Can a vacuum become a conductor? OR The physics of electron flow

by Stephanie Chasteen on June 11, 2009

From Nessman on Flickr

From Nessman on Flickr

Here was an interesting discussion on a science teacher’s listserv, which came down to the question — can a vacuum become a conductor?  What is it that we really need in order for charge to flow between two points?  What is the physics of electron flow?  The physics teacher in question wrote:

The Paul Hewitt book has a test question that reads:  In order for charge to flow from one place to another, there must be a
A. Potential difference between the two places.
B. Conductor, such as a wire, connecting the two places.
C. both A and B.
D. none of these.

The book’s answer is C (both A and B).  I’m wondering why A isn’t the answer. In the case of the van de Graff or lightening you create a potential difference between two locations (say me and the van de Graff) and the charge will eventually flow … I guess the air acts as the conductor from van de Graff to me? But is there a situation where there is enough of a potential difference between to places and charge doesn’t flow? Would the van de Graff not work in a vacuum?

A veteran physics teacher (Al Sefl, who always knows more physics than moi) responded:

The key to the Hewlett question is that it asks about charge flow.  Current cannot flow through an insulator until the point of breakdown is reached.  If you have a sphere X with a positive charge and a sphere Y with a negative charge there will be no flow between them until a conductor connects them.  Before that conductor is there, there will be lines of electrostatic force but no flow.  If the potential were great enough the air would break down to become a conductor and you would have flow.  So, C is the correct answer, you must have a potential difference AND a conductor to conduct the charges.

Yes, a Van de Graaff generator will work in a vacuum.  High voltage capacitors used in broadcast transmitters and radar units are vacuum capacitors where the charge is stored between two plates surrounded by an evacuated space.  The electrostatic lines of force do not need air to exist.

That’s all fine and dandy, but most of the people on the listserv didn’t understand that second paragraph (and neither did I), not knowing enough about broadcast transmitters and such.  A teacher asked, do you mean that the Van de Graaff will store charge in a vacuum, but not throw off sparks?  After all, what would the charge be flowing through if the Van de Graaff is in a vacuum?  There’s no air to ionize (or “break down”).

Al responded with a clarification:

A vacuum may also be a conductor.  The old cathode ray tube of years past sends a beam of electrons from a gun through a high vacuum to impact the phosphor screen.  So, when the potential becomes high enough current will flow through a vacuum.  In a CRT it does get an assist from thermionic emission in the gun.

The miniature lightning bolts we see from a Van de Graaff are really the paths of broken down insulator air that has become conductive and ionized.  You would not see that in a vacuum.  If you put a sharp point on the negative terminal then the charge concentration will push off electrons that will travel to the positive terminal.  The vacuum will become a conductor.

So, C is still the correct answer.  If charges FLOW they must do so through a conductor.  ANYTHING will become a conductor if the electrostatic charge exceeds its dielectric.  If electrons are flowing through something it *is* a conductor.

Perhaps where the Hewlett presentation becomes unclear is the definition of what a conductor is.  Most of us immediately think of a piece of copper wire *but* it can be anything if the potential is high enough.

So, a vacuum can become a conductor, even though there’s nothing to ionize (and thus you won’t see the glow from the electrons as they travel through a vacuum, as you do in the air).  But by definition, if charge is flowing, it’s flowing through a conductor!  Paul Doherty explained that when there is an electric field that is large enough (it has to be very very large), then it will produce electron/positron pairs in the vacuum.  Those electrons and positrons are what flow to conduct electric current.

On a side note — the charged particles given off by the Sun aren’t visible as they pass through the vacuum of space… but they are visible when they hit our magnetosphere as the aurora borealis.

And another teacher offered a clarifying comment:

I was taught to get over the idea of being protected by an insulator. We were told that an insulator is a bad conductor. My trade teacher felt that insulator was a weak word and preferred to talk about everything being a conductor, just good conductors (copper) or bad conductors (glass).

So, the discussion got interestingly esoteric here.  The original questioner then posited:

If any space can be considered a conductor given a high enough potential difference, then I think the answer to Hewett’s question should be we just need a potential difference to get a flow of charge.  After all, he didn’t explicitly state that we need to have charged particles, which I think would be necessary to have a flow of charge.  So why state that an omnipresent conductor is necessary?

Also, if a vacuum has charged particles moving through it, is it still a vacuum?

Paul Doherty emphasized that the correct answer to the question is still “C.”  You can have a potential difference and no flow of charge, because the voltage may not be low enough to create its own conductor out of the insulator between the two places.  With enough potential difference an insulator is turned into a conductor, but you STILL need both a potential difference and a conductor for charge to flow.

{ 21 comments }

Chris Goedde June 11, 2009 at 12:54 pm

Consider the photo-electric effect, where you shine light on a metal and (if the light has a short enough wavelength), electrons will be ejected from the metal. In that case, you get charge flow between two points without a potential difference between those points (in fact, the charge can flow against a potential difference) and without a conductor. So I would argue that the correct answer is really D.

Assaf June 11, 2009 at 6:18 pm

All this lawyer-speak makes me dizzy :). Every person reading the original question will interpret ‘conductor’ slightly differently, and the answer will depend on the interpretation.

