Why is the E-field inside a conductor zero?

[Idoubt]

Let's try to see this using atoms.

Some atoms have excess e^-, some don't.

Now I am guessing those electrons that aren't bounded to their nucleus gets to move around and neutralize other atoms? And this supposingly happens instantaneously?

atoms of a neutral conductor won't have excess e^-s, only a charged conductor will have excess or shortage of e^-s

For conductors the electrons in the outermost shell are extremely loosely bound and they move from atom to atom randomly through thermal agitation.

so put a charge somewhere inside the conductor and the loosely bound e^-s come running towards/away from it

How fast it does this can be figured out if you know the charges and distance between them, ( to actually calc this is kind of complicated cos there are a lot of factors which are unimportant to our discussion ) and as your text says it takes 10^-16 seconds on an average, this for all practical purposes is instantaneous.

To give you a feel for that time, light that travels 300,000 km per second, will travel 30 nano meters in that time.
That's like 300 atoms in a line

So the time is absurdly small.

 

[flyingpig]

I know none of you are in the position to answer this but do you thinks I finally understand it? I think I do, but it's just that in class my professor and many others often confuse me when they don't understand the concept of "net charge" as opposed to "nothing (charges) inside"

 

[flyingpig]

Also, now that I jussst started on surface integrals, is this the REAL Guass's Law?

$$\varepsilon_0\;\iint_S\!\!\!\!\!\!\!\!\!\!\!\!\!\!\bigcirc \mathbf{E}\; \cdot\; \mathbf{dS}= \sum Q_{enclosed}$$

I am really opposed to people using "q_inside" and "dA"

 

[diazona]

I know none of you are in the position to answer this but do you thinks I finally understand it? I think I do, but it's just that in class my professor and many others often confuse me when they don't understand the concept of "net charge" as opposed to "nothing (charges) inside"

They probably do understand the distinction. It's just that you're not using the same definitions they do. That's fine as long as you can "translate" what other people are saying from their terminology to yours. When someone talks about a conductor with no charge inside it, for example, you should know that they mean no net charge, because if there were no charged material at all, it wouldn't be a conductor.

Also, now that I jussst started on surface integrals, is this the REAL Guass's Law?

$$\varepsilon_0\;\iint_S\!\!\!\!\!\!\!\!\!\!\!\!\!\!\bigcirc \mathbf{E}\; \cdot\; \mathbf{dS}= \sum Q_{enclosed}$$

I am really opposed to people using "q_inside" and "dA"

That is one way to write the integral form of Gauss's law, but only if the enclosed charges are all point charges (and only in a vacuum). The most general form, using your notation, would be
$$\epsilon\iint_S\!\!\!\!\!\!\!\!\!\!\!\!\!\!\bigcirc \mathbf{E}\cdot\mathrm{d}\mathbf{S} = \iiint_V\rho\,\mathrm{d}V$$
Here $\epsilon$ is the permittivity of the material that fills the volume $V$. If the region is filled with nothing (vacuum), it's equal to the electric constant $\epsilon_0$.

 

[Idoubt]

well my advice is to keep thinking about it till you yourself are satisfied, look at the applications of the law to get a better feel for it. Sometimes you have to understand the same thing from 2 or 3 different angles before you are fully satisfied :)

 

[flyingpig]

Now that I think I understand conductors here are my new questions

1) If a conductor has a net "positive" or "negative" charge, it just means an excess of protons or electrons right? Or a dominance of one over the other.

2) If a conductor has a net charge is 0 then

2i) There is an equal amount of protons and electrons

2ii) The conductor inside still has a net charge of zero and there isn't enough electrons or protons to go out on the surface. In other words, if a conductor has a net charge of 0, then you can't conclude whether there are electrons and protons on the surface.

3) What exactly does it mean for a 'charged' and an 'uncharged' conductor mean?

4) Remember when we were talking about resistance in the conductor and how only ideal conductors have 0 field inside? Does that mean real conductors (non-ideal) does NOT take 10^-16 seconds to reach electrostatic equilibrium? Is that what it means for real conductors?

