How much charge is there on an conductor

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  • #1
usljoo
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im asking this because, if one has a conductor and there is a extern electric field then all the charge in the conductor redistributes itself so as to cancel out the external field.
but lets suppose that there is not enough charge in the conductor to cancel out the electric field, would this situation even be possible
 

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  • #2
tiny-tim
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hi usljoo! :smile:

a conductor is defined as something with charges which can move …

if the electric field is so strong, and the number of charges so small, that all the available charges pile up at one end, then it will cease to be a conductor, and the usual rule about cancelling the external field will no longer apply :wink:
 
  • #3
usljoo
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aah ok i see so its a thing of definition ok thank you much ;)

ok ok so if they are all stuck and cant move its not a conductor but who says that they cant move if one applys another field afterall charge doesnt move in any conductor if there is no electric field
 
  • #4
usljoo
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ok ok so if they are all stuck and cant move its not a conductor but who says that they cant move if one applys another field afterall charge doesnt move in any conductor if there is no electric field
 
  • #5
tiny-tim
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well, you know what the material is, so you know whether it has mobile charges …

you don't need to apply a field to find out! :wink:
 
  • #6
usljoo
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but if you define a conductor as something that has mobile charges then anything thats follows this definition is a conductor and when they first used copper as a conductor they didnt know beforehand if it has mobile charges instead they applied fields and watched charges move

and if you define a conductor as something that has moving charges then the conductor witch has all charge at one end and has not a zero field inside does not necessarily fall out of this definition because the charges can move, they are still free electrons and not bound through nuclear forces to the material or whatever so it doesnt make quite sence
 
  • #7
tiny-tim
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… so it doesnt make quite sence

yes, but your question is highly unrealistic …

a field so strong that it forces all the free electrons in a copper plate to bunch at one end just isn't going to happen
 
  • #8
usljoo
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hmm i dont know about this, why not do you have anything to back this up
 
  • #9
ednobj
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tiny-tim is right.

If all the free electrons of a conductor were at one end, the localized charge and thus the corresponding energy of the configuration would be absolutely enormous. The energy required would strongly hinder anyone trying to attempt to create a field that strong. This is all aside from the fact that a conductor with 0 electric field inside and freely mobile electrons is an idealized model which breaks down long before that field strength is approached. Often you will find texts that explicitly say for conductor thought experiments that the E-field is assumed to be relatively weak.

Not to be too much of a nit picker, but it is kind of hard to follow your questions when you forgo a lot of punctuation.
 
  • #10
Delta2
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Well physics are also about some extra hidden or "silent" assumptions that go along with the basic definitions and theory. In this case of the electrostatic equilibrium of a conductor you make the assumption that the conductor has enough free charges to cancel out the external field. If this assumption doesnt hold then simply the conductor wont reach equilibrium. This assumption is very logical to make cause as the others said it would be extremely rare case for the external field to be so big that the free charges arent enough to cancel it out.
 
  • #11
usljoo
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tiny-tim is right.

If all the free electrons of a conductor were at one end, the localized charge and thus the corresponding energy of the configuration would be absolutely enormous. The energy required would strongly hinder anyone trying to attempt to create a field that strong. This is all aside from the fact that a conductor with 0 electric field inside and freely mobile electrons is an idealized model which breaks down long before that field strength is approached. Often you will find texts that explicitly say for conductor thought experiments that the E-field is assumed to be relatively weak.

Not to be too much of a nit picker, but it is kind of hard to follow your questions when you forgo a lot of punctuation.

the model is idealized because first of the charge is not continuous in the conductor nor is it on the conductor and so there will sure be oscillations caused by electric fields about some region and so in general gausses rule cannot apply a zero charge within the conductor but i guess it is a very very good approximation.

but how do you figure that the energy would be enormous and that the charge on one side would be enormous.

after all electrons have a very very small charge, i dont remember how low it was exactly and also there should be some data for the number of free electrons in copper for example, so if one multiplies these two one should get the charge that could be concentrated at some smaller volume than the conductor itself or even at a point.

a point would require enormous amounts of energy to achieve, if not infinite dont know right now.

but any other volume should not be that unrealistic as for example lets say 10% less than the volume of the conductor.
 
