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Trilobyte
06-03-05, 07:17 AM
Simple question but I can't find a satisfactory answer:

Why does negative static charge (or positive?) involve the build up of electrons on the surface of a material instead of throughout the volume?
And therefore why is it larger surface area means higher capacitance?

Billy T
06-03-05, 11:26 AM
Why does negative static charge (or positive?) involve the build up of electrons on the surface of a material instead of throughout the volume?

Electons mutually repel. They move to be as far from each other as possible. On a sphere, well separated from every thingelse, this results in uniform surfac charge density.

Electrons are also attracted to positive charge, which in most cases is the absence of some electons. Thus if two plates are close to each other, one with excess and one with difficiency of electrons, the electons will be mainly on the side of their plate near the plate with positive charge.

And therefore why is it larger surface area means higher capacitance? Consider the isolated sphere again. if there are a million excess electrons on its suface each is repeled by others more if the sphere is small than if it is large. that is the potential of each will be higher on the smaller sphere. I.e the surface will be at a higher voltage on the smaller sphere. V = Q/ C thus for V to be larger on small sphere with the same Q, C must be smalller.

cato
06-03-05, 12:29 PM
nice explaination.

Trilobyte
06-03-05, 02:00 PM
If electrons accumulate on the surface they will be at the lowest charge density but there will be a potential difference between the surface and the centre that will force some electrons back towards the centre to neutralise it. If not why not?

Also would there not be some central point that is the same distance from the surface at any angle and so be able to accommodate some electrons as well as the surface (like zero g at the centre of the earth)?

kevinalm
06-03-05, 02:39 PM
This is a little counter intuitive. Given a uniform spherical shell of charge the force on a small test charge anywhere inside the shell is zero. Works for gravity/mass as well. Newton was the first to prove this mathematically (calculus).

Trilobyte
06-03-05, 02:53 PM
If there is a solid metal sphere with a small spherical void or hollow in the centre of it (like a bubble) would more charge gather on the outer surface than the inner surface (since the outer surface has a larger area) or would no charge at all gather on the inner surface?

Billy T
06-03-05, 03:49 PM
If there is a solid metal sphere with a small spherical void or hollow in the centre of it (like a bubble) would more charge gather on the outer surface than the inner surface (since the outer surface has a larger area) or would no charge at all gather on the inner surface?All charge is on the outside. Read about Van De Graph high voltage generators (Charge is carried inside a sphere on a belt, stripped off the belt by fine metal points that are in electrical contact with the sphere. This charge migrates to the sphere surface, despite fact it may already be charged to million volts.)

It is easy to see, without calculus, that any inverse square law force (graviational of electrostatic) has no force /field inside a spherical shell. Take any point inside the spherical shell and imagine that it is the apex of two coaxial cones of the same internal angle. If the part of the shell closest to the chosen apex point is r distant and the shell on the more distant side is R away then each small area of the shell will effect the apex point as the inverse square of R or r. That is each square unit of area (e.g a square mm of the shell) of the closer shell will have (R/r)^2 more effect than the symetricly opposite equally large unit of area that is R distant. Exactly counter balancing this is the fact that the total area of the section of the shell R distant and intercepted by the larger cone is (R/r)^2 larger. That is each piece of the nearer shell has force exactly cancelled by the larger, symetrically opposite, area more distant at R from the apex point.

Specifically at no point in the interior of a shell of uniform thickness (for gravity case) or uniform charge density (for electrostatic case) is there any net force.

Very easy to see in the 2D case: Draw a circle. Chose ANY point inside. Draw two line thru this point. Note that the larger "far" shell area exactly compenstates for the stronger effect of the smaller "near" shell contained within your lines, if and only if the force is inverse square.

Trilobyte
06-03-05, 04:45 PM
If no force is acting on electrons near the centre (or anywhere off centre) why are they forced to the outer surface ("migrate")?

Billy T
06-03-05, 06:41 PM
If no force is acting on electrons near the centre (or anywhere off centre) why are they forced to the outer surface ("migrate")? A single charge might just drift in zero force field inside the shell until it hits the shell, but two or more would rapidly mutually repel. Both (or all) expand away from each other , rapidly reaching the shell.