5) Also $$\varepsilon_0\;\iint_S\!\!\!\!\!\!\!\!\!\!\!\!\!\! \bigcirc \mathbf{E}\; \cdot\; \mathbf{dS}= \sum Q_{enclosed}$$, I just want to ask. In my physics class, we say that Gauss's Law only holds for symmetric surfaces, but this integral doesn't seem to suggest it only works for symmetric objects. Isn't this just a vector field dotting a surface area element?

 

[Idoubt]

1) If a conductor has a net "positive" or "negative" charge, it just means an excess of protons or electrons right? Or a dominance of one over the other.

This is fundamentally correct. It might be worth remembering however that in a metal it is always an excess or deficiency of electrons that decide the charge. ( because you can't remove protons without nuclear reactions )

2) If a conductor has a net charge is 0 then

2i) There is an equal amount of protons and electrons

yes that is correct.

2ii) The conductor inside still has a net charge of zero and there isn't enough electrons or protons to go out on the surface. In other words, if a conductor has a net charge of 0, then you can't conclude whether there are electrons and protons on the surface.

A better way to say it would be that there would be an equal number of electrons and protons EVERYWHERE, so that of course includes the surface.

3) What exactly does it mean for a 'charged' and an 'uncharged' conductor mean?

Could you explain this question some more?

4) Remember when we were talking about resistance in the conductor and how only ideal conductors have 0 field inside? Does that mean real conductors (non-ideal) does NOT take 10^-16 seconds to reach electrostatic equilibrium? Is that what it means for real conductors?

Ideal conductors have no RESISTANCE. This means that an electron if it is moving, will keep moving forever ( like frictionless surface). I don't think ideal conductors are relevant to our discussion.

Real conductors will have a 0 field once they reach electrostatic equilibrium ( not before) and it takes about 10^-16 seconds for them to reach electrostatic equilibrium.

5) Also $$\varepsilon_0\;\iint_S\!\!\!\!\!\!\!\!\!\!\!\!\!\! \bigcirc \mathbf{E}\; \cdot\; \mathbf{dS}= \sum Q_{enclosed}$$, I just want to ask. In my physics class, we say that Gauss's Law only holds for symmetric surfaces, but this integral doesn't seem to suggest it only works for symmetric objects. Isn't this just a vector field dotting a surface area element?

You are absolutely correct, Gauss's law is once of the four fundamental laws of electromagnetic theory ( there are 4, called Maxwell's equations) and they are true everywhere.

The reason we are concerned with symmetry is that, Guass's law is only useful for finding the electric field if we have some sort of symmetry, otherwise the surface integral is very complicated ( you have to take the dot product of the electric field and the surface element for every point on the surface, this is complicated for an irregular surface )

But if we take our Gaussian surface in such a way that the electric field is constant for parts on the surface and usually parallel or perpendicular to the surface elements, then we can forget about the dot product and take the electric field outside the integral, and just integrate the surface.

If you look at Guass's law problems you will see that we always choose a Gaussian surface such that the electric field is parallel or perpendicular at all points of our surface, and usually a constant too.

 

[flyingpig]

This is fundamentally correct. It might be worth remembering however that in a metal it is always an excess or deficiency of electrons that decide the charge. ( because you can't remove protons without nuclear reactions )

Is it equally correct to say "an excess of charged atoms over neutral atoms"

A better way to say it would be that there would be an equal number of electrons and protons EVERYWHERE, so that of course includes the surface.

But is my deduction right?

Could you explain this question some more?

what is a charged conductor? What is an uncharged conductor?

Ideal conductors have no RESISTANCE. This means that an electron if it is moving, will keep moving forever ( like frictionless surface). I don't think ideal conductors are relevant to our discussion.

Real conductors will have a 0 field once they reach electrostatic equilibrium ( not before) and it takes about 10^-16 seconds for them to reach electrostatic equilibrium.

Yeah, I mean can we take the Earth as a conductor?

I mean electrons in ideal conductor only stop moving when they reach equilibrium which takes place almost instantaneously.

Real conductors also take place instantaneously

But I am asking whether if Real Conductor's "instantaneous time" > Ideal Conductor's "instantaneous time"

The reason we are concerned with symmetry is that, Guass's law is only useful for finding the electric field if we have some sort of symmetry, otherwise the surface integral is very complicated ( you have to take the dot product of the electric field and the surface element for every point on the surface, this is complicated for an irregular surface )

It isn't that complicated, it's just tedious sometimes.