  • #12
usljoo
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Well physics are also about some extra hidden or "silent" assumptions that go along with the basic definitions and theory. In this case of the electrostatic equilibrium of a conductor you make the assumption that the conductor has enough free charges to cancel out the external field. If this assumption doesnt hold then simply the conductor wont reach equilibrium. This assumption is very logical to make cause as the others said it would be extremely rare case for the external field to be so big that the free charges arent enough to cancel it out.

i guess too that it is only an assumption but i have heard and read somewhere that ABSOLUTELY NO stationary charge outside the conductor can ever produce an el. field inside it
witch would be very funny :)
 
  • #13
ZapperZ
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im asking this because, if one has a conductor and there is a extern electric field then all the charge in the conductor redistributes itself so as to cancel out the external field.
but lets suppose that there is not enough charge in the conductor to cancel out the electric field, would this situation even be possible

I'm a bit puzzled by the questions and responses in this thread that started with this question. I'm also not sure if I missed something here when I consider that this is a rather "simple" problem.

1. We all know what happens when, say, we have a semiconductor in an electric field. Such a material is exactly what you are describing, i.e. it has a limited amount of charge carrier, and at some point, you'll have an electric field strong enough that it can no longer effectively do the necessary "shielding". So in that case, you have electric field penetrating the bulk of the material. That is why it has a dielectric constant value! You then solve Maxwell Equation using that dielectric constant with appropriate boundary conditions, etc.

2. If you insist on using a conductor, then you need to consider if you are dealing with the ideal situation, or do you want to consider a "real" situation. This is important, because you cannot swing back and forth between the two, certainly not without telling us when. When you deal with an ideal situation, then the conductor has an infinite amount of charge carrier to draw upon and will do whatever necessary to satisfy a "perfect metal" boundary condition.

Now, if you don't like that and want to deal with "real" situation, then you must also consider the reality that there's no such thing as a perfectly smooth metallic surface and there is no such thing as a metal with perfect crystallographic arrangements. What this means is that at some point, as you are increasing the external E-field, you will start getting field emission off the surface, and at some point, the breakdown will occur. This is all BEFORE you get to the point of having insufficient amount of charges on the metal for effective shielding!

So be careful when you deviate from the path of the ideal situation. Reality is not simple, and it is certainly not pretty.

Zz.
 
  • #14
ednobj
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To give you an example of how much energy it takes to come close to depleting a conducting material of it's internal mobile charge, consider a 10cm ^3 block of copper (this is a relatively small bit of copper, actually). Copper has a free electron density of 13.6*10^9 C/m^3 remembering that 1 electron = 1.6*10^-19 C to give you an idea of how large that is. Copper is also one of the best conductors of electricity so more often than not, people consider Copper to be very close to an idealized conductor. Now let's say only 10% of the available free electrons are moved to expel an approximately uniform electric field normal to one of the faces of the copper cube. That's 13.6*10^5 Coulombs moved to the surface of the copper cube. This corresponds to an electric field of E = Q/(A epsilon_0) [A = 10^-2 m^2]. The energy of the electric field expelled by this charge distribution is U = epsilon_0/2 * Vol * E^2 = Vol * Q^2 /(2 * A^2 * epsilon_0^2). Plugging in these numbers, the energy of the field within the measly 10^-3 m^3 volume is 1.04*10^24 Joules. For a comparison, you would need the energy from a typical US nuclear fission reactor running for millions of years to achieve that level of energy or the total amount of energy from the sun that strikes the Earth in one year. Note also how this energy goes as Vol/A^2 so as you approach the shape of a cylinder, wire, etc., this value climbs higher and this is only considering the field expelled and not through all space.
 
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  • #15
usljoo
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I'm a bit puzzled by the questions and responses in this thread that started with this question. I'm also not sure if I missed something here when I consider that this is a rather "simple" problem.

1. We all know what happens when, say, we have a semiconductor in an electric field. Such a material is exactly what you are describing, i.e. it has a limited amount of charge carrier, and at some point, you'll have an electric field strong enough that it can no longer effectively do the necessary "shielding". So in that case, you have electric field penetrating the bulk of the material. That is why it has a dielectric constant value! You then solve Maxwell Equation using that dielectric constant with appropriate boundary conditions, etc.

2. If you insist on using a conductor, then you need to consider if you are dealing with the ideal situation, or do you want to consider a "real" situation. This is important, because you cannot swing back and forth between the two, certainly not without telling us when. When you deal with an ideal situation, then the conductor has an infinite amount of charge carrier to draw upon and will do whatever necessary to satisfy a "perfect metal" boundary condition.

Now, if you don't like that and want to deal with "real" situation, then you must also consider the reality that there's no such thing as a perfectly smooth metallic surface and there is no such thing as a metal with perfect crystallographic arrangements. What this means is that at some point, as you are increasing the external E-field, you will start getting field emission off the surface, and at some point, the breakdown will occur. This is all BEFORE you get to the point of having insufficient amount of charges on the metal for effective shielding!