Trilobyte
06-03-05, 08:53 PM
However if this is the case it is a dynamic equilibrium. Electrons that repel from the negatively charged surface will be forced back in towards the centre in which no force is acting on them.

MacM
06-04-05, 12:34 AM
However if this is the case it is a dynamic equilibrium. Electrons that repel from the negatively charged surface will be forced back in towards the centre in which no force is acting on them.

You will saturate at some point and no further electrons will migrate to the sphere. But if you crank up the speed (raise the rate electrons are being stripped off of the belt (hence increasing the voltage), then more will begin to flow again.

PS: I have stood on a wooden chair and used a VG to charge my body to 250,000 volts. Your hair and clothes move around like you were standing in the wind. It is wierd. I know, I know. I'm wierd. :D

Trilobyte
06-04-05, 07:11 AM
However if this is the case it is a dynamic equilibrium. Electrons that repel from the negatively charged surface will be forced back in towards the centre in which no force is acting on them.

If this is right then a large number of electrons that do not have enough kinetic energy from mutual repulsion will end up accumulating throughout the volume and not on the surface, so the idea is that the density of electrons tends to decrease (steadily/exponentially?) as the distance from the surface of the material increases .

If the surface is "fully charged"/"saturated" and you add more electrons to the volume via an inner surface presumably no more electrons can be held on the surface and so they will start to radiate - ignoring constant radiation of electrons for simplicity.(yes/no?)

Billy T
06-04-05, 03:21 PM
If this is right then a large number of electrons that do not have enough kinetic energy from mutual repulsion will end up accumulating throughout the volume and not on the surface, so the idea is that the density of electrons tends to decrease (steadily/exponentially?) as the distance from the surface of the material increases .

If the surface is "fully charged"/"saturated" and you add more electrons to the volume via an inner surface presumably no more electrons can be held on the surface and so they will start to radiate - ignoring constant radiation of electrons for simplicity.(yes/no?)Nonsense. You do not seem to understand that there is no electric field inside a spherical metal shell.
Except for any preexisitng momentum the electron may have, one electron inside a spherical shell will not move. It will not be repelled by even a very large number of electrons on the skin of the spherical shell. If it has zero momentum, it will not drift (neglecting Browian motion drift and thermal convection air current etc.) and it certainly will not be acted upon by shell electrons as they add up to symetrically cancell or make zero electric field inside the shell. In basic physics classes this "zero field in side a conducting shell" is known as Gauss's law. For the general case proof of any shape closed metal surface, I think calculus is required.
I gave you a non calculus math proof of its truth for the special case of a spherical shell, as I assume you do not understand calculus. That proof showed that for ANY point inside the spherical shell there is exactly zero total force from the electrons on both sides of the shell inside small (say of 1 degree internal angle) co-axial cones with their mutual apex at the chosen point. Recall that the number of electrons inside the cone, on the shell R away from the common apex point, is proportiona to R^2 but each of them has only (r/R)^2 of the effect that an electron inside the cone, on the shell r away from the apex has. That is, the greater number of electrons R away produce exactly the same repulsion force at the apex point that the fewer number electrons r away from the chosen point produce so the net repulsion is zero.

This is just another way to say (actually to prove) that the electric field inside the shell is every where zero. (Electric field is by definition the electric force on a unit test charge placed at the point the field is to be measured.)

The change in potential energy between points a&b, Vab, is the work done as this test charge is moved between a and b. Thus everywhere inside the shell the electric potential is the same as no work is done when moving a test charge in a region free of electric field. Thus, if the shell is at a million volts, so is the interior, but it is free of field. Outside the spherical shell the field decreases as the square of the distance for the center of the sphere and the potential inversely with this distance. This is all very simple freshman physics, if not high-school physics. (Your ignorance of these facts and proofs is why I assumed that you do not know much about calculus. If I can teach some of them to you I am happy and certainly do not mean any offense - at some time years ago I did not know either also - that is no sin.)

