side of the conductor.

The repulsive force of the infinite plane is constant, but the attraction between the point charge and its image goes as 1/r^2. Therefore, no matter how strong the repulsion is between the charge and the plane, there is some distance r within which the attraction between the charge and its image is stronger. Call this distance D.

Therefore, to fully remove an electron from a conducting surface, it must be removed at least a distance D, or it will actually be attracted back to the surface! To remove the charge a distance D against this attractive force requires some energy; calculation will show that this energy is equal to the "work function" of the material in question (the same work function used to calculate the photoelectric effect, where electrons are knocked out of a metal by energetic photons).

A great post! I think this answers the original question, and is also related to one other problem I was thinking about some time earlier. Doesn't this also explain why electrons don't form gas in normal circumstances? Free electrons always get attracted to the solids for this reason? I might guess that this effect is present also with insulators, because you don't need macroscopic currents for this.

 

[Ulysees]

But it's not obvious why the redistribution of charge is equivalent to a mirror image of equal and opposite charge appearing on the opposite side of the mirror, where the mirror is the metal surface. What is the proof?

 

[ZapperZ]

But it's not obvious why the redistribution of charge is equivalent to a mirror image of equal and opposite charge appearing on the opposite side of the mirror, where the mirror is the metal surface. What is the proof?

Have you ever looked at the Method of Images before? It is a standard technique in solving for the fields in electrostatic problems, i.e. it is in textbooks on classical E&M.

Zz.

 

[Ulysees]

Thanks, I'm sure it is a standard method, but does that satisfy your curiosity? It is so much more fun to look for the fundamentals of things, otherwise we'd just say:

1. metal, good conductor
2. air, bad conductor
3. therefore current does not flow
4. end of story

Isn't the root of things in terms of basic, basic elements, so much more satisfactory and profound understanding?

 

[ZapperZ]

Thanks, I'm sure it is a standard method, but does that satisfy your curiosity? It is so much more fun to look for the fundamentals of things, otherwise we'd just say:

1. metal, good conductor
2. air, bad conductor
3. therefore current does not flow
4. end of story

Isn't the root of things in terms of basic, basic elements, so much more satisfactory and profound understanding?

But shouldn't you also follow your own advice and investigate what is already well-known and see if the answers to your questions are already contained in there. I mean, this "curiosity" works both ways, don't you think? Aren't you curious as to why such Method of Images actually works?

The fields in E&M are nothing more than solution to Laplace equation. If you know anything about differential equation, you then will realize that if you find the solution to Laplace equation using whatever means that satisfies the boundary condition, then there is a Uniqueness theorem that says that that solution is the ONLY solution possible, other than those that differ by an additive constant. That is why one can replace all that surface charge with the image charge!

I can point to you several experimental work that not only detect, but also make use of the image charge state. So I don't just "accept" things.

Zz.

 

[peter0302]

I don't think that considering the bound states of electrons on the gas molecules was helpful either. If the electrons cannot get onto the bound states of molecules, shouldn't the free electrons then just behave as a gas as if the electrons were like other molecules?

In a metal the electrons roam freely throughout the metal. Current begins to flow because electrons are being "stolen" from one end of the conductor and so electrons from the other end move toward the newly created charge imbalance. This doesn't happen easily in insulators because in those substances the electrons are much more tightly bound to the individual moleucles and therefore it requires a much higher potential to liberate them. You don't need quantum mechanics to understand this.

 

[Ulysees]

> But shouldn't you also follow your own advice and investigate what is already well-known

I don't remember giving such advise to anyone but surely we don't need to go into the unknown, just go to 1940's technology for amplification. What were those lamps called? Electrons jumped off a metal and into thin air, inside the lamp. This should be well understood by now!

> Aren't you curious as to why such Method of Images actually works?

You want my honest answer? I'm not so curious because it's only a mathematical trick that saves you time, it's not fundamental. I doubt it would be any useful in a quantum simulation of electrons, nuclei and E-M fields. I'd love to see a full simulation at the atomic level:

Imagine electrons interacting electromagnetically with each other, and with thermally vibrating nuclei.

And then seeing an electron wavepacket leave the metal, and then collide with an air molecule.

Another electron misses all air molecules and reaches the other end, the anode.

That's what I call fundamental understanding.

-----------

Does anybody know software for this?

 

[ZapperZ]

> But shouldn't you also follow your own advice and investigate what is already well-known

I don't remember giving such advise to anyone but surely we don't need to go into the unknown, just go to 1940's technology for amplification. What were those lamps called? Electrons jumped off a metal and into thin air, inside the lamp. This should be well understood by now!

Er... classical E&M even predates 1940! This is not going into the "unknown". It is a very well known

> Aren't you curious as to why such Method of Images actually works?

You want my honest answer? I'm not so curious because it's only a mathematical trick that saves you time, it's not fundamental. I doubt it would be any useful in a quantum simulation of electrons, nuclei and E-M fields. I'd love to see a full simulation at the atomic level:

Wait a second. You were the one who said just now that we don't need to go into the unknown and 1940's technology. And now you're invoking quantum mechanics?

Imagine electrons interacting electromagnetically with each other, and with thermally vibrating nuclei.

And then seeing an electron wavepacket leave the metal, and then collide with an air molecule.

Another electron misses all air molecules and reaches the other end, the anode.

That's what I call fundamental understanding.

-----------

Does anybody know software for this?

