Electrostatic field inside a conducting shell

[asmani]

Hi all

Suppose we have a conducting shell (which is not necessarily a sphere) and no electric charge inside.
Is the electrostatic field inside the conducting shell always zero? (even if there is some charges outside the shell)
What if the shell is grounded?
If the answer is yes, how it can be proved?

(It's not a homework or something.)

Thanks

 

[shoestring]

I think it has to do with the uniqueness of solutions to Laplace's equation.

The potential in the conductor has to be constant, because otherwise there'd be a field and a current in the conductor. A constant potential in the volume enclosed by the shell will satisfy both Laplace's equation in the charge-free interior, and the boundary condition of being constant at the conductor. Because of the uniqueness of solutions, there's no need to look for other possible solutions, because there won't be any.

The field is then calculated as the negative gradient of the potential. The gradient of a constant function is zero, so the field will be zero.

 

[asmani]

Thanks a lot, I forgot about the Laplace's equation.

For someone who don't know the proof of the uniqueness theorem, isn't there a simpler solution rather than proving the uniqueness theorem?

 

[shoestring]

I think there's also a rule that says that the potential can't have any local maximum or minimum in a charge-free domain. Any maximim or minimum, if they exist, must be on the boundary of the domain. But we already know that the potential is constant there. That proves that the potential is constant throughout the interior, and from that follows that the field must be zero.

It's still kind of a mathematical argument, based on properties of solutions to Laplace's equation. A more physical approach is to use Gauss's law:

A local maximum of the potential anywhere in the interior means that there'll be a field radiating out from that area, because the field is given as the negative of the gradient of the potential, and the potential decreases in every direction away from the local maximum. Applying Gauss's law to a small closed surface surrounding a local maximum gives a positive value of the integral, indicating the presence of a charge, but that contradicts the assumption that there's no charge in the interior. So the potential must be constant, and the field zero.

 

[yungman]

If you have the Griffiths "Introduction To Electrodynamics" 3rd Edition. P 100 to 101. It explain why if there is no charge inside the hollow conductor, then there would be no field in the cavity.

Basically is the inner walls has to have equal potential difference or else a field will be present and cause the electrons on the surface to move so it cancel any potential. If no potential difference inside the cavity, there will be no electric field.

The book can say it much better, I am no expert but I happened to write down in my notes with page number.

 

[asmani]

Thanks a lot.

2 related questions:

1. Can we enclose a charge with a charge-free material to reduce the intensity of electrostatic field outside the material? Is there exist such material?
I think theoretically the answer is negative, but in practice, does electrostatic field pass any material with no reduction?
I mean we found a way to block electrostatic field on a finite and closed space, can we do that on infinite space?

2. We concluded the electrostatic field inside a charge-free conducting shell is zero.
Is there exist such thing like that shell to make magnetostatic field zero?

 

[yungman]

Thanks a lot.

2 related questions:

1. Can we enclose a charge with a charge-free material to reduce the intensity of electrostatic field outside the material? Is there exist such material?
I think theoretically the answer is negative, but in practice, does electrostatic field pass any material with no reduction?
I mean we found a way to block electrostatic field on a finite and closed space, can we do that on infinite space?

2. We concluded the electrostatic field inside a charge-free conducting shell is zero.
Is there exist such thing like that shell to make magnetostatic field zero?

1) With a conducting shell, the total charge will appear on the outer surface of the shell. With a lossy dielectric, varying E field supposed to be attenuated when passing through, it is freq. dependent. The total charge is still zero in the dielectric and the charge still reflect at the outer surface. In practical situation, we grounded the out conducting shell and the problem solved. But I take that you don't mean that. You just want some material that by nature block the static E without hooking to anything.

2) Not much knowledge on this one. I heard on some kind of material call "nu" metal ( not even sure about the spelling) that we used in CE testing before to cut down emmition. Other than that, I have no comment on this.

Sorry can't be of any help.

 

[shoestring]

2. We concluded the electrostatic field inside a charge-free conducting shell is zero.
Is there exist such thing like that shell to make magnetostatic field zero?