As far as concept of "Saturation" is concerned, the electric field strength is maximium just outside the shell (and zero inside as proven). Dry air will only support a certain electric field before some atom of the air ionize. (Moist air supports even less.) The electron from that ionized atom, now an ion, will be replled by the external field (I am assuming we charged the shell up negativtive as that is typical for a Van de Graph generator.) and the positive ion will strike the shell, become neutral atom again. It is if one electron has been removed from the total on the shell. This "bleading away" of shell charge is called corona discharge and the rate increses as the shell voltage is increased. In this sense there is a saturation voltage where the rate that the V de G belt delivers electrons to the shell is equal to the rate that the coronal discharge is removing them. (If you want higher voltage, put the whole machine in vacuum or at least a gas hard to ionize, like Helium.)

You may not have personal experience with Van de Graph machines, but if you look at a high voltage transmission line on a very dark, windless, night you may see the a bluish light produced by the coronal discharge. Also there is coronal discharge from a lightning suppression rod on your house when lightning is a serious possibilility. You may not ony see this blue light, but hear the hiss as the air continusously breaks down /ionizes.

I was once sailing thru the night and for quite some time there was a natural cornoal discharge from the tip of the alumunum mast whichwas electrically connected to the keel. (Sailors call this coronal discharge "St. Elmo's fire.") I knew what it was and was appropriately scared, but being far from land there was nothing to do but pray. (Something I almost never do, but as they say: "There are no atheists in fox holes when mortars are incoming.")

Trilobyte
06-05-05, 07:45 AM
The only reason I made the assumption that the electrons will be repelled from the surface that already has excess electrons is because of MacM's statement "You will saturate at some point and no further electrons will migrate to the sphere". The purpose of my last post was to probe exactly this ambiguity having already understood exactly what you previously said. You'll notice I said "if this is right" referring back to MacM's statement. The second statement is also a question that came about because of his explanation. (To be more clear admittedly I probably should have quoted MacM instead of myself )

I was merely presenting the implications of the above statement to increase the chance that someone or himself will more easily be able to distinguish between what he said and what he should have said. Also I do understand calculus, but I would not ask about fields if I already knew. If I were you I wouldn't try to seem too knowledgeable when you yourself already said you are not certain how calculus is involved with non-spherical charged objects.

I have another question to add (not a sin I hope). If electrons are removed, making a similar sphere positively charged, will the electrons be distributed evenly throughout the volume then?

MacM
06-05-05, 08:27 AM
The only reason I made the assumption that the electrons will be repelled from the surface that already has excess electrons is because of MacM's statement "You will saturate at some point and no further electrons will migrate to the sphere". The purpose of my last post was to probe exactly this ambiguity having already understood exactly what you previously said. You'll notice I said "if this is right" referring back to MacM's statement. The second statement is also a question that came about because of his explanation. (To be more clear admittedly I probably should have quoted MacM instead of myself )

I was merely presenting the implications of the above statement to increase the chance that someone or himself will more easily be able to distinguish between what he said and what he should have said. Also I do understand calculus, but I would not ask about fields if I already knew. If I were you I wouldn't try to seem too knowledgeable when you yourself already said you are not certain how calculus is involved with non-spherical charged objects.

I have another question to add (not a sin I hope). If electrons are removed, making a similar sphere positively charged, will the electrons be distributed evenly throughout the volume then?

If the surface did not saturate then you are saying you can continue to increase the charge to an infinite voltage :confused:

Billy T
06-05-05, 09:06 AM
The only reason I made the assumption that the electrons will be repelled from the surface that already has excess electrons is because of MacM's statement "You will saturate at some point and no further electrons will migrate to the sphere". .I have another question to add (not a sin I hope). If electrons are removed, making a similar sphere positively charged, will the electrons be distributed evenly throughout the volume then?I don't want to say MacM is wrong because he often means something different with a word than I understand by it. If he realy means that a bunch of electrons released inside a spherical metal shell will not move to the surface of the shell, he is wrong. They will mutually repel and feel no force from the electrons already on the shell surface. As I explained, depending on the coronal discharge characteristic of the gas surrounding the shell, there is a vaild "saturtion" concept, but it sets a limit on the charge that will accumulated on the surface. I.e. all the electrons released inside the shell will migrate under mutual repulsion to the surface and the same number of electrons will leave in coronal discharge from the surface.

