tolove
If we have charge in an isolated conductor, is there any way to get it off without physical contact or arcing?

Gold Member
The presence of a free charge indicates that there are extra electrons, or an electron deficit.

If the charge is negative (electrons) the photo-electric effect plus a nearby anode would lead to the emission of electrons ... you would need to know when to stop the process!

tolove
The presence of a free charge indicates that there are extra electrons, or an electron deficit.

If the charge is negative (electrons) the photo-electric effect plus a nearby anode would lead to the emission of electrons ... you would need to know when to stop the process!

Hmm, I suppose this counts. It's not quite what I was thinking, though. In the photoelectric effect, we're sending energy to the conductor, right?

I suppose in my original question, I was thinking in terms of fields. If an electric field is caused by charge, then could a charge experiencing an external field be drained away? But as you reminded me, charge is the result of mass carrying particles. I guess a chunk of mass just can't up and disappear... unless it obtains some energy and jumps off.

...it can also convert to energy? Are electrons allowed to convert into energy under any normal circumstances? Is anything?

When I look at the definitions of Charge, Magnetic Field, Photon, Electric Field, Photoelectron, etc.. there are clear and obvious differences. But conceptually, these concepts keep bleeding together in my head.

Sorry for the unclear questions, and thanks for your time!

Gold Member
Electrons can be annihilated by positrons ... but this is a particle bombardment. Other than that electrons don't just disappear except in some nuclear reactions. But charge is absolutely conserved.

If you impose a negative electric field the unbound charges of the isolated conductor will quickly be rearranged so as to cancel the external field on the interior of the conductor. There is a vast "sea of free electrons" within any conductor, and they can be rearranged within a few femtoseconds (10^-15 s). This is why charge appears to act as a fluid (currents, voltage "pressure") at the macroscopic level.

Gold Member
tolove
Electrons can be annihilated by positrons ... but this is a particle bombardment. Other than that electrons don't just disappear except in some nuclear reactions. But charge is absolutely conserved.

If you impose a negative electric field the unbound charges of the isolated conductor will quickly be rearranged so as to cancel the external field on the interior of the conductor. There is a vast "sea of free electrons" within any conductor, and they can be rearranged within a few femtoseconds (10^-15 s). This is why charge appears to act as a fluid (currents, voltage "pressure") at the macroscopic level.

When you say that 'charge' is absolutely conserved, can I replace 'charge' with 'electrons and protons are...' and keep the entirety of your meaning? I've been thinking of the subject in terms of fields and flow, but it can also be thought of in terms of individual mass caring particles?

Gold Member
No ... protons are not conserved (they can be smashed, or modified by nuclear reactions, or annihilated), nor are electrons (which can be annihilated). But the net charge is always conserved.

Thus positron (+1) and electron (-1) annihilate to generate two or three photons (zero charge). +1 + -1 = zero.

There are other conserved quantities.

tolove
Electrons can be annihilated by positrons ... but this is a particle bombardment. Other than that electrons don't just disappear except in some nuclear reactions. But charge is absolutely conserved.

If you impose a negative electric field the unbound charges of the isolated conductor will quickly be rearranged so as to cancel the external field on the interior of the conductor. There is a vast "sea of free electrons" within any conductor, and they can be rearranged within a few femtoseconds (10^-15 s). This is why charge appears to act as a fluid (currents, voltage "pressure") at the macroscopic level.

A little more thought.. I guess a better way to ask my question would be, Can charge be thought of strictly as a property of mass? Further, strictly as a property of electrons and protons? (treating the components of the proton as a single entity, I'm not quite there yet)

edit: Or positrons/antiparticles, apologies for leaving those out. haven't quite countered the anti particles yet, is there anything unique to the theory of charge that they add? Just opposites with all the same laws of physics?

dauto
There are lots of different particles that carry electric charge. Charge is not a property of mass. It is a separate property on its own right.

tolove
There are lots of different particles that carry electric charge. Charge is not a property of mass. It is a separate property on its own right.

Hm, maybe if I open my question up a little more. How many different ways can an isolated conductor lose charge?