All I can think of is a superconductor, which behaves like a perfect diamagnet and because of that expels a magnetic field. Otherwise, diamagnetism is usually quite weak.

 

[shoestring]

Here's a Q&A about magnetic field blocking materials that I found online: http://www.Newton.dep.anl.gov/askasci/phy00/phy00508.htm".

It mentions coaxial cables as a way to shield magnetic fields from a current, but that works only because the shield is part of the circuit. I.e. it's not the magnetic field itself that induces the outer current that creates the opposing magnetic field, but rather one of Kirchhoff's laws, and there's no guarantee that the magnetic field in the outer condctor will be zero. In fact, t's only at the outer boundary of the outer conductor, and outside, that the enclosed net current will zero, so the magnetic field won't be zero anywhere inside the coaxial cable (well, at the very center and at the boundary, but not throughout any measurable volume in the coaxial cable). If the outer conductor blocks the magnetic field it's not because of any material properties other than being a conductor.

 

[Born2bwire]

Here's a Q&A about magnetic field blocking materials that I found online: http://www.Newton.dep.anl.gov/askasci/phy00/phy00508.htm".

It mentions coaxial cables as a way to shield magnetic fields from a current, but that works only because the shield is part of the circuit. I.e. it's not the magnetic field itself that induces the outer current that creates the opposing magnetic field, but rather one of Kirchhoff's laws, and there's no guarantee that the magnetic field in the outer condctor will be zero. In fact, t's only at the outer boundary of the outer conductor, and outside, that the enclosed net current will zero, so the magnetic field won't be zero anywhere inside the coaxial cable (well, at the very center and at the boundary, but not throughout any measurable volume in the coaxial cable). If the outer conductor blocks the magnetic field it's not because of any material properties other than being a conductor.

The shield is only effective for an AC signal, best if it is in the microwave frequencies. Something like 60 Hz off of the mains can still bleed through but it is much much better than a simple twisted pair (God... I remember an audio amplifier where they ran the mains power by a signal tube to the power switch in front. I had that replaced with a coax to get rid of the induced noise.). In general, RF and higher can be efficiently blocked using an appropriate shield made of a good conductor like copper. This works by being able to induce currents on the surface of the shield to cancel out the incident electromagnetic radiation.

A coax cable does have a good geometric arrangement in that it can still provide good shielding for low frequency or static currents if you use the outer shield as the return line. In this case, the inner conductor and outer shield have equal and opposite total currents and from Ampere's Law we would know that the net magnetic field is zero. This is a no brainer for static currents but if you have an AC signal then you can still have problems if the outer conductor is used as a return line since the phases of the currents can be off and due to how the current flows on the outer conductor.

However, for static or very low frequency fields, the magnetic field becomes decoupled from the electric field and a good conductor no longer provides any worthwhile shielding. Usually we can try a large mass of a high permeability material. This could be ferromagnetic (iron) or a composite of a class of materials called high-mu materials. These do not completely remove the field but at best concentrate and draw the field away from the target volume. Generally though this requires a large amount of material surrounding the object for it to work. A more convenient solution is to generate another magnetic field that cancels out the undesirable field. This is often done in speakers that are touted as being magnetically shielded.

 

[yungman]

The shield is only effective for an AC signal, best if it is in the microwave frequencies. Something like 60 Hz off of the mains can still bleed through but it is much much better than a simple twisted pair (God... I remember an audio amplifier where they ran the mains power by a signal tube to the power switch in front. I had that replaced with a coax to get rid of the induced noise.). In general, RF and higher can be efficiently blocked using an appropriate shield made of a good conductor like copper. This works by being able to induce currents on the surface of the shield to cancel out the incident electromagnetic radiation.