Chris, I’m not an expert on light-matter interaction but I’d bet there is a “potential” difference involved as well. The electrons are ejected from the metal due to the atom-photon-lattice interaction. The photons themselves constitute an electromagnetic field, which is basically a potential difference (albeit in a time-dependent framework there is also a vector potential involved).

sciencegeekgirl June 11, 2009 at 8:31 pm

Chris, you make a good point, but I still disagree with you — the answer is C.

In that case, you get charge flow between two points without a potential difference between those points

When you hit an electron with a photon, you have given it kinetic energy (joules). That *is* a voltage difference from where it was before. You’ve lifted it out of its potential well, and you’ve also given it a different energy/voltage than the other electrons just below the surface that were not hit by the photon. So, in a way, there is a potential difference between the electrons on the surface (hit by the light) and those not on the surface (which are still in their potential wells). So the electrons are accelerated away from the surface.

But, and this is where your question becomes really interesting, they’re being accelerated away from the surface in all directions. Is this a current? Lets use an analogy, where electrons are golf balls. The classical definition of a current is that you have a bunch of golf balls on a slab, you tilt the slab, and they all roll downhill. If you put a line across the path of the golf balls, you can measure how many go across that line per unit time = current.

Now, instead, you have a pile of golf balls and you come along with a golf club (the light) and start hitting the balls in all directions. Is this a current? You can draw a line around the pile of balls and there is a net flow of golf balls away from the original pile, but it is hard to talk about a total current flow. By following the formal definition of current, you could say it’s a current, but it’s certainly a very different beast than the standard “golf balls falling along a slope”.

As to whether there’s a conductor, I think that question got kind of beat to death in the original discussion in the blog post. As the previous commenter suggested, it is somewhat lawyerly semantics.

Assaf June 11, 2009 at 9:41 pm

Oddly enough when you tear the electron out of its orbit you’re moving it against the nucleus’s potential (i.e. from low to high potential) – that is, opposite the direction you’d envision current in (in a typical circuit electrons flow from high to low potentials). It’s “anti current”! 🙂

sciencegeekgirl June 11, 2009 at 10:05 pm

This conversation just goes to show that all definitions have exceptions!

Stich June 12, 2009 at 3:59 pm

As I understand it, electron flow through conductors is more of a bumping of electrons from atom to atom in the conductor medium. In the case of say copper wiring, they bump very nicely and readily, so it is a great conductor. In air, there aren’t many atoms near each other, so they have to bump under a lot more amperage (pressure).

A vacuum by definition is devoid of atoms, so no bumping could occur. Instead, a free flow of electrons would have to propagate through the vacuum, and as such it doesn’t make a lot of sense to me to call those electrons “the conductor.” It would more be electron flow without a conductive medium, which would be a very uncommon case of electrical flow and require special conditions. Do electrons really ever like to hang out without some sort of atom to orbit?

Of course you could eject matter and make electrons conduct along the matter stream. That is how a new military plasma weapon system works. A laser creates a straight path of photons, and then high power plasma is injected into the laser path, creating a directed energy flow similar to lightning.

Ben June 12, 2009 at 10:57 pm

It is D. Vacuum conduction usually occurs by two mechanisms, “glow discharge” or ballistic carrier transport. Most of the time, vacuum conduction is glow discharge, which is really the ionization of trace gasses in the imperfect vacuum. This occurs when the electric field derivative gets high enough. It’s kind of a funny process in that it occurs most easily at intermediate vacuum levels, typically around 1 Torr. In that range, even a typical 9 volt battery can arc between its electrodes. This is a major design problem in building electronics for operation in vacuo, since it allows components to short out in strange and unexpected ways. In some sense, the trace gases become conductive, but they only remain conductive as long as you keep dumping a sufficient amount of energy into it (called “simmering” in glow discharge lamp jargon). It’s called “glow discharge” because it’s usually accompanied by a glowing arc, so it can be quite useful for making light sources.

Ballistic carrier transport is rarer in nature, but it does happen. In that case, the electron (usually) is ejected out of the cathode and hurries on its merry way over to the anode. There is no conducting medium, just an electron and a lot of nothingness for it to not run into. This mainly occurs in vacuum electronic devices like photomultipliers (the photoelectric effect) or cathode ray tube electron guns (thermionic emission). Ballistic transport can also occur in solids, but that’s a very weird and unusual thing that you usually only see on microscopic scales. I should note that this won’t happen without a potential difference. This ballistic transport mechanism can give rise to charge transport without or against an imposed potential, in which case what you’re actually doing is converting the energy that produced the emission into electrical power.

There’s also another kind of vacuum breakdown that can theoretically occur at very high field strengths, but I don’t know if it’s ever been demonstrated in the lab. Basically, if the field gets intense enough, it can dissociate particle-antiparticle pairs generated in the vacuum.

It’s certainly true that in most practical cases, you need a conductor, but if you make the field strength high enough, those charges are going to find a way to get to each other come hell or high water.