Gold Member
Thermionic emission (heat the conductor to high temperature), photoelectric effect (bombard conductor with photons of high enough frequency), secondary emission (bombard conductor with electrons), cold field emission (place the conductor in a very high electric field in a vacuum, by putting a high p.d. between it and a nearby electrode). That's all I can think of, though bombardment by particles other than electrons will no doubt also cause electrons to be emitted.

tolove
Thermionic emission (heat the conductor to high temperature), photoelectric effect (bombard conductor with photons of high enough frequency), secondary emission (bombard conductor with electrons), cold field emission (place the conductor in a very high electric field in a vacuum, by putting a high p.d. between it and a nearby electrode). That's all I can think of, though bombardment by particles other than electrons will no doubt also cause electrons to be emitted.

Cold field emissions sounds like what I'm looking for! Could you explain it to me like I'm 5?

Gold Member
You could use a needle-like electrode,N, and place another electrode, A, of any shape very close to the needle point in a vacuum. If you put a high enough voltage between the two electrodes, with N made negative and A positive, then a very high electric field will exist near the needle point and electrons will be sucked out of the metal N and across the vacuum gap to A.

This happens, but, according to pre-quantum Physics, it shouldn't. This is because the electron needs energy (the work function before it can get out of the metal. The fact that afterwards it can pick up a bonanza of energy from the electric field won't help it to escape in the first place. If a marble is just below the top of a rounded hill, it can't by itself start rolling towards the top and then right down the other side, however far it could roll down on the other side. But Quantum Physics, which came properly on stream in the 1920s, solved the problem by predicting the phenomenon of tunnelling through the hill. The chances of this happening for the marble are utterly negligible, but for the electron coming out of the metal they are not negligible (largely because the distance an electron near the metal's surface has to travel in order to escape is very small).

tolove
You could use a needle-like electrode,N, and place another electrode, A, of any shape very close to the needle point in a vacuum. If you put a high enough voltage between the two electrodes, with N made negative and A positive, then a very high electric field will exist near the needle point and electrons will be sucked out of the metal N and across the vacuum gap to A.

This happens, but, according to pre-quantum Physics, it shouldn't. This is because the electron needs energy (the work function before it can get out of the metal. The fact that afterwards it can pick up a bonanza of energy from the electric field won't help it to escape in the first place. If a marble is just below the top of a rounded hill, it can't by itself start rolling towards the top and then right down the other side, however far it could roll down on the other side. But Quantum Physics, which came properly on stream in the 1920s, solved the problem by predicting the phenomenon of tunnelling through the hill. The chances of this happening for the marble are utterly negligible, but for the electron coming out of the metal they are not negligible (largely because the distance an electron near the metal's surface has to travel in order to escape is very small).

Alright, these feel like stupid questions, but I can't really form an answer to them on my own.

Why can't an electron pick up energy from an electric field? I guess I'm wanting to ask, what is an electric field?

Gold Member
It can - once it's outside the metal. The electric field is outside the metal. The electron has first to get out of the metal in order to experience the field. [An electric field is a region in which a charged particle experiences a force proportional to its charge. We create an electric field in the region between N and A by placing the voltage between them.]

tolove
It can - once it's outside the metal. The electric field is outside the metal. The electron has first to get out of the metal in order to experience the field. [An electric field is a region in which a charged particle experiences a force proportional to its charge. We create an electric field in the region between N and A by placing the voltage between them.]

Oh wow, alright. And inside the conductor, the electron is still experiencing that force, right? This is the reasoning behind electrostatic pressure?

If that force becomes strong enough, what happens? Will the conductor itself rip apart, or will the electron be pulled out of the conductor?

Gold Member
Your first point. No: not the force from the applied external field. The external field doesn't penetrate inside. That's why tunnelling is required.

tolove
A free electron in an electric field will experience a force. But when that electron goes into a conductor, what happens to the force? Wouldn't it still have to be acting on the electron, just spread over the conductor?

What's going with electrostatic pressure?

$P = \frac{e_0}{2}E^2$

Gold Member
"Wouldn't it still have to be acting on the electron, just spread over the conductor?"

I don't think that that is at all the right picture....

