Undergrad The 'outer surface' should be redefined when applying Gauss' law?

[feynman1]

Gauss' law dictates that charge will only appear on the outer surface of a conductor. But if there's charge in a conducting cavity, the inner surface of the conductor will accumulate induced charge. So what's outer or inner should be redefined?

 

[troglodyte]

there will be no field inside a closed cavity!The charge densities lies on the surfaces on every layer and have the same sign(Why?).The net flux through this cavity should be zero.Its on you what you want to measure with the Gaussian theorem.For example,if you wanted to measure the field contribution of a charge density on the surface where the electric field pointed radial outwards in all direction then the E-field decreases with increasing distance.So,find a way in your integral representation where this facts pop out.
One remark to the Gaussian theorem: When you measure with Gaussian theorem you create imaginary ,closed ,oriented Gaussian surfaces layers to measure the net flux which goes through a surface .

https://www.feynmanlectures.caltech.edu/II_05.html

 

[vanhees71]

Of course, if there is a charge inside the cavity there will be a non-zero field inside the cavity. There's no field inside the conductor (for the static case, i.e., if no currents flow within the conductor), and this is because the charge inside the cavity induces a surface-charge distribution on the inner surface of the conductor such that the field of the charge is canceled precisely by the field due to this induced surface-charge distribution. It is given by the jump of the normal component of the electric field at this surface.

 

[feynman1]

there will be no field inside a closed cavity!The charge densities lies on the surfaces on every layer and have the same sign(Why?).The net flux through this cavity should be zero.Its on you what you want to measure with the Gaussian theorem.For example,if you wanted to measure the field contribution of a charge density on the surface where the electric field pointed radial outwards in all direction then the E-field decreases with increasing distance.So,find a way in your integral representation where this facts pop out.
One remark to the Gaussian theorem: When you measure with Gaussian theorem you create imaginary ,closed ,oriented Gaussian surfaces layers to measure the net flux which goes through a surface .

https://www.feynmanlectures.caltech.edu/II_05.html

Thanks. I actually meant there was a charge artificially put in the hollow region of the cavity. Then there'll be induced charge on the inner surface of the cavity to keep the electric field to be 0 within the conducting cavity.

 

[feynman1]

Of course, if there is a charge inside the cavity there will be a non-zero field inside the cavity. There's no field inside the conductor (for the static case, i.e., if no currents flow within the conductor), and this is because the charge inside the cavity induces a surface-charge distribution on the inner surface of the conductor such that the field of the charge is canceled precisely by the field due to this induced surface-charge distribution. It is given by the jump of the normal component of the electric field at this surface.

That's right. But I was asking to redefine 'charge only on outer surface', as the inner surface has charge.

 

[Dale]

Gauss' law dictates that charge will only appear on the outer surface of a conductor

Where did you read that? Gauss’ law does not dictate that in general. Under specific additional assumptions plus Gauss’ law you can get that result, but not just from Gauss’ law.

 

[feynman1]

Where did you read that? Gauss’ law does not dictate that in general. Under specific additional assumptions plus Gauss’ law you can get that result, but not just from Gauss’ law.

Sorry I meant to refer to isolated conductors. So isolation+Gauss' law give charge only on 'outer' surface. Though If there's charge in the hole of a cavity, the inner surface of the cavity, in the eye of the charge in the hole, could also be called an 'outer' surface, in my opinion?

 

[etotheipi]

the inner surface of the cavity, in the eye of the charge in the hole, could also be called an 'outer' surface, in my opinion?

It's best to stick to less ambiguous terms in communications; I don't think I would call the inner surface that. What's wrong with just saying charges reside on the surface of a conductor in equilibrium? This covers both the outer and inner face.

 

[Dale]

Sorry I meant to refer to isolated conductors. So isolation+Gauss' law give charge only on 'outer' surface. Though If there's charge in the hole of a cavity, the inner surface of the cavity, in the eye of the charge in the hole, could also be called an 'outer' surface, in my opinion?

This is not correct. Can you please post your source for this? Either the source is wrong or you are misunderstanding the source.

The correct claim should be that in an electrostatic situation all charge resides on the surface of a conductor. There should be no involvement of isolation and no outer or inner.

 

[etotheipi]

This is not correct. Can you please post your source for this? Either the source is wrong or you are misunderstanding the source.

I think OP understands but the 'outer'/'inner' terms he uses are just slightly misleading.

I suspect he meant something along the lines of a charge placed in the cavity of a conductor will result in a surface charge on the inner surface of the conductor around the cavity, but no charge in the cavity means no charge on that inner surface.