But my question concerns whether it still holds for non-symmetric surfaces.

 

[Idoubt]

Is it equally correct to say "an excess of charged atoms over neutral atoms"

No, because only the charged atoms matter, how many neutral atoms there are are irrelevant.

But is my deduction right?

Well i think your deduction kind of has the cart in front of the horse, If there is no net charge, there is no electric field and we don't even have to think about electric fields pushing charges to the surface.

what is a charged conductor? What is an uncharged conductor?

A conductor is a material whose valance electrons are extremely loosely bound to the atoms ( so much so that even thermal agitation can free it) and hence it conducts electricity with ease.

A charged conductor is one which has excess or insufficient electrons compared to protons.
( from our discussion we know that such conductors will have all the charge at the surface. The surface will have an excess of e^-s or a deficiency of e^-s )

An uncharged conductor or neutral conductor is one in which there are an equal number of electrons and protons

Yeah, I mean can we take the Earth as a conductor?

No i believe the resistance of the Earth is very high compared to typical conductors say copper or silver, so it cannot be called a conductor

I mean electrons in ideal conductor only stop moving when they reach equilibrium which takes place almost instantaneously.

Real conductors also take place instantaneously

But I am asking whether if Real Conductor's "instantaneous time" > Ideal Conductor's "instantaneous time"

I would say yes, because for the same charge configuration ( hence the same electric field strength) more charge will flow in the case of an ideal conductor due to zero resistance

It isn't that complicated, it's just tedious sometimes.

But my question concerns whether it still holds for non-symmetric surfaces.

hmm I don't know, trying to find the electric field due to a charged sphere with a square Gaussian surface seems pretty complicated to me.

But to answer your question yea it holds for ANY closed surface.

 

[flyingpig]

No, because only the charged atoms matter, how many neutral atoms there are are irrelevant.

So a conductor would still be considered as charged if # charged atoms < # neutral atoms

Well i think your deduction kind of has the cart in front of the horse, If there is no net charge, there is no electric field and we don't even have to think about electric fields pushing charges to the surface.

So I am right...?

An uncharged conductor or neutral conductor is one in which there are an equal number of electrons and protons

Is there actually such thing as atoms with only the neutron?

I would say yes, because for the same charge configuration ( hence the same electric field strength) more charge will flow in the case of an ideal conductor due to zero resistance

Alright, here is what I want to really set the definitions here now.

Does that mean "real conductors" (as one with resistance) that

1. Insulators is a conductor with infinite (or very big) resistance
2. Semi-conductors are conductors with a moderate resistance

In other words, everything is a conductor.

 

[Idoubt]

So a conductor would still be considered as charged if # charged atoms < # neutral atoms

Yes. The easiest way to define it would be to say that a conductor is charged when

# protons is not equal to # electrons.

So I am right...?

If you are saying there is no net charge on the surface on an uncharged conductor you are right. I'm still not sure why you are asking this though.

Is there actually such thing as atoms with only the neutron?

No. Neutrons are just that... neutrons. To my knowledge they never bond without protons.

Alright, here is what I want to really set the definitions here now.

Does that mean "real conductors" (as one with resistance) that

1. Insulators is a conductor with infinite (or very big) resistance
2. Semi-conductors are conductors with a moderate resistance

In other words, everything is a conductor.

Yes to all the above. But 'everything' cannot be called a conductor, a material with charged particles can be called a conductor in the way you defined ( ie insulators being conductors with high resistance) and most things we encounter do fall into that category.
But I think there can be exceptions eg: a neutron star ( made up neutrons )

But keep in mind conductors are defined in reality as materials with a low resistance.

 

[flyingpig]

Yes. The easiest way to define it would be to say that a conductor is charged when

# protons is not equal to # electrons.

Just want more ways to explain one thing because that's physics!

If you are saying there is no net charge on the surface on an uncharged conductor you are right. I'm still not sure why you are asking this though.

I just think it is important for that clarification because I think most people wouldn't be able to realize that. It's like saying if the flux is 0, that means there is no presence of field, which is totally wrong.