The absence of some electron from the shell (a positively charged shell) does not produce any electric field inside. The same proof holds reguarless of there being an excess or difficiency of electrons on the shell. Answering directly - No a positively charged shell will not have electons inside it.

MacM
06-06-05, 12:08 AM
I don't want to say MacM is wrong because he often means something different with a word than I understand by it. If he realy means that a bunch of electrons released inside a spherical metal shell will not move to the surface of the shell, he is wrong. They will mutually repel and feel no force from the electrons already on the shell surface. As I explained, depending on the coronal discharge characteristic of the gas surrounding the shell, there is a vaild "saturtion" concept, but it sets a limit on the charge that will accumulated on the surface. I.e. all the electrons released inside the shell will migrate under mutual repulsion to the surface and the same number of electrons will leave in coronal discharge from the surface.

The absence of some electron from the shell (a positively charged shell) does not produce any electric field inside. The same proof holds reguarless of there being an excess or difficiency of electrons on the shell. Answering directly - No a positively charged shell will not have electons inside it.

No, we are in agreement. By saturation I simply mean unless you generate a higher driving voltage at the belt the migration will stop. The sphere will not fill with mutually repelled electrons. Any released from the belt will migrate to the spherre.

Trilobyte
06-06-05, 03:50 AM
Originally posted by MacM:
"If the surface did not saturate then you are saying you can continue to increase the charge to an infinite voltage :confused: "

The maximum voltage is determined by the surrounding resistor ie air (which is not infinite but very high) as well as the discharging effect caused by the creation of ions and the fact that even in a vacuum the electrons will start to radiate at a certain voltage. So it is not infinite but there will not be any point at which the flow of electrons into the sphere in a Van de Graph will stop. The rate of loss will be the same as the replacement rate and a high voltage will be maintained.

Billy T
06-06-05, 07:00 AM
Originally posted by MacM:
"If the surface did not saturate then you are saying you can continue to increase the charge to an infinite voltage :confused: "

2inquisitive
06-06-05, 12:30 PM
by Billy T:
"Third example: Electrons racing around in an accelerator are accelerated even if their speed is constant. Their radiation is called Shirnkoff radiation (Don't think I have it spelled correct.) "
================================================== ============

So Cherenkov radiation only comes from accelerated electrons? Now I am confused as
to how muons entering our atmosphere are said to reach the Earth's surface by SRT's
time dilation, an inertial frame of reference. The muons give off cherenkov radiation,
which implies they are in a non-inertial frame of reference, undergoing acceleration
(deceleration) while traveling through the atmosphere. Or Earth's increasing
gravitational potential.

Billy T
06-06-05, 01:13 PM
by Billy T:
"Third example: Electrons racing around in an accelerator are accelerated even if their speed is constant. Their radiation is called Shirnkoff radiation (Don't think I have it spelled correct.) "
================================================== ============

So Cherenkov radiation only comes from accelerated electrons? Now I am confused as
to how muons entering our atmosphere are said to reach the Earth's surface by SRT's
time dilation, an inertial frame of reference. The muons give off cherenkov radiation,
which implies they are in a non-inertial frame of reference, undergoing acceleration
(deceleration) while traveling through the atmosphere. Or Earth's increasing
gravitational potential.
Thanks for correcting my spelling. Unfortunately my name for the radiation that electrons in a circular orbit create was entirely wrong. That radiation is called cyclotron radiation. Cherenkov radiation is the name of radiation that any charged particle makes even when traveling in a straight line with velocity greater than speed of light in the medium it is passing thru. It is sort of an electromagnetic shock wave, much like a bullet traveling faster than speed of sound in air creates.

I have seen chrenkov radiation produced (mainly) by beta rays (fast electrons traveling at almost c thur water where the speed of light is approximately 2c/3.) It is a lovely blue and to me, standing above the "swimming pool reactor", it was quite reasuring as I could see that its range was much less thant the depth of the water between me and the small research reactor.

