Electric field in metal cavity

[radiogaga35]

Suppose one has a solid metal body (in static eq'm) with a cavity inside it.

Assume the cavity has no charge in it. This is how one could prove that there are no fields inside the cavity: choose two points A and B, both in the solid part of the metal body. Follow a path through the metal (but not through the cavity) - since the electric field is zero throughout the metal, the path integral of the field (i.e. the potential difference) must be zero. Now take a path from A to B that goes THROUGH the cavity -- using path independence of potential diff., the p.d. must for this path also be zero. Hence, no fields in the cavity.

My question is this: why does this same argument not apply to a cavity when you DO have a charge inside it (i.e. you place a charge inside the cavity?). I know that using Gauss' Law, it is easily shown that there will be an equal and opposite charge distributed on the walls of the cavity (though perhaps this is irrelevant?). However, there are fields within the cavity...so where does the "path independence of p.d" fall apart?

Thanks

 

[radiogaga35]

(Apologies - in retrospect I realize that this post will have been better suited to the "homework questions" board...it's not a homework question though, so I didn't think to post it on that board)

 

[mgb_phys]

If you have a charge on the walls of cavity and they are connected so they form an equipotential then there isn't a field inside the cavity. It's exactly the logic you followed.

 

[radiogaga35]

Ok, so there is a point charge inside the cavity. Now suppose the cavity is spherical and the charge is distributed uniformly on its walls - shell thorem/Gauss's Law says that there is no field due to this shell of charge, inside the cavity. However, from an observation location inside the cavity, surely there will exist a field due to the point charge?

(Sorry, perhaps I'm missing something obvious here) Thanks for the reply...

 

[Doc Al]

My question is this: why does this same argument not apply to a cavity when you DO have a charge inside it (i.e. you place a charge inside the cavity?). I know that using Gauss' Law, it is easily shown that there will be an equal and opposite charge distributed on the walls of the cavity (though perhaps this is irrelevant?). However, there are fields within the cavity...so where does the "path independence of p.d" fall apart?

It's not the path independence that falls apart, it's the argument that that implies no field that falls apart. With no charge within the cavity, the field lines--if they exist--must go from positive to negative charges on the inner surface. If there were such surface charges, then you would fail path independence because the potential difference must be non-zero going from + to - charges. Thus there's zero field within the cavity.

But with charges within the cavity, you can have path independence even with inner surface charges because the charge within the cavity contributes to the field.

Ok, so there is a point charge inside the cavity. Now suppose the cavity is spherical and the charge is distributed uniformly on its walls - shell thorem/Gauss's Law says that there is no field due to this shell of charge, inside the cavity. However, from an observation location inside the cavity, surely there will exist a field due to the point charge?

The charges would only be uniformly distributed on the inner surface if the point charge is centered in the spherical cavity. In which case, no problem: within the cavity you'll have the field from a point charge. (Realize that in going from point A to point B, where both points are equidistant from the point charge, the potential difference is zero.)

But if the point charge is not centered, things are more interesting. The charges on the inner surface are not uniformly distributed, but path independence still holds due to the field contribution of the point charge. (And as you know, the field at points outside the inner surface due to the inner surface charge plus the point charge must be zero.)

Let me know if it's making sense.

 

[zoobyshoe]

My question is this: why does this same argument not apply to a cavity when you DO have a charge inside it (i.e. you place a charge inside the cavity?). I know that using Gauss' Law, it is easily shown that there will be an equal and opposite charge distributed on the walls of the cavity (though perhaps this is irrelevant?). However, there are fields within the cavity...so where does the "path independence of p.d" fall apart?

Thanks

The notion of a charge in a cavity in a metal object is new to me. My understanding was that charge always moves to the outer surface in a conductor:

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ShellsConductors.xml

Under electrostatic conditions, field lines must terminate or begin on the surface of a conductor - that is, there is no electric field within a conductor. If any field line penetrated into the conductor, then electrons would respond to its presence and be accelerated within the conductor. If that happened, the conductor would no longer be under electrostatic conditions.

So, I'm curious to learn under what circumstances one can charge the walls of a cavity in a metal object.

 

[radiogaga35]

Ah, ok, thanks a lot Doc Al that makes sense!

@zoobyshoe: I never really gave that much thought (guess it would be ironic to call this a "thought experiment" then!). I've seen the idea mentioned in many texts (e.g. Feynman Lectures Vol. II, Chapter 5 I think). This is how I think it may be possible: have some charge, say a point charge, and "build" the metal body (and cavity) around it.

Now, consider a Gaussian surface that has edges only inside the metal, but one that completely encloses the cavity. Since metal is in electrostatic eq'm ==> no fields inside the metal itself ==> zero flux in/out of surface ==> zero charge enclosed in the surface. But, since there was charge there to begin with, one can deduce that there must be an equal and opposite charge distributed on the cavity walls.

If the metal body was initially neutral, then an equal and opp. charge to that on the cavity walls (i.e., a charge equal to the original charge in the cavity) will reside on the outer surface of the metal.