If you take the enclosed net charge inside a conductor and it turns out to be 0, you really can't make any conclusion that there are charges that will go to the surface. For all we know, it could just so happen that there is an equal amount of electrons and protons inside the conductor to make that 0 net charge.

Yes to all the above. But 'everything' cannot be called a conductor, a material with charged particles can be called a conductor in the way you defined ( ie insulators being conductors with high resistance) and most things we encounter do fall into that category.
But I think there can be exceptions eg: a neutron star ( made up neutrons )

Ohh so there is such thing as "atoms with only neutrons"...?

 

[Idoubt]

I just think it is important for that clarification because I think most people wouldn't be able to realize that. It's like saying if the flux is 0, that means there is no presence of field, which is totally wrong.

If you take the enclosed net charge inside a conductor and it turns out to be 0, you really can't make any conclusion that there are charges that will go to the surface. For all we know, it could just so happen that there is an equal amount of electrons and protons inside the conductor to make that 0 net charge.

If the net charge of a conductor is zero, then we can form the conclusion that there is NO charge on the surface ( as long as there is no external field ), by using the word 'net' charge we already take into account the no of protons and electrons so we needn't mention them separately.

Ohh so there is such thing as "atoms with only neutrons"...?

No they aren't atoms. It's just neutrons. Atoms by definition are positive nuclei surrounded by negative charge, and atomic bonds are electrostatic.

Neutrons are uncharged so they cannot be involved in forming molecules etc so they are not atoms.

 

[flyingpig]

If the net charge of a conductor is zero, then we can form the conclusion that there is NO charge on the surface ( as long as there is no external field ), by using the word 'net' charge we already take into account the no of protons and electrons so we needn't mention them separately.

Nah I just mean the inside

 

[Idoubt]

Nah I just mean the inside

I'm still not sure what distinction or definition you are trying to make here. Are you saying that with a Gaussian surface inside the conductor you cannot say whether there are charges on the surface?

 

[flyingpig]

Yeah a Gaussian surface inside a conductor encloses a net charge of 0, but that's all we know. We cannot make any conclusion whatsoever that there are charges on the surface.

 

[Idoubt]

ok I think I finally get what you are saying.

Let us now consider a charged conductor.

Also we accept that in an equilibrium state there can be no electric field inside a conductor.

Now we consider a Gaussian surface inside the conductor in the exact shape of the conductor so that it encloses every part of the conductor but the surface.

Since this Gaussian surface is inside the conductor the total electric flux through it is zero. So by Gauss's law, the charge enclosed is also zero. So we conclude that there is no charge in any interior part of the conductor.

Now consider a second Gaussian surface one that encloses the whole conductor (including the surface) . Now since there is a net charge on the conductor, there will be an electric flux though this Gaussian surface. Now we conclude that there has to be charge somewhere on the conductor.

The only way for both these conditions to be satisfied is for all the charge to be on the surface.

 

[flyingpig]

ok I think I finally get what you are saying.

Let us now consider a charged conductor.

Also we accept that in an equilibrium state there can be no electric field inside a conductor.

Now we consider a Gaussian surface inside the conductor in the exact shape of the conductor so that it encloses every part of the conductor but the surface.

Since this Gaussian surface is inside the conductor the total electric flux through it is zero. So by Gauss's law, the charge enclosed is also zero. So we conclude that there is no charge in any interior part of the conductor.

Now consider a second Gaussian surface one that encloses the whole conductor (including the surface) . Now since there is a net charge on the conductor, there will be an electric flux though this Gaussian surface. Now we conclude that there has to be charge somewhere on the conductor.

The only way for both these conditions to be satisfied is for all the charge to be on the surface.

But unless you do that second Gaussian surface you would never know

I was thinking last night about static equilbrium again.

I was imagining what's actually happening to the electrons. I just want to know, do atoms move when their electrons leave them?

Then I thought about once the equilibrium is reached what happens? I mean there still could be electrons inside the conductor, but you might not be able to see them because the protons' charge is canceling them out, is that right?

Then I also looked at the periodic table. I mean just ignoring the gas part, and focusing on the metals side. Is it true that "odd" valence shells on the periodic table makes good conductors?