So be careful when you deviate from the path of the ideal situation. Reality is not simple, and it is certainly not pretty.

Zz.

ok, when you say field emision you mean like breakdown into air right and that there would be charge lost

and how would in an ideal situation a conductor have infinite amount of charge, as i figure the free charge would only be continuous and i dont see a reason to make that idealization you mentioned

i rather think that in both situations ideal or real, if the e field is strong enough than there will be charge inside the conductor and therefor an e field inside it
 
  • #16
ZapperZ
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ok, when you say field emision you mean like breakdown into air right and that there would be charge lost

Field emission is tunneling of charge exiting the material.

and how would in an ideal situation a conductor have infinite amount of charge, as i figure the free charge would only be continuous and i dont see a reason to make that idealization you mentioned

When you solve Maxwell equation, is there a limit on the amount of charge in a conductor?

i rather think that in both situations ideal or real, if the e field is strong enough than there will be charge inside the conductor and therefor an e field inside it

OK, solve Gauss's law for me and show where, at some electrostatic E-field strength, there's a charge inside the conductor.

Zz.
 
  • #17
usljoo
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To give you an example of how much energy it takes to come close to depleting a conducting material of it's internal mobile charge, consider a 10cm ^3 block of copper (this is a relatively small bit of copper, actually). Copper has a free electron density of 13.6*10^9 C/m^3 remembering that 1 electron = 1.6*10^-19 C to give you an idea of how large that is. Copper is also one of the best conductors of electricity so more often than not, people consider Copper to be very close to an idealized conductor. Now let's say only 10% of the available free electrons are moved to expel an approximately uniform electric field normal to one of the faces of the copper cube. That's 13.6*10^5 Coulombs moved to the surface of the copper cube. This corresponds to an electric field of E = Q/(A epsilon_0) [A = 10^-2 m^2]. The energy of the electric field expelled by this charge distribution is U = epsilon_0/2 * Vol * E^2 = Vol * Q^2 /(2 * A^2 * epsilon_0^2). Plugging in these numbers, the energy of the field within the measly 10^-3 m^3 volume is 1.04*10^24 Joules. For a comparison, you would need the energy from a typical US nuclear fission reactor running for millions of years to achieve that level of energy or the total amount of energy from the sun that strikes the Earth in one year. Note also how this energy goes as Vol/A^2 so as you approach the shape of a cylinder, wire, etc., this value climbs higher and this is only considering the field expelled and not through all space.

thanks this is the best answer so far i guess, ill lock at it more closely
 
  • #18
usljoo
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Field emission is tunneling of charge exiting the material.



When you solve Maxwell equation, is there a limit on the amount of charge in a conductor?



OK, solve Gauss's law for me and show where, at some electrostatic E-field strength, there's a charge inside the conductor.

Zz.

ok so by field emision you mean that charge follows the field in vacuum.

and how am i supposed to solve maxwells equations if i dont know the charge nor the e field inside the conductor.
i rather have to assume that the e field is zero inside to conclude that no charge is inside.

but let me solve it for you by assuming that there is an e field inside then easy like 2+2 it follows that there is charge enclosed by my chosen closed surface, and lets suppose for simplicity that it is a point charge to avoid considerations about density
 
  • #19
ZapperZ
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ok so by field emision you mean that charge follows the field in vacuum.

and how am i supposed to solve maxwells equations if i dont know the charge nor the e field inside the conductor.
i rather have to assume that the e field is zero inside to conclude that no charge is inside.

but let me solve it for you by assuming that there is an e field inside then easy like 2+2 it follows that there is charge enclosed by my chosen closed surface, and lets suppose for simplicity that it is a point charge to avoid considerations about density

This is getting even stranger. You claim that even under the IDEAL conditions, you can get charge in a conductor under an applied electrostatic E-field.

I would like you to know that this is a COMMON question being solved in standard E&M classes. In fact, it is also common at the graduate level where you use various geometries to apply the Dirichlet and Neumann boundary conditions to solve the Poisson's equation.

But even without that, are you telling me that if I have a uniform field, and I put a solid spherical conductor in that field, you cannot use simple Gauss's law to argue if there is a charge or not inside the conductor? Or are you claiming that Gauss's law isn't valid and cannot be used in this case?

Zz.
 
  • #20
usljoo
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Field emission is tunneling of charge exiting the material.



When you solve Maxwell equation, is there a limit on the amount of charge in a conductor?



