Waste less time on Facebook — follow Brilliant.
×

Need Help: Gauss' Law: Net Charge Inside a Conductor

I'm really lost on this topic, and my book stinks at conceptual explanations; my confusion is with the net charge of zero inside a conductor. Why is this so? Those lame explanations don't ring any bell; sounds more like biology to me ("this is so because we see it as so and because it is good as so"). Also, the book keeps reasserting its statements assuming that I already agree with it that E=0. That's like trying to prove a constant using the constant in the proof. Gee.

I get the mutual repulsion thing; if there's charge, it will spread out. But why not leave some charge inside too? After all, if you're gonna have all that charge on the outside, eventually it will make more sense for the charge to occupy a less pressurized space.

Also, imagine we have an electrostatic equilibrium. Now what if I put a strong charge opposite to that of what lies on the outside of that copper conductor. Now the net electric field will be all messed up, won't it? And what if I put the same charge? How are any of those going to 'calibrate' and form a net zero field?

And for the cavity one, this rings even less bells. Why not gather on the inside of the cavity?

Furthermore, on that last one, won't the charges just fly apart?? What will make them all just magically stand in place? Or is it assuming that 'plastic coating' shell?

Any clues? I'm tired thinking about this, I need to move on. Any help would be greatly appreciated. Thanks!

Note by John Muradeli
2 years, 1 month ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Thank you all for responding!

@Soumo Mukherjee So it seems to me ultimately that the underlying assumption is the mere fact that the conductor is ALREADY in an electrostatic equilibrium. Right?? Then everything becomes crystal clear! You see, what I was (or maybe still am) having trouble with is, that I thought the law stated that it is impossible to have an extended period of time over which there is a net charge inside a conductor. I figure not. As soon as charges re-distribute, we're good to go.

@Agnishom Chattopadhyay

Here let me draw this:

See now? If we equally space the force due to charge area B to the left, and call it B', they cancel. But there is nothing to cancel A - A' is out of bounds! And same goes for the rest of the area of charge distribution within the shell; thus the net force is rightward.

But as of right now I don't even see how this is relevant - not every point is charged, since each atom within the conductor forms a net force of zero with its elemental counterpart. But as far as free electrons go, I can see this happening easily; if we charge the conductor, that is. And THEN, I'm assuming, THIS is EXACTLY why the electron would be driven to the surface!!! But now I have proven the complete opposite of what the example aimed to prove; I've proved that there IS an electric field, and it drives the charge outward.

Any counter-arguments? John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli The (free) electrons in conductor are movable.

Suppose two equal and opposite charges (\(e\) & \(-e\)) are uniformly distributed over two insulated spheres (balls) respectively. Then they are clamped at a separation. There exists an electric field between them.

Now release any one of them, it will move towards the other due to attractive electric force.

Same things happens inside a conductor.

Take a neutral conducting solid sphere. There are free electrons but the net charge on the conductor is still zero. now suppose due to some reason \(n\) electrons moves to a place \(B\) from place \(A\) within the conductor. The electron density increases at \(B\) and decrease at the site from where the \(n\) electrons got moved (i.e \(A\)). Creating two oppositely yet equally charged 'sites'. But will the \(n\) electrons stay there? No, they will immediately move to the electron deficient site \(A\) just like the 'balls'.

There is nothing at \(A'\) (in your figure) to exert a force. But lets imagine we place an ( arbitrary ) charge \(+Q\) there. Then the face facing the charge \(+Q\) will acquire \(-Q\) charge (due to induction). The opposite face however will acquire \(+Q\) charge. Due to \(+Q\) and \(-Q\) there will be an electric field inside the conductor? No.

The \(-Q\) charge on the opposite face of the conductor results due to accumulation of electrons there. And \(+Q\) due to deficiency of the electrons.The accumulation is, however, on the surface and the inside remains neutral. If there exists an electric field, there will also exist a corresponding electric force. This electric force will drive electrons. But why would it after the redistribution? This is similar to the "the conductor having a charge within a cavity" case in one of my previous comments. I am trying to find another link which can explain it more vividly. I will notify when I find one. Soumo Mukherjee · 2 years, 1 month ago

Log in to reply

@Soumo Mukherjee Hm I really liked everything you said before "There is nothing at A' ... But 'let's imagine'." Okay, we imagine, but that's beyond my point; my point is that the charge that is supposed to balance out the distribution in area A does not exist - hence the leftward charges prevail and generate a net electric force field line toward the right. Am I missing something?