 

[flyingpig]

Also for a semi-conductor, is it true that it can have a E-field of 0 and it can suddenly become non-zero? How do we determine it? Is it usually E(t)? That is Electric field as a function of time or E'(t)? How E-field changes with time because it is a semi-conductor?

 

[Idoubt]

I was imagining what's actually happening to the electrons. I just want to know, do atoms move when their electrons leave them?

There is probably a small motion because of the other charged particles in the atom, but if you compare the masses of the atom and an electron, you can see that the force that moves the electron is too low to seriously move the atom at any great speed.

Then I thought about once the equilibrium is reached what happens? I mean there still could be electrons inside the conductor, but you might not be able to see them because the protons' charge is canceling them out, is that right?

Yes there will be electrons "hidden" by protons or you can just call these neutral atoms.

Then I also looked at the periodic table. I mean just ignoring the gas part, and focusing on the metals side. Is it true that "odd" valence shells on the periodic table makes good conductors?

To my knowledge this is not true. As you should know all atoms with the exclusion of hydrogen and helium, try to get 8 electrons in their outer most orbit ( apparently this many give a very stable configuration - this is why noble gases are very stable )

atoms do this either by accepting more electrons into their outermost orbit until there are 8 electrons ( eg: chlorine has 7 valance electrons and it accepts one to become stable)

Now conductors are usually atoms with 1-3 electron in the outermost orbit and 8 electrons in the second outermost shell.

So instead of accepting 7-5 more electrons it simply loses 1-3 electron to attain the stable config.

Also for a semi-conductor, is it true that it can have a E-field of 0 and it can suddenly become non-zero? How do we determine it? Is it usually E(t)? That is Electric field as a function of time or E'(t)? How E-field changes with time because it is a semi-conductor?

Perhaps someone else can answer this.

 

[flyingpig]

There is probably a small motion because of the other charged particles in the atom, but if you compare the masses of the atom and an electron, you can see that the force that moves the electron is too low to seriously move the atom at any great speed.

F = ma

$$F_1 = F_2$$

$$m\vec{a}_1 = M\vec{a}_2$$

If M>>m

$$\vec{a}_1 = M\vec{a}_2$$?

Yes there will be electrons "hidden" by protons or you can just call these neutral atoms.

Okay so that's interesting, but what if I make a Gaussian Surface just small enough and just big enough to only enclose that electron? Doesn't that make it so that I don't have a E-field of 0?

To my knowledge this is not true. As you should know all atoms with the exclusion of hydrogen and helium, try to get 8 electrons in their outer most orbit ( apparently this many give a very stable configuration - this is why noble gases are very stable )

Gas can't be conductors right...?

Perhaps someone else can answer this.

With at least 8 pages and 2600 views, I think I drove off everyone in this forum, right Sammy...?

 

[Idoubt]

F = ma

$$F_1 = F_2$$

$$m\vec{a}_1 = M\vec{a}_2$$

If M>>m

$$\vec{a}_1 = M\vec{a}_2$$?

If m is very small you can't just not write it like 0.0001* x is not the same as x

The last step will be a₁ = Ma₂ / m

or a₁ = K * a₂, where K=M/m which is very large since M>>m, so

a₂ is very small compared to a₁

Okay so that's interesting, but what if I make a Gaussian Surface just small enough and just big enough to only enclose that electron? Doesn't that make it so that I don't have a E-field of 0?

yes that's absolutely correct. It is this electric field that attracts the proton to the electron But from a very large distance ( large enough so that a Gaussian surface encloses the whole atom ) the two equal and opposite charges cancel and there is no net E field.

Gas can't be conductors right...?

They can conduct if you ionize them, this is how a fluorescent light bulb works. But noble elements do not undergo any reactions under normal conditions unlike all the other elements.

With at least 8 pages and 2600 views, I think I drove off everyone in this forum, right Sammy...?

I doubt that ;)

 

[flyingpig]

I will fetch SammyS if he will even come back...

 

[SammyS]

I've been away from the computer for most of the last 10 days !

Sure, I'll come back. (Really, I never left.) I knew that you didn't like a remark of mine - I didn't mean to offend, but yes - sometimes my humor ain't too funny to others - so I've been reading PF posts & responding to some & I've been waiting for a good situation to leave a post to you that is obviously very constructive.