OK, solve Gauss's law for me and show where, at some electrostatic E-field strength, there's a charge inside the conductor.

Zz.

This is getting even stranger. You claim that even under the IDEAL conditions, you can get charge in a conductor under an applied electrostatic E-field.

I would like you to know that this is a COMMON question being solved in standard E&M classes. In fact, it is also common at the graduate level where you use various geometries to apply the Dirichlet and Neumann boundary conditions to solve the Poisson's equation.

But even without that, are you telling me that if I have a uniform field, and I put a solid spherical conductor in that field, you cannot use simple Gauss's law to argue if there is a charge or not inside the conductor? Or are you claiming that Gauss's law isn't valid and cannot be used in this case?

Zz.

oooh come on youre just laughing at me

you dont need to solve poisons equation for any boundary conditions because you dont even get that far, youre stuck one step behind that, you dont have the el potential not even the e field.

and im not saying that gausses law isny valid hehe you make me laugh.

what im saying is that the configuration of the system is unknown and cannot be subjected to any kind of analysis especially not to maxwell equations.
 
  • #21
usljoo
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To give you an example of how much energy it takes to come close to depleting a conducting material of it's internal mobile charge, consider a 10cm ^3 block of copper (this is a relatively small bit of copper, actually). Copper has a free electron density of 13.6*10^9 C/m^3 remembering that 1 electron = 1.6*10^-19 C to give you an idea of how large that is. Copper is also one of the best conductors of electricity so more often than not, people consider Copper to be very close to an idealized conductor. Now let's say only 10% of the available free electrons are moved to expel an approximately uniform electric field normal to one of the faces of the copper cube. That's 13.6*10^5 Coulombs moved to the surface of the copper cube. This corresponds to an electric field of E = Q/(A epsilon_0) [A = 10^-2 m^2]. The energy of the electric field expelled by this charge distribution is U = epsilon_0/2 * Vol * E^2 = Vol * Q^2 /(2 * A^2 * epsilon_0^2). Plugging in these numbers, the energy of the field within the measly 10^-3 m^3 volume is 1.04*10^24 Joules. For a comparison, you would need the energy from a typical US nuclear fission reactor running for millions of years to achieve that level of energy or the total amount of energy from the sun that strikes the Earth in one year. Note also how this energy goes as Vol/A^2 so as you approach the shape of a cylinder, wire, etc., this value climbs higher and this is only considering the field expelled and not through all space.

look man its not 13.6*10^5 Coulombs its 13.6*10^5 electrons so a big mistake no wounder you get such a large energy needed to move this.

13.6*10^5 should be multiplied by the charge of electron and then the calculation continued

and then you would find that the energy needed to move the 10% is very reasonable
 
  • #22
ZapperZ
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oooh come on youre just laughing at me

you dont need to solve poisons equation for any boundary conditions because you dont even get that far, youre stuck one step behind that, you dont have the el potential not even the e field.

and im not saying that gausses law isny valid hehe you make me laugh.

what im saying is that the configuration of the system is unknown and cannot be subjected to any kind of analysis especially not to maxwell equations.

Huh? I GIVE you an E-field! A Uniform E-Field! Isn't this a common problem in textbooks the way it is given?

I don't need to know the "configuration of the system" to use Gauss's Law! Put the sphere inside this field, construct a gaussian sphere just inside the surface of the conductor, and solve it! What do you get? Why is this not applicable here?

Zz.
 
  • #23
usljoo
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Huh? I GIVE you an E-field! A Uniform E-Field! Isn't this a common problem in textbooks the way it is given?

I don't need to know the "configuration of the system" to use Gauss's Law! Put the sphere inside this field, construct a gaussian sphere just inside the surface of the conductor, and solve it! What do you get? Why is this not applicable here?

Zz.

the e field inside the conductor is the unknown ok?

and that i need and that i assumed to be zero and not and thats didnt answer my question at all but ok

your claim that an ideal conductor is supposed to have infinite free charge is what bothers me
 
  • #24
ZapperZ
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the e field inside the conductor is the unknown ok?

and that i need and that i assumed to be zero and not and thats didnt answer my question at all but ok

Er...w hy is the E-field not known? Have you ever solved a simple Gauss's law problem with a charged conductor? This is very puzzling.

your claim that an ideal conductor is supposed to have infinite free charge is what bothers me

It is there simply to indicate that it can shield all E-field and that the dielectric constant doesn't come into play! I work with high resistive material where both the conductivity and dielectric constant can't be ignored since these things do not sit in a static E-field. But even at resistance of the order of tens of megaOhm, by the time scale of microseconds after one switch on the external E-field, ALL E-field inside the conductor is essentially zero!