And ultimately this should resolve it all; a sample problem:

Yeah a lot of reading but bear with me. Here's what I don't get: Why do the electrons 'have to' move out to compensate with the protons being distributed inside the shell? Is that one electron with charge -5\(\mu C\) insufficient? And how come the "positive charge on the inner wall cannot produce an electric field in the shell to affect the distribution of charge on the outer wall"? This makes no sense; they are as close as anything, so why wont the protons disturb the electric field to move electrons somehow? I wouldn't be complaining much if they were paired as dipoles - toe-to-toe - but protons are not uniformly distributed whereas electrons are, and they are not paired. And Coulomb's law tells me they must be paired!

Tackle THAT! John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli Okay, yet imagination could have helped. :)

Protons don't get redistributed. Only the electron does. Where the electron accumulates there is a site of negative charge. The site of deficiency of electrons gets positive charge. In short the electrons have to take the burden to move.

The positive charge on the inner surface equals the charge within cavity. So it's busy fighting with that.

We have to assume uniform distribution of mass. Hence of the protons (nucleus). The electrons are always in a state of motion. However, (if you imagine :D come on try at least this on) the no. of electrons crossing an (cross-sectional) area from one side equals the no.of electron crossing from other side in any giving time interval , giving no net current. Soumo Mukherjee · 2 years, 1 month ago

Log in to reply

@Soumo Mukherjee Sure. I can imagine that.

Though I still can't understand why electrons must move out in response to the positive charge in the cavity, and, why the inner-shell protons have no effect on rearranging the outer-shell electrons. John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli I could have replied to you then but my broadband link failed. So, I utilized my time in MS-paint. Lets see if me drawings can help or not. I wouldn't mind to try again. So if doubt persists feel free to tell.

  • If all electrons from B move to A, there will be an electric field between and hence the electrons will then try to go back to the previous configuration.

  • The "negative charge" must move out in response to the "positive charge" on the inner surface. Note that the "charge withing cavity is a negative charge not positive". The "negative charge" inside cavity induces equal "positive charge" on the inner surface by repelling the the electrons from there to the outer surface.

The net charge inside the green box is zero. The inner surface "positive charge" is busy balancing/canceling/nullifying/fighting the charge q inside cavity. That is why it will "have no effect on rearranging the outer-surface charge"

For a moment try thinking in terms of "charges"(positive charge & negative charge). May be the electron-proton model is causing a hindrance to understand.

"...if doubt persists feel free to tell." Soumo Mukherjee · 2 years, 1 month ago

Log in to reply

@Soumo Mukherjee WOah nice illustrations! And yes, some doubt yet persists; so I perfectly understood the concept in the first drawing. And I also perfectly understand what happens under an influence of an external electric field. And I really like the 'Neutral object with a cavity' drawing. But all that follows knocks me off a bit: so basically, you're telling me, that if we place a negative charge in the middle, in the cavity, then the free negative charge in the metal will be repelled to the surface, thus generating positive regions, which are then attracted to the negative charge in the middle. Right?

So now I guess the tougher question is, why don't the outer electrons merely fly out, if they are on the outer surface due to repulsion in the first place? And, why aren't the positive charges just bonding with the negative charge in the middle? What causes them align so perfectly as to generate a net electric field of zero inside?

And what if there was no cavity, in the exact same example? Oh, and that last issue isn't resolved - those positive charges on the inner shell - why cannot they influence the electric field on the outer electrons? They seem more than capable to do so.

Thanks and sorry for teh buggin! John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli Okay I am working on it. But I won't include any explanation which is beyond high school (or UG). Why electron's don't fly out can be (more satisfactorily) answered with Solid-State physics. We would require Mr. Mendrin for that. Soumo Mukherjee · 2 years, 1 month ago

Log in to reply

@John Muradeli

why don't the outer electrons merely fly out, if they are on the outer surface due to repulsion in the first place?

Do you mean why the electrons don't fly off the sphere? You have to remember that the air around the sphere is an insulator. For the electrons to pass through an insulator (air), it will need a huge amount of energy, which the electron doesn't have. The only way this could could happen is if there was an insane amount of charge on the sphere. This would lead to a large amount of electric potential energy on the electron which would allow it to escape.

And, why aren't the positive charges just bonding with the negative charge in the middle?