A coax cable does have a good geometric arrangement in that it can still provide good shielding for low frequency or static currents if you use the outer shield as the return line. In this case, the inner conductor and outer shield have equal and opposite total currents and from Ampere's Law we would know that the net magnetic field is zero. This is a no brainer for static currents but if you have an AC signal then you can still have problems if the outer conductor is used as a return line since the phases of the currents can be off and due to how the current flows on the outer conductor.
I think the phase on the shield is equal and opposite. This is not like traveling wave that it travel all the way forward and reflect back. I believe the charge on the outer shield of the coax is induced directly by the inner conductor and is always equal and opposite ideally.
However, for static or very low frequency fields, the magnetic field becomes decoupled from the electric field and a good conductor no longer provides any worthwhile shielding. Usually we can try a large mass of a high permeability material. This could be ferromagnetic (iron) or a composite of a class of materials called high-mu materials. These do not completely remove the field but at best concentrate and draw the field away from the target volume. Generally though this requires a large amount of material surrounding the object for it to work. A more convenient solution is to generate another magnetic field that cancels out the undesirable field. This is often done in speakers that are touted as being magnetically shielded.

I am no expert, In all your cases, I think the shield serve as the return. theoretically, it should cancel out the fields of the inner conductor, but in real life, there are leakages. maybe you are onto something about the low frequency leakage. I remember when I design femto (10EE-15) amp Faraday Cup amplifier which is very low frequency, we have to use double shielded coax with one shield made of foil or something.

regarding to my assertion that the return current on the shield is actually induced by the inner conductor, I reference from "Field and Wave Electromagnetics" by david K Cheng 2nd edition, page 430 on the explanation of the parallel plate transmission lines. It explain how the E and H wave look like and the induced surface charge and surface current density. They are equal and opposite phase ideally. That is the reason even in the case of microstrip, 95% or so of the image(return) current is right underneath the signal trace within 4 trace width from each side of the trace. We use this characteristic to perform a lot of the layout in high signal ingrity circuit layout.

 

[Born2bwire]

I am no expert, In all your cases, I think the shield serve as the return. theoretically, it should cancel out the fields of the inner conductor, but in real life, there are leakages. maybe you are onto something about the low frequency leakage. I remember when I design femto (10EE-15) amp Faraday Cup amplifier which is very low frequency, we have to use double shielded coax with one shield made of foil or something.

It doesn't have to. You can do a balanced wiring configuration where the shield is set to a ground but you run two signal wires and take the differential to find your signal (this also has the added bonus of being very good at rejecting common mode noise, especially if you run the balanced lines close to each other). Or you could run a coax with the shield set to Earth ground (or floating), the inner core to signal, and run this above a plate that is set to circuit ground as your return. This version of course purposely negates any shielding properties of the coax but I merely suggest it as an example where one can easily defeat the shield.

You can also run into problems even if you are propagating an electromagnetic wave. For example, in an ideal infinite coaxial waveguide, the guided wave is fully contained between the outer conductor (shield) and the inner conductor (core). However, problems arise when you have a finite coaxial cable because of the termination of the waveguide. What can happen is that the induced currents on the inner surface of the outer conductor can leak out onto the outer surface of the outer conductor. Just simply cut the end off of a coax cable and now currents can run from the inner side of the shield out onto the outer side of the shield at the termination. This causes a lot of problems (one of the reasons why you need a balun when using a coaxial connection to various antennas) and I can imagine it as giving rise to some radiation as once it travels away from the termination it is no longer linked to the reflected wave on the interior of the coaxial cable.

For lower frequency signals you need to have a better shield on the coax due to the increase in the skin depth. A simple and cheap coax usually just has a thin foil as the shield. This is flexible and provides good coverage around the inner conductor. However, the shield is very thin and has finite conductivity. To counteract lower frequencies of operation, you can improve the shielding by using a braid. A braid allows for a physically thicker shield (generally of better material than say aluminum (copper)) while maintaining flexibility. Of course this comes at a compromise of having air gaps and thus not being a perfectly solid shield. Better cables will be a combination of foil and braid or multiple layers of braiding.

EDIT: Oh yeah, mentioning baluns, one could also cause problems by introducing say a shunt to the coaxial shield. This could (roughly) cause a phase shift in the exterior currents vis-à-vis the interior currents due to the difference in the electrical lengths travelled. Of course when we make a balun out of coaxial cables the length of the shunt is designed to prevent this from happening.