I don't know if you're just wanting an argument, or you simply don't understand basic electrostatics. Why you do not want to accept E-field inside a conductor being zero under electrostatic conditions is beyond me. I can understand not understanding something, but when these are stuff right out of standard textbooks, i.e. you can check those if you don't believe me, then it is highly puzzling.

I've offered my help. You can take it or leave it. I don't expect expression of gratitude, but I certainly draw the line when you disregard such help in this manner.

I'm done.

Zz.
 
  • #25
usljoo
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lock thank for ur help

of course it is puzzling to solve it with charged conductors when you dont know the charge distribution but that aside.

so an ideal conductor is supposed to be able to cancel out all ext e fields ok thank you very much

and in your work with these resistive materials you should maybe try an very large e field, uniform or not and then observe what happends.

and i dont know who dosnt want to accept thing, me or you that might be a nice subject to some psychiatrist or something :)

what exactly do you do, are you a teaching or what
 
  • #26
ednobj
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look man its not 13.6*10^5 Coulombs its 13.6*10^5 electrons so a big mistake no wounder you get such a large energy needed to move this.

13.6*10^5 should be multiplied by the charge of electron and then the calculation continued

and then you would find that the energy needed to move the 10% is very reasonable

Um...http://hyperphysics.phy-astr.gsu.edu/hbase/tables/fermi.html#c2. A simple google would've shown that the numbers are fine and the energy is right within an order of magnitude.

I really think you're just getting confused with definitions. In E&M, we start by defining an ideal case of a conductor to be a material with sigma=infinity that expels all electric fields so by necessity it must have infinite free electrons. We use it to model conductors to get a good guess at what happens because the ideal conductor approximation works in low energy, high conductivity limits. The reality is that no conductor truly has infinite free electrons or infinite conductivity. If that were true, we would never be able to pass current through a wire.

What you've been asking us is what happens at high electric field, i.e. high energies. Well, as I've shown, for the best of conductors, you won't get to the point you are considering because other effects happen at more reasonable energies such as arcing or field emission as ZapperZ said that will break the model of a stable conductor. For other materials with fewer free electrons like semi-conductors, high electric fields tend to rip rip apart molecules or effect the substance on an atomic level changing its properties and/or destroying it completely. Thus you are not actually looking at the conductor/semi-conductor; it's something else.

In summary, perfect conductors expel all electric fields, real conductors fight electric fields within their capabilities and tolerances resulting in properties such as conductivity and dielectric constants, and relatively high electric fields tend to destroy what you're trying to study.
 
  • #27
usljoo
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Um...http://hyperphysics.phy-astr.gsu.edu/hbase/tables/fermi.html#c2. A simple google would've shown that the numbers are fine and the energy is right within an order of magnitude.

I really think you're just getting confused with definitions. In E&M, we start by defining an ideal case of a conductor to be a material with sigma=infinity that expels all electric fields so by necessity it must have infinite free electrons. We use it to model conductors to get a good guess at what happens because the ideal conductor approximation works in low energy, high conductivity limits. The reality is that no conductor truly has infinite free electrons or infinite conductivity. If that were true, we would never be able to pass current through a wire.

What you've been asking us is what happens at high electric field, i.e. high energies. Well, as I've shown, for the best of conductors, you won't get to the point you are considering because other effects happen at more reasonable energies such as arcing or field emission as ZapperZ said that will break the model of a stable conductor. For other materials with fewer free electrons like semi-conductors, high electric fields tend to rip rip apart molecules or effect the substance on an atomic level changing its properties and/or destroying it completely. Thus you are not actually looking at the conductor/semi-conductor; it's something else.

In summary, perfect conductors expel all electric fields, real conductors fight electric fields within their capabilities and tolerances resulting in properties such as conductivity and dielectric constants, and relatively high electric fields tend to destroy what you're trying to study.

i see, ok then ill take it that an ideal conductor is supposed to have infinite charge because it is the only thing that makes sense with the statement that no electric charge distribution (stationary) can ever produce an e field inside everything else was not good.

and zapper z gave me maybe the best answer but then he blew it by telling me to solve maxwell equations to conclude that the charge on a conductor is infinite hahaha like i dont expect this from an expert
 
  • #28
Delta2
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Yes in my opinion one has to say that an ideal conductor has infinite mobile charge in order to be able to state that any external static E field will eventually (after short time) be zero inside the ideal conductor. But thats for an ideal conductor i.e a conductor which will not break down due to other phenomena like field emission.
 
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