Electrostatics for dummies - Faraday Cage

[JamesC]

Hi all, there is recent talk on a faraday cage on a couple of posts and I wanted to firm up my understanding.

In the attached diagram there are 6 ions, 2 on either side and 2 inside a faraday cage. The red lines represent the coulomb repulsion between the ions. Which is the correct diagram A,B,C? or ( D - none over the above )

Should the faraday cage be grounded? If not then if it is grounded do it's properties disappear