The "positive charge" is just the absence of electrons which were repulsed by the fixed negative charge. Since there are no electrons near the fixed negative charge, all that remains are the protons, which have a positive charge. Now the protons are present in the nucleus of an atom. Since the sphere is solid, the atoms/nucleus are fixed in their positions, due to intermoleculer bonds.. So the protons can't move towards the fixed negative charge. However, if the negative charge is extremely large, and the force it applies on the proton is larger than the bond forces between the atoms, the proton might break away. But this of course means that the negative charge must be abnormally large.

What causes them align so perfectly as to generate a net electric field of zero inside?

Are you talking about the protons? Nothing. They are fixed. It is the electrons that move. As to how they align so perfectly, I'll explain this point further in the next answer.

And what if there was no cavity, in the exact same example?

Are you talking about the negative charge example? Where would you place the negative charge if there is no cavity?

those positive charges on the inner shell - why cannot they influence the electric field on the outer electrons?

When we talk about the electric field on the outer electrons, we talk about net electric field. Granted, the positive charge influence the electric field, but you're forgetting that the outer electrons also influence the electric field. These cancel each other out.

The proof that the fields cancel each other out is in the images you give. I'll try to explain it. In a conducting body, we have two cases. Either the body eventually reaches electrostatic condition, i.e., no charge is moving (other than the random motion of electrons, but we ignore this since this cancels out.), or there is some circular flow of current. Experimentally, it has been observed that the second case doesn't happen (cop out, I know), so we are left with the first case. So there is no moving of charge. If there was an electric field \( E \), it would exert some force on the electrons, which would cause them to move. But we know that the electrons aren't moving. Therefore, there can't be any electric field \( E \) inside the conductor. Siddhartha Srivastava · 2 years, 1 month ago

Log in to reply

@Siddhartha Srivastava Woah I love these explanations! The quote-answer method is great! So, let me make minor qualifications on some of these:

"why don't the outer electrons merely fly out, if they are on the outer surface due to repulsion in the first place?"

So, what if, instead of air, we assume a vacuum? Will they not fly out because they are somewhat bonded with the conductor?

"And, why aren't the positive charges just bonding with the negative charge in the middle?"

Oh so you're saying the positive charge concentrations are induced by the absence of electrons, right? It's not that the protons aren't there, but it's that they weren't moved into that place - rather, electrons were forced out. Right?

"And what if there was no cavity, in the exact same example?"

Well, just assuming theoretically, say a large negative charge simply 'quantum leaped' or something - appeared inside a section of the sphere.

And the last one:

"those positive charges on the inner shell - why cannot they influence the electric field on the outer electrons?"

What do you mean by 'outer electrons'? The intermediate electrons between inner and outer shells? In that case, aren't those regions supposed to be electrically neutral? If it's the 'outer electrons' that are due to the excess charge - THE electrons - then well so they are being influenced by the protons. By generating a net zero field they are affected, and thus should be dragged inward. But I assume this is not so, and the electric field generated by the protons cancels out prior to reaching the outer shell. If so, I'm still not quite clear on how it occurs. A bit more detail, please?

Oh and that last paragraph is just what the book keeps tossing at me - experiment to explain theory. I personally prefer theory to explain experiment. Yeah sure I agree if charge is not moving electric field has to be zero - that's simply a consequence of the situation. But the mechanisms behind this equilibrating system is yet to be cleared up, and you're doing a great job at it so far ;D John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli

So, what if, instead of air, we assume a vacuum? Will they not fly out because they are somewhat bonded with the conductor?

When we talk about free electrons in a conductor, it doesn't mean that the electrons are completely free. We just mean that in the conductor, the electrons act like free electrons. They are still bonded to the all the atoms as a whole. Try reading Band Theory Simplified .

Oh so you're saying the positive charge concentrations are induced by the absence of electrons, right? It's not that the protons aren't there, but it's that they weren't moved into that place - rather, electrons were forced out. Right?

I'd use "caused" instead of "induced" there. But yes. The protons are fixed, the electrons are mobile. In normal conditions, the negative charge on the electrons an positive charge on the protons cancel out. But for example when a negative charge is brought near, the electrons are forced away while the protons stay, which causes a net positive charge.

Photo

Photo

... Link if pic doesn't show

The red is the protons. The blue are the electrons. That big blue dot is a negative charge. The postion of the red stays roughly the same. Ignore the stuff coming out of the lines. I suck at paint.