Sorry that I confused you, but glad you were confused as that implies you understand that to the extent that the muons are going faster than the speed of light in air, they will lose energy by chernkov radiation. (There is currently a big experiment now about to start up in the very dark "Patogona" region of Argintina that will detect both the cosmic ray showers, measuring their direction, which is generally the in a narrow cone with axis of the initial cosmic ray and the chernkov radiation with wide view angle telescopes. When it is fully operations, a lot should be learned about cosmic rays.)

I think it is better to understand the deacceleration of the muon (when emitting chernkov radiation) as a consequence of this energy loss mechanism instead of the decelaration as the cause of the chernkov radition; but as is the case of a knife passing down thru butter, one has a certain freedom - perhaps the knife is going down because the butter is moving out of the way. Although one can not be sure what is cause and what is effect, the two facts are consistent with each other.

2inquisitive
06-06-05, 03:20 PM
Yes, I knew synchrotron radiation was given off in circular particle accelerators. That
is the reason they keep building particle accelerators larger, the synchrotron radiation
limits the increase in kinetic energy given to accelerated particles. A larger radius for
the tube equals more kinetic energy for the particles, since less energy is lost to
synchrotron radiation. The particles can be accelerated to speeds greater than the
speed of light in atmosphere, so cherenkov radiation is produced. Do anyone happen
to know the difference in synchrotron radiation and cherenkov radiation? I don't
remember reading anything explaining the exact difference.

Billy T
06-06-05, 04:06 PM
....Do anyone happen to know the difference in synchrotron radiation and cherenkov radiation? I don't remember reading anything explaining the exact difference. I tried to tell how they differed, in last post but can add a little more. Just like the shock wave that forms from the nose of supersonic plane, Chernkov radiation is a "trailing" cone of radiation with its apex attached to the particle producing it. At creation, this cone axis identical to the trajectory of the charged particle that is traveling faster than light can in the medium it is passing thru. (I said "at creation" as if the charged particle is being curved by a magnetic field, the "cone" would be "warped" just as the shock wave from a supersonic jet is "warped" if the plane is making a turn.) Likewise, if the particle is currently in oil floating on water, there are refractive effects at the oil-water interface, so Chernkov radiation is not always a perfect cone.

If your are looking at a it you do not see these cones because there are so many. What you see is just a steady soft blue light that decreases in intensity with distance from the reactor producing the greater than light speed in the medium charged particles. When they have traveled far enough to have their speed reduced to that of light in the medium, there is no more Chernkov radiation. In vacuum, Chernkov radiation does not/ can not occur.

Cyclotron radiation is almost always produce in high vaccum. It is also roughly a cone, actually many cones, all of which have axis tangential to the curved trajectory of the electons being accelerated. That is, in the orbital plain of the circulating electrons and diverging from their circumference is a 360 degree narrow band of radiation diverging from the outer edge of the circulating beam.

If the accelerator is not a cyclotron but a large circle of acceleration sections (with magnetic focusing to keep the beam bunched) then the "cyclotron radiation" comes in pulses at any station as the bunch of electrons/charged particles more generally speaking/ (the packet) passes that station.

I am not sure about the name, but reasonably sure that a bunch of charged particles being accelerated in a long linear accellerator also radiate in a "forward cone" with axis identical to the linear accellerator axis. Perhaps this radiation is also called cyclotron radiation - it is produced in vacuum, so it certainly is not called Chernkov radiation. Perhaps it has a different name. Surely linearly accellerated charged particles make radiation. In the reference frame traveling with them, this radiation may be more omidirectional and the forward conical nature of it in the lab frame is due mainly to SRT effects.

Summary: Cernkov radiation is never produced in vacuum. Cyclotron radiation almost always is produce in high vacuum, but can be allowed to escape thru windows and used to treat tumors, or do physics experiments. As far as I know, Chrenkov radiation has no pratical applications, but is is very pretty to look at as I have done.