Image charge isn't just a mathematical trick. If it is, then everything is a mathematical trick. When you simulate charges in a particle accelerator, there are "wake fields" in the metal walls as the beam passes by. These ARE the image charges!

Note that I'm not tackling the issue you have with the OP in this thread. I'm tackling your disbelief in the image charge method. This is classical E&M that is well-established, whenever you accept it or not. In fact, even in quantum mechanics, the "work function" that we know and love contains the image charge potential! I work at a particle accelerator where we try to produce photoelectrons "cold", meaning with very little kinetic energy to minimize its emittance. But because it has almost no KE upon emission, we need to quickly accelerate it away from the photocathode (a metal) before the image charge forces pull it back into the cathode.

Zz.

 

[peter0302]

Um, he is the OP. And many people have answered his questions and he's not listening.

 

[ZapperZ]

I'm familiar with lightning rods taking advantage of the mutual repulsion of charges to shoot off a corona discharge off the sharp end and start a thunder, but why doesn't corona discharge happen to all charged metals? What makes air such a good insulator, when it's just gases, relatively few molecules moving all over the place bouncing on each other, how can this be a good insulator?

Since this is something that I am also working on (breakdown effects), I'll tackle this finally since I'm already involved in this thread anyway.

The process of breakdown is not trivial. It involves a number of factors with a number of different models, depending on what scenario is involved.

1. Presence of field-enhance regions. These sharp, pointy metals in an electric field can be enhanced to a large value;

2. The emission of field-enhance current. This is the Fowler-Nordheim description of field-enhancement emission due to tunneling of electrons out of the metal;

3. The ionization of the neutral gases in the vicinity of the pointy object by the field-emitted electrons.

4. The presence of positive ions enhances the gradient further and causes more electrons to go further into the neutral gases. At some point a "cascade" of ions and electrons are produced and causes a catastrophic "discharge".

Zz.

 

[ZapperZ]

Um, he is the OP. And many people have answered his questions and he's not listening.

I know. However, I was not tackling the OP (original post)... till now.

Zz.

 

[Ulysees]

Sorry Zapper, you're not reading my posts.

Read again the last one please.

It says we do NOT need to go into the unknown, just go into 1940's (instead). How you thought 1940's technology would be presented as an example of... the unknown, is beyond me.

I guess you're reading lots as a mentor and therefore reading fast. No problem, just don't accuse others of not reading your posts when in fact it's you that doesn't read read theirs.

Anyway.

 

[peter0302]

Oh, hehe, I thought OP meant "original poster." Still learning the acronyms. :)

[Edit]
Oh and for the record, Ulysses, I was the one accusing you of not reading posts.

 

[ZapperZ]

Sorry Zapper, you're not reading my posts.

Read again the last one please.

It says we do NOT need to go into the unknown, just go into 1940's (instead). How you thought 1940's technology would be presented as an example of... the unknown, is beyond me.

Er... then you have misread what I wrote.

I reread what you wrote, and I understood you just fine. You wanted us to not go into the "unknown", but instead, use the 1940's technology, whatever that is. Isn't that what you said?

And I counter the fact that classical E&M that I have been mentioning even PREDATES the 1940's technology. I would even say that such technology USES classical E&M. That is why I questioned why you have such distastes towards method of images, when it is simply nothing more than a way to solve the Laplace equation of E&M. Would you consider the Green's function method also nothing more than a mathematical trick?

I guess you're reading lots as a mentor and therefore reading fast. No problem, just don't accuse others of not reading your posts when in fact it's you that doesn't read read theirs.

Anyway.

If you're trying to be cute, it isn't working.

Zz.

 

[ZapperZ]

Oh, hehe, I thought OP meant "original poster." Still learning the acronyms. :)

Unfortunately, it can mean "original poster" as well. It's one of those short-handed notation on here that can be ambiguous as far as which one it is referring to. It isn't the best short-hand notation.

Zz.

 

[Ulysees]

> You wanted us to not go into the "unknown", but instead, use the 1940's technology, whatever that is. Isn't that what you said?

Now you say it right, yes. But look what you wrote before:

> You were the one who said just now that we don't need to go into the unknown and 1940's technology.

Why the "and" between "unknown" and "1940's"? No to the unknown, yes to 1940's. Why the "and"?

-----------Anyway, thanks for the bit about discharge, it's what I was looking for.

By the way, I never said your image method is wrong. It's just not intuitive, therefore not a satisfactory visualisation to me when I'm trying to imagine "wavy little things" interacting.

------------------

Any simulation software? Or a description how to make my own?

 

[ZapperZ]

By the way, I never said your image method is wrong. It's just not intuitive, therefore not a satisfactory visualisation to me when I'm trying to imagine "wavy little things" interacting.

You should know by now, especially in a "Quantum Physics" forum, that your "intuition" is not something that is dependable. I can show you many more simple "classical" situation where your intuition can be very wrong. That's why we have mathematics and that's why we describe physics with mathematics.

And I am quite sure that anyone who has formally studied classical E&M, there's nothing not intuitive about method of images, especially after one sees how well it does. If you have a problem with that method, then you have a problem with all of E&M because of what I said about the solution to Laplace equation. To me, that is even less intuitive.

Zz.

 

[Ulysees]

Also, if there's only 5 extra electrons in the charged metal (as in an initial simulation), I think the image charge would be a very inaccurate model. Yes?

 

[Ulysees]

And what if we heat a sphere with 5 extra electrons. I think they will leave eventually, and form air ions like in the discharge.