The force on the electron (charge –e) is –eE, in which E is the local electric field strength, that is the electric field strength at the point where the electron is. A simple large-scale picture, good enough for most purposes, is that E abruptly changes at the surface of the conductor, from being very high outside the surface (due, in this case to the high voltage between A and N) to being zero inside the metal. This has been discussed in Physics Forum in the past.

When it comes to the details of electron emission, the picture I've just given isn't good enough. In particular we can't just say that E inside the metal is zero. We need to take account of the array of ions which gives rise to a periodically varying potential, to a band structure of energy levels, and to the existence of a work function. We're now into a quantum-mechanical picture. I believe that cold field emission (or, as, I think, it's usually called these days, field emission) is hard to model in detail, even using the quantum mechanical picture. Too hard for me, anyway. So let's hope someone more knowledgeable takes up the thread...

Last edited:
1 person
tolove
"Wouldn't it still have to be acting on the electron, just spread over the conductor?"

I don't think that that is at all the right picture....

The force on the electron (charge –e) is –eE, in which E is the local electric field strength, that is the electric field strength at the point where the electron is. A simple large-scale picture, good enough for most purposes, is that E abruptly changes at the surface of the conductor, from being very high outside the surface (due, in this case to the high voltage between A and N) to being zero inside the metal. This has been discussed in Physics Forum in the past.

When it comes to the details of electron emission, the picture I've just given isn't good enough. In particular we can't just say that E inside the metal is zero. We need to take account of the array of ions which gives rise to a periodically varying potential, to a band structure of energy levels, and to the existence of a work function. We're now into a quantum-mechanical picture. I believe that cold field emission (or, as, I think, it's usually called these days, field emission) is hard to model in detail, even using the quantum mechanical picture. Too hard for me, anyway. So let's hope someone more knowledgeable takes up the thread...

We can leave the field emission thing alone if you'd like to. I'm still confused with the classical rules.

I'm still confused with the force.

An electron experiences F=qE in any field because it is a point charge, and a sphere experiences the same F=QE in a uniform field because it might as well be a point charge? Since the surface of a conductor is an equipotential, should I consider a conductor to be a "charge" with extra properties, as opposed to an electron being a charge with no properties (at this level at least). I'm not sure how to relate polarization to this, though.

What do I do with this concept of an electron being ripped away from the conductor via an E field? Is this concept completely meaningless in the classical sense?

Gold Member
A very small conducting sphere would experience almost the same force as an equal point charge at its centre, but a larger sphere wouldn't, if it were in a non-uniform field or if the charges induced on its surface altered the distribution of charge elsewhere, and therefore the field in which it is sitting! The behaviour of conductors in electrostatics is a mini-topic in itself. Do you have access to a textbook from which you can learn electrostatics systematically?

1 person
tolove
A very small conducting sphere would experience almost the same force as an equal point charge at its centre, but a larger sphere wouldn't, if it were in a non-uniform field or if the charges induced on its surface altered the distribution of charge elsewhere, and therefore the field in which it is sitting! The behaviour of conductors in electrostatics is a mini-topic in itself. Do you have access to a textbook from which you can learn electrostatics systematically?

I have been using Griffiths Intro to EM text as part of a course.

I'm confused about what you mean by a larger sphere. As long as the field is uniform over the entire sphere, its size shouldn't matter, right?

Help me set up an example! Two infinite charged sheets are at -d and d with surface charges +/- σ respectively. Uniform electric field E exists everywhere between -d..d, right?

Is the force experienced on both a free electron and sphere be found by F=qE, with no adjustments other than for the value of q?

If the sphere is floating in space, will it become polarized, or just accelerate?

If the sphere is somehow fixed, it will become polarized. How does the now polarized sphere affect the electric field around itself?

Gold Member
I'm confused about what you mean by a larger sphere. As long as the field is uniform over the entire sphere, its size shouldn't matter, right?

Help me set up an example! Two infinite charged sheets are at -d and d with surface charges +/- σ respectively. Uniform electric field E exists everywhere between -d..d, right?

Is the force experienced on both a free electron and sphere be found by F=qE, with no adjustments other than for the value of q?