Well, just assuming theoretically, say a large negative charge simply 'quantum leaped' or something - appeared inside a section of the sphere.

Kind of the same as a cavity the size of the fixed negative charge. It pushes away electrons, leaving only the protons, which leaves a net positive charge near the negative charge.

What do you mean by 'outer electrons'?

What I meant by "outer electrons" was the extra electrons which gather on the surface.

Suppose we have a neutral conducting sphere. Then the number of electrons = number of protons = \( x \). The configuration is basically the first one I gave in the photo. Since the protons and electrons are equally distributed here, the filed due to the protons and the field due to the protons cancel out.

Now, suppose you add \( y \) more electrons to the sphere. Then the \( y \) extra electrons accumulate on the surface of the sphere (I'll prove this below.). So, in the end, the sphere has the same number of electrons and protons inside it as before (the surface doesn't count as inside the sphere).. Since initially the sphere exerted no field, in the end, the sphere will exert no field. So any field exerted is exerted only by the outer shell of electrons on the surface. Now, using this argument (taking the circle things A and B to be on the outer shell of surface electrons), we can prove that the field by the surface electrons inside the sphere is also zero.

Now, the proof that all extra charge accumulates on the surface is in principle the same as the surface tension proof. Assume all the \( y \) extra electrons are placed at random in the sphere. Total electrons = \( x + y \).Now, there will be random motion. For any electron which reaches the surface, it only experiences forces from inside the sphere.Since there are more electrons than protons, it experiences a net repulsive force, which causes the electron to stay there. For any electron not on the surface, (as long as the number of electrons in the sphere (not on surface) is greater than the number of protons \( x \)), the electron must be experiencing some repulsive force and thus is moving around.

So in effect, as soon as a electron reaches the surface, it is trapped. Any electron inside the sphere is constantly moving to different parts of the sphere, including the surface. So considering a single electron, moving over a long distance (not displacement), at some point it will reach the surface and get stuck and since the movement of electrons is at 75% of the speed of light, it happens fairly quickly.

But that would imply that every electron moves to the surface. This doesn't happen due to the qualifier I wrote, i.e (as long as the number of electrons in the sphere (not on surface) is greater than the number of protons \( x \)). Because as soon as the number of electrons in the sphere \( <= x \), the electrons stop experiencing a net repulsive force, which causes them to stop their movement. Since the movement is stopped, the electrons won't hit the surface and get stuck anymore. When is the first value \( <= x \)? Obviously \( x \). Now since \( x \) electrons are in the sphere, the remaining \( (x+y) - x \) electrons must be on the surface. Siddhartha Srivastava · 2 years, 1 month ago

Log in to reply

@Siddhartha Srivastava Ow dang one more thing: I don't think we ever fully covered this issue:

"those positive charges on the inner shell - why cannot they influence the electric field on the outer electrons?"

You've stated that the electrons also influence the field and thus the fields cancel. But the protons are thus in turn influencing the electrons' electric field now, aren't they? Oh and probably the biggest bugger of all, why aren't those electrons attracted to the protons?? They have all the reason to do so, after all. Oh wait let me guess - it's because there's that negative charge in the cavity (or wherever inside the conductor), isn't it? Just confirming. John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli

it's because there's that negative charge in the cavity

Exactly. Siddhartha Srivastava · 2 years, 1 month ago

Log in to reply

@Siddhartha Srivastava Perfect. Thanks for all your help. You've earned yourself a follower ;) John Muradeli · 2 years, 1 month ago

Log in to reply

@Siddhartha Srivastava Woah I think we should like synthesize all this info into one well-written article on net charge in a sphere being zero. This is great! But you see, to be honest, I really had no problem with the sphere to begin with. I had a problem with the fact that this reasoning applies to ALL shapes of conductors - hence the irregular shape used in the original note. But I can see how this could apply to all the conductors now - and just to confirm, it would, wouldn't it? Someone should really animate all those scenarios for my questions you've quoted - that would be great.

Now one last question with Gauss' Law. So, flux is a measure of how much 'stuff' enters and leaves a particular region of space. But one 'fact' using Gauss' Law makes no sense to me, and all the conductor net zero confusion arose from this: If net flux is zero, electric field is zero.