I'll look at this thread to see if I can help.

 

[flyingpig]

I've been away from the computer for most of the last 10 days !

Sure, I'll come back. (Really, I never left.) I knew that you didn't like a remark of mine - I didn't mean to offend, but yes - sometimes my humor ain't too funny to others - so I've been reading PF posts & responding to some & I've been waiting for a good situation to leave a post to you that is obviously very constructive.

I'll look at this thread to see if I can help.

No they were funny because it showed me that you did care about me...

 

[SammyS]

...
With at least 8 pages and 2600 views, I think I drove off everyone in this forum, right Sammy...?

I have continued to read it from time to time. Idoubt seems to be keeping up very well.

Now on the 9th page & 130 posts in this thread... What's the question you fetched me for??

 

[flyingpig]

Well first I said

Alright, here is what I want to really set the definitions here now.

Does that mean "real conductors" (as one with resistance) that

1. Insulators is a conductor with infinite (or very big) resistance
2. Semi-conductors are conductors with a moderate resistance

In other words, everything is a conductor.

Then I wanted to ask

For a semi-conductor, is it true that it can have a E-field of 0 and it can suddenly become non-zero? How do we determine it? Is it usually E(t)? That is Electric field as a function of time or E'(t)? How E-field changes with time because it is a semi-conductor?

 

[flyingpig]

Please come back

 

[flyingpig]

IDoubt, what exactly happens if you apply a non-zero e-field to a conductor? Does it make the field inside non-zero again?

 

[flyingpig]

Now I am alone...

 

[diazona]

I suppose I can take this up again...

what exactly happens if you apply a non-zero e-field to a conductor? Does it make the field inside non-zero again?

For an ideal conductor: no it does not.

For a real (non-ideal) conductor: yes, but only for a very short time, until the conductor returns to electrostatic equilibrium. Then the field inside is zero again.

 

[flyingpig]

I suppose I can take this up again...

For an ideal conductor: no it does not.

For a real (non-ideal) conductor: yes, but only for a very short time, until the conductor returns to electrostatic equilibrium. Then the field inside is zero again.

So you have to keep up the field to make it non-zero? What happens if you keep it at an alternating frequency? Like pull it in and out?

 

[SammyS]

Well first I said

"Alright, here is what I want to really set the definitions here now.

Does that mean "real conductors" (as one with resistance) that

1. Insulators is a conductor with infinite (or very big) resistance
2. Semi-conductors are conductors with a moderate resistance

In other words, everything is a conductor."​

Then I wanted to ask

"For a semi-conductor, is it true that it can have a E-field of 0 and it can suddenly become non-zero? How do we determine it? Is it usually E(t)? That is Electric field as a function of time or E'(t)? How E-field changes with time because it is a semi-conductor?"​

Although the definitions you give for 'real conductors', 'insulators' and 'semi-conductors' can be useful in some situations, they're not very helpful in discussing the 'E-field problem' you're dealing with here. For instance, conductivity is the reciprocal of resistivity, so an insulator (by the above definition) has zero (or very small) conductivity.

For electrostatics problems, the conductor is usually a metal. Metals have the property of having one or more electrons in the 'conduction band' per atom at ordinary temperatures. Even for the case of a metal with high resistivity, these electrons (in the 'conduction band') are relatively free to move, not being associated with any individual atom in particular.

A complete answer for these three types of materials would be extremely long.

 

[diazona]

So you have to keep up the field to make it non-zero? What happens if you keep it at an alternating frequency? Like pull it in and out?

Not quite. Even if you keep the external electric field up at a constant level, the field inside the conductor will drop away to zero. But if you have a constantly changing electric field, such as an EM wave (or AC current), if the frequency of the change is high enough then I suppose you could keep the electric field inside the conductor from settling down to zero.

 

[flyingpig]

Although the definitions you give for 'real conductors', 'insulators' and 'semi-conductors' can be useful in some situations, they're not very helpful in discussing the 'E-field problem' you're dealing with here. For instance, conductivity is the reciprocal of resistivity, so an insulator (by the above definition) has zero (or very small) conductivity.

And that definition is wrong...?