Capacitors and charge

[Trilobyte]

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]

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]

nice explaination.

[Trilobyte]

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]

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]

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]

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]

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]

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]

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]

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.

[Trilobyte]

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]

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]

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]

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

[Billy T]

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]

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]

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 "

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]

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 "

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.

Well done trilobyte. I have been ignoring your comments about electrons "radiating." (Mainly because I did not know to what you were referring.) It is true that electrons can radiate, but only when they are accelerated. For example the acceleration of electrons up and down in the vertical antenna of your local radio station is what makes the radio waves. Second example: the acceleration (actually deacceleration or negative acceleration) of a beam of electrons stricking a tungeston target in your denist's X-ray tube makes the X-rays. (This closely resembles Bremstrallum and Compton scattering radiation.) 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.) It is very usefull well defined tangential beam with both medical and research applications. I can not think of any significant radiation from electons just sitting on the Van de Graph sphere, but don't want to be too categorical and state that there is none, but that is what I think.

[2inquisitive]

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.) "
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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.