Now that's really bugging. Basically what this is saying is, that if I draw a Gaussian surface just outside an electron, then NO NET ELECTRIC FIELD EXISTS THERE! But that's donkey! Of course there is! And the worst part is, the equation confirms this reasoning:

Here's that whole argument below, especially Surface 3:

The flux part of it makes sense, but not the electric field part. Nope. Not at all.

Any clue? I would greatly appreciate some input. John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli OOOOOOOOOOOOOOOOOOOH!!! Whoa! Just noticed something: "Gauss law' requires that the net \(\huge FLUX \) of the electric field through the surface of this field be zero." Ah-ha!! Now it all makes sense!!! Missed ONE word and the entire concept got destroyed. Alright wow haha okay. This is so much better now.

And all our above discussions explained the mechanics of why the excess charge re-distributes itself along the surface. And once it's on the surface it's already crystal clear why all excessive charge is on the surface and why not only net electric field FLUX inside is zero, but why the net electric field is zero. Good stuff. John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli @John Muradeli I highly recommend reading Ed Purcell's intro to Electro Magnetism(The Morin edition) along with the standard text you are reading. It is a beautiful book. Thaddeus Abiy · 2 years, 1 month ago

Log in to reply

@Thaddeus Abiy Thanks, I shall look at it when I'm done with the course (right now I'm as packed as possible with what I have at hand!) John Muradeli · 2 years, 1 month ago

Log in to reply

If a conducting material is placed in an external electric field, the free electrons respond to the field. There is a charge redistribution due to the field. The free electrons inside the conductor move opposite to the direction of the external field. As a result one face gets negatively charged and the opposite face positively charged. Such charge redistribution produces an extra electric field ( inside the conductor) which is equal and opposite to the external field. The net electric field inside the conductor is then zero. Thus, a steady state is reached in which some positive and negative charge appear at the surface and there is no electric field inside the plate.

Now we can say 'According to Gauss's law since the electric field inside the conductor is zero, the charge enclosed by the Gaussian surface should also be zero."

A) If charge is injected anywhere in the conductor

B) Conductor has a cavity

C) A charge is placed within the cavity

for A and B we can argue similarly.

for C : if there is cavity, then there will be a inner surface. If a charge is placed within that cavity, the inner surface cannot be charge free. The inner surface would acquire charge equal but opposite to the one placed within the cavity. Now the net charge within any Gaussian surface enclosing the the cavity and a part of the conductor will be zero. Soumo Mukherjee · 2 years, 1 month ago

Log in to reply

I've seen this argument most convincing.

And yet, I can easily counter-argue:

Why not consider an area towards the left just as far away as it is on the right? Then cancellation occurs right away. Furthermore, all that leftward charge area that remains will not only cancel the charge on the right, but entirely overpower it and create a net electric field line directed rightward. John Muradeli · 2 years, 1 month ago

Log in to reply

@John Muradeli Have you proved the fact by yourself that electric field due to uniformly charged spherical shell inside the shell is zero.

What arguments do you use to prove it. Ronak Agarwal · 2 years, 1 month ago

Log in to reply

@John Muradeli " A net electric field line directed rightward" would exert force on the electrons leftward until a charge redistribution is achieved giving zero net electric field inside it. Soumo Mukherjee · 2 years, 1 month ago

Log in to reply

@John Muradeli Yep, that is the most convincing argument. We use this argument in the gavitation wiki too .

There is an argument using Gauss Law, however it requires knowing that the field is spherically symmetrical inside the conductor Agnishom Chattopadhyay · 2 years, 1 month ago

Log in to reply

@Agnishom Chattopadhyay I agree that Gauss's law is a good way to look at it. If there is one thing that is always reliable in Electrodynamics, it is Gauss's law. If you accept that the charges accumulate on the surface then drawing a spherical Gaussian surface inside the conductor will tell you that there is zero flux. Why? Because the area of the sphere is obviously not zero then the field has to be zero. Thaddeus Abiy · 2 years, 1 month ago

Log in to reply

@Agnishom Chattopadhyay So, what about my counter-argument? This is far more convincing if you ask me. John Muradeli · 2 years, 1 month ago

Log in to reply

Log in to reply

@John Muradeli I do not understand your counter-argument. Agnishom Chattopadhyay · 2 years, 1 month ago

Log in to reply

Charge spreads out to minimize the potential energy of the system as any system wants to have minimum potential energy So all charge spreads out at the surface Akshit Srivastava · 1 year, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...