I think you're right. [Caveat: the charges need to be locked in place on the sheets, otherwise induced charges on the sphere would cause the charges on the sheet to distribute themselves unevenly.] If the sphere were uncharged, free electrons would redistribute themselves on the sphere, forming a dipole. This will experience no net force in a uniform field. But if there is a charge q on the sphere, this will, I believe, result in a force qE, just as if we had a point charge.
If the sphere is floating in space, will it become polarized, or just accelerate?
Both, I think.
If the sphere is somehow fixed, it will become polarized. How does the now polarized sphere affect the electric field around itself?
As they approach the sphere, field lines will curve towards it and end on the sphere, meeting it normally. I don't possess Griffiths, but there must surely be a diagram of this.

Last edited:
Philosophaie
The force induced is due to the charge and the Electric Field. The Electric Field is depends on the radius and, for a dipole, the length of charge separation in the core of the planet or the star. The Q and radius of the dipole are illusive because of the rotating mass of the iron core and other metals at the center.

tolove
I think you're right. [Caveat: the charges need to be locked in place on the sheets, otherwise induced charges on the sphere would cause the charges on the sheet to distribute themselves unevenly.] If the sphere were uncharged, free electrons would redistribute themselves on the sphere, forming a dipole. This will experience no net force in a uniform field. But if there is a charge q on the sphere, this will, I believe, result in a force qE, just as if we had a point charge.

Both, I think.

As they approach the sphere, field lines will curve towards it and end on the sphere, meeting it normally. I don't possess Griffiths, but there must surely be a diagram of this.

Ok, I think I'm starting to get a picture of this. If a net neutral conducting sphere is placed in this uniform field, it will become polarized, and also remain motionless with no net force being experienced. But now that the sphere is polarized, it will alter the uniform field around it a little.

Now, in this polarized sphere... I guess I should ask, why can't the electrons just hop out of the conductor and follow the E field? If the potential difference increases between two planes, will the electron eventually hop out of the conductor (in a vacuum)?

Gold Member
Now, in this polarized sphere... I guess I should ask, why can't the electrons just hop out of the conductor and follow the E field? If the potential difference increases between two planes, will the electron eventually hop out of the conductor (in a vacuum)?

Oh no! We're back where we started - with (cold) field emission! As I tried to explain earlier, the external field doesn't penetrate the metal, and can't supply the electrons with the energy they need to overcome the work function - at least not without 'tunnelling'. I've done my limited best to explain this. Someone else needs to have a go!

tolove
Oh no! We're back where we started - with (cold) field emission! As I tried to explain earlier, the external field doesn't penetrate the metal, and can't supply the electrons with the energy they need to overcome the work function - at least not without 'tunnelling'. I've done my limited best to explain this. Someone else needs to have a go!

Hahah, thank you very much. You've helped a lot!

The electric field doesn't penetrate the sphere, but it does cause the sphere to polarize. Isn't there some kind of force associated with this polarization? Aren't the electrons trying to go somewhere?.. or.. it's entire atoms trying to go somewhere? If it's entire atoms trying to move, then the question of an electron leaving the conductor in the presence of an electric field is meaningless.

Gold Member
The 'free' electrons in the metal can move about inside the metal. The atoms (or, strictly ions, because the atoms have lost one or two electrons each to the 'sea' of free electrons) can't move; they're bonded into the lattice.

Now I have to make an apology... I said that the external field doesn't penetrate the metal. This is not really true, though it seems as if it is. I'll explain. Imagine you put a metal sphere (or whatever shape you like) into an electric field. The field does penetrate and will exert forces on the free electrons, causing them to redistribute. They will redistribute in such a way that when the equilibrium distribution is reached (in a split second after the sphere is put in the field) there is no resultant field inside the metal (not on a large scale, anyway, though on the level of atoms there are electric fields). How do we know that at equilibrium there can be no resultant field? It's because, if there were, the electrons would still be experiencing forces and redistributing!

It's quite easy to show that no resultant field inside the metal implies no large-scale regions of charge inside the metal (though obviously there are local charges - on the electrons and protons). The charge resides on the surface.