MacM
06-06-05, 10:33 PM
Originally posted by MacM:
"If the surface did not saturate then you are saying you can continue to increase the charge to an infinite voltage :confused: "

The maximum voltage is determined by the surrounding resistor ie air (which is not infinite but very high) as well as the discharging effect caused by the creation of ions and the fact that even in a vacuum the electrons will start to radiate at a certain voltage. So it is not infinite but there will not be any point at which the flow of electrons into the sphere in a Van de Graph will stop. The rate of loss will be the same as the replacement rate and a high voltage will be maintained.

I concur. There will be some leakage and current does not stop entirely but the voltage will saturate and current goes to a minimum unless you increase the rate of electrons being stripped from the belt.

2inquisitive
06-06-05, 11:21 PM
Cherenkov radiation is used to determine the velocity of charged particles in particle accelerators. They measure the angle of the cone, which varies with the particle's
velocity. A cut and paste from one quick source. I had read of it many times in reading
high energy physics papers.

Max-Planck-Institute
für Festkörperforschung
in Stuttgart

The light emitted by charged
particles propagates with a distinctive
cone-shaped wave
front, and the axis of this wave
front is determined by the particle
velocity . The angle of the
cone varies with the velocity of
an important experimental tool
in high-energy physics. Particles
created in collisions are counted,
identified and analyzed,
with regard to their speed,
using this method.
http://www.mpg.de/english/illustrationsDocumentation/multimedia/mpResearch/2001/heft02/mpr01_2_014_015.pdf

Also, I believe clycotron radiation is the term used when the radiation is from the old
clycotron particle accelerators or medical devices. In modern particle accelerators
where the particles are accelerated with pulses, a synchrotron accelerator, and some
modern x-ray devices, they use the term synchrotron radiation as I mentioned earlier.
I only mention this because you said you weren't sure of the term used today. However, I don't know offhand what the radiation is called if it can be produced in a linear
accelerator. As far as I know, there is no similar radiation produced in a linear accelerator, just cherenkov radiation. There is no 'bending' acceleration in a linear accelerator.

Billy T
06-07-05, 02:39 PM
Cherenkov radiation is used to determine the velocity of charged particles in particle accelerators. They measure the angle of the cone, which varies with the particle's
velocity. A cut and paste from one quick source. I had read of it many times in reading
high energy physics papers.

Max-Planck-Institute
für Festkörperforschung
in Stuttgart

The light emitted by charged
particles propagates with a distinctive
cone-shaped wave
front, and the axis of this wave
front is determined by the particle
velocity . The angle of the
cone varies with the velocity of
an important experimental tool
in high-energy physics. Particles
created in collisions are counted,
identified and analyzed,
with regard to their speed,
using this method.
http://www.mpg.de/english/illustrationsDocumentation/multimedia/mpResearch/2001/heft02/mpr01_2_014_015.pdf

Also, I believe clycotron radiation is the term used when the radiation is from the old
clycotron particle accelerators or medical devices. In modern particle accelerators
where the particles are accelerated with pulses, a synchrotron accelerator, and some
modern x-ray devices, they use the term synchrotron radiation as I mentioned earlier.
I only mention this because you said you weren't sure of the term used today. However, I don't know offhand what the radiation is called if it can be produced in a linear
accelerator. As far as I know, there is no similar radiation produced in a linear accelerator, just cherenkov radiation. There is no 'bending' acceleration in a linear accelerator.
Chernkov radiation is when the charged particle is moving faster than the speed light can move in the medium the charged particle is moving in. (I think we all agree with this.) I only want to note that the difference between perfect vacuum and the volume in which the charged particles are being accelerated is extremly small - the accelerator has large vaccum pumps is thermally "out gassed" and then often its walls are cooled to get as few particles (gas atoms) in the way of the beam as possible.

Thus the speed of light in the volume of the acclerator is very nearly the same as in vacuum. (Even at one atmosphere the difference is small.) Thus in the acceleration region there must be very little if any Cherinkov radiation. The beam is, I think, typically deflected at some circumferncial experimental port and then may pass thru a gas before striking a denser target etc. Perhaps Chernkov cone and a known gas optical density are a useful way to confirm that beam energy. I don't know, as never have worked with big accelerators, but feel quite sure that Chernkov radiation in the main part of the accelerator is made as small as possible - it steals energy from the beam and thus works against the acceleration.