 

[hutchphd]

Gauss' law dictates that charge will only appear on the outer surface of a conductor.

This premise is simply not true. Why are we discussing its ramifications?
.

 

[vanhees71]

That's right. But I was asking to redefine 'charge only on outer surface', as the inner surface has charge.

I don't understand the question then. There's of course also induced charge on the outer surface and you can of course also put additional charge on it too, but that won't change the field inside the cavity (in the conductor it stays 0 anyway).

 

[vanhees71]

This is not correct. Can you please post your source for this? Either the source is wrong or you are misunderstanding the source.

The correct claim should be that in an electrostatic situation all charge resides on the surface of a conductor. There should be no involvement of isolation and no outer or inner.

If there's a charge inside the cavity it's there and not on the surface of the condutors though!

 

[Delta2]

Gauss' law dictates that charge will only appear on the outer surface of a conductor.

What you saying here is not 100% correct. You must remove the word "outer" and add "in electrostatic equilibrium" at the end. So the correct statement is:
"Gauss's law dictates that charge will only appear on the surface of a conductor in electrostatic equilibrium"

 

[feynman1]

Thanks a lot everyone. Let me clarify. I meant to say:
'Gauss' law says under electrostatic equilibrium charge only appears on the surface of a conductor and that charge only appears on the outer surface of an isolated conducting cavity'.

 

[Delta2]

Thanks a lot everyone. Let me clarify. I meant to say:
'Gauss' law says under electrostatic equilibrium charge only appears on the surface of a conductor

Fine so far

..and that charge only appears on the outer surface of an isolated conducting cavity'.

You don't have to say the above. By not putting the word "outer" or "inner" in the first half (before the "and") of the complete quoted sentence, you incorporate the case of the isolated conducting cavity, because the outer surface of the conducting cavity, is the inner surface of the conductor.

 

[feynman1]

Please find references:
www.quora.com/What-do-people-mean-by-an-isolated-conductor

But now I realize I missed saying another overlooked condition: curlE=0. So I hereby clarify that under electrostatic equilibrium the conditions for charge only residing on the outer surface of an isolated conducting cavity are Gauss' law+curlE=0.

 

[feynman1]

Fine so far

You don't have to say the above. By not putting the word "outer" or "inner" in the first half (before the "and") of the complete quoted sentence, you incorporate the case of the isolated conducting cavity, because the outer surface of the conducting cavity, is the inner surface of the conductor.

Nice. So in your opinion, inner surface of cavity is actually outer in the eye of its surroundings?

 

[Delta2]

Nice. So in your opinion, inner surface of cavity is actually outer in the eye of its surroundings?

Yes you can say that.

 

[Delta2]

Please find references:
www.quora.com/What-do-people-mean-by-an-isolated-conductor

But now I realize I missed saying another overlooked condition: curlE=0. So I hereby clarify that under electrostatic equilibrium the conditions for charge only residing on the outer surface of an isolated conducting cavity are Gauss' law+curlE=0.

I am not so sure here but I think if $curlE\neq 0$ then there would be a time varying magnetic field, hence the electric field would also be time varying, hence the conductor can't reach electrostatic equilibrium.

 

[Dale]

Please find references:
www.quora.com/What-do-people-mean-by-an-isolated-conductor

So it seems that you correctly understood your source. However, this source is incorrect. I would recommend using a different source, such as a good textbook.

 

[feynman1]

I am not so sure here but I think if $curlE\neq 0$ then there would be a time varying magnetic field, hence the electric field would also be time varying, hence the conductor can't reach electrostatic equilibrium.

I'm only considering electrostatics, so curlE=0 is automatic

 

[Delta2]

I'm only considering electrostatics, so curlE=0 is automatic

Again , I am not 100% sure , but since we say that the conductor is under electrostatic equilibrium we don't need to include that $Curl(E)=0$ it is redundant.

 

[vanhees71]

Thanks a lot everyone. Let me clarify. I meant to say:
'Gauss' law says under electrostatic equilibrium charge only appears on the surface of a conductor and that charge only appears on the outer surface of an isolated conducting cavity'.

I thought you consider the case of a cavity, containing some charge. Then you have an induced surface charge on the inner boundary surface as well as on the outer. You can also put extra charges on the outer surface.

The one very illuminating example you can quite easily solve analytically is a spherical shell of finite thickness. Here you can use the method of image charges (with only one image charge) to calculate the Green's function, which then covers all cases of given charge distributions inside and outside the cavity as well as arbitrary additional surface charge on the outer boundary. It's very helpful to study this example carefully though it's a bit of work to get all the algebra right!