The potential difference is required in order to get charge transport without putting any energy into the system, but if you do put energy into the system, it’s certainly possible to make a current flow without having imposed an electrical potential. Now, you could argue that there’s necessarily a potential difference created when you make a current flow, so there is a chicken and egg issue – you can’t separate charges without a potential difference, and you can’t have an (electrical) potential difference without separated charges.

Captain Skellett June 16, 2009 at 8:37 am

Very interesting post! I can’t believe that a vacuum can be a conductor and still a vacuum, but I’m not going to argue!!! Physics, I have found, often does this sort of thing – states something simple, and then has to explain it in a very complex way once you start really thinking about it. It’s a problem with most of the sciences, I think, and probably all things in general.

Squark June 16, 2009 at 12:57 pm

Firstly, the electric field strength at which electron-positron pair creation begins is of the order of magnitude of

http://www34.wolframalpha.com/input/?i=(electron+mass)^2+*+(speed+of+light)^3+%2F+(electron+charge+*+Planck%27s+constant)

2.1E17 V/m OUCH!!!

Secondly, even if the electric field is not that strong, a capacitor disconnected from a voltage source will discharge (albeit very slowly). This is because thermal fluctuations will occasionally give an electron enough energy to cross the potential barrier and reach the other electrode.

Thirdly, this will happen even at zero temperature because of quantum tunneling!

It should be fun to compute the current induced by the above two effects.

Squark June 16, 2009 at 2:07 pm

Sorry, I was inaccurate. At first, I thought dimensional considerations only are sufficient to given an order of magnitude estimate of the critical electric field strength. However, dimensional considerations are not enough since we can introduce an arbitrary function of the fine structure constant.
Now, all tree level Feynman diagrams for electron-positron emission in a macroscopic external field depend on alpha and E only through the product alpha E. Hence, my previous estimate was in fact an estimate of alpha E. E itself is

http://www34.wolframalpha.com/input/?i=(electron+mass)^2+*+(speed+of+light)^3+%2F+(electron+charge+*+Planck%27s+constant+*+fine+structure+constant)

2.9E19 V/m

Tom June 16, 2009 at 3:53 pm

Paul Doherty’s justification at the end is wrong. The presence of a potential difference is a necessary but insufficient condition for current flow, but the question asks what is necessary, not what is sufficient.

That materials, and even a vacuum, may be considered conductors under certain conditions makes this a trick question, IMO, especially so if this an introductory book/course, as it is drawing on advanced concepts to justify that answer; one must consider the definition of “conductor” presented in the book or course. So one must be prepared to accept a “wrong” answer as being right — it’s not like this is an unknown phenomenon in physics.

Stich June 17, 2009 at 2:35 pm

Well put, Tom. That’s what I was getting at.

sciencegeekgirl June 18, 2009 at 7:23 pm

I can’t believe the depth of conversation that was generated by this post!

I don’t think there can be a final arbitration of the answer, since some of it depends upon semantics (is a vacuum still a vacuum if it has electrons in it?), to name just one persnickety bit.

For myself, I still believe the answer is C, if we allow ourselves to define a vacuum as a conductor.

Charles July 8, 2010 at 5:02 am

This is an interesting experiment of vacuum tube. I believe it can be used as an conductor. But not a replacement, Just IMHO.

rebeca December 3, 2010 at 2:20 am

I really want to become more intrested in science and get straight A’s in it. I really need some help!!!!!!!

Glenn travis December 7, 2010 at 5:25 am

Just a thought.
Get on the wayback machine, say before all this solid state stuff
Does anyone remember electric circuits or even computers that used Vacuum Tubes? Diodes, Triodes and even moreodes?
High voltage, red hot heater/cathodes and anode plates? Those “cooked off” electrons zoomed right across that vacuum.
So yes I guess in a way, a vacuum is a conductor. Of course there are wires on either end of the tube.
Then what about lightning? It does not need a conductor, and no the air ain’t one.
Been zapped by static electricity, seen the arc, and not touching anything?
The real absolute answer is A. If a potential can be created, and it is great enough, electrons will flow.

awsm1 November 2, 2013 at 1:16 pm

vacuum is just space where it is zero density pressure etc.
So even if it has electrons its density and pressures will increase very slightly ase electrons has very very less mass compared to atomes (lets say air or something).

In middle of outer space there is zero density because the particles are attracted towards nearest planet.

awsm1 November 2, 2013 at 1:27 pm

And in reality A is the answer ! because if everything is a conducter because it is.composed of atoms.

But due to their different properties (metals nonmetals and so on) they have different critical potentials at which they start conducting a current. This is something we know as resistivity.

So no need to really mention a conductor because there is nothing that cannot pass a current.

Just you if you want to make a current flow, you dont need to rush to a store purchase some copper wire, neither rush to get air because its already there.

So A is correct option.

If object me if u can

Raju November 19, 2013 at 11:33 am

its possible to have electron flow by forced air supply? sorry….

Miguel Dimase March 17, 2015 at 3:44 am

¿Might be you meant “voltage may not be HIGH enough” instead “low enough”?

mophat francis June 8, 2015 at 5:12 pm

answer is D due to absences of air that cloud collied with electrons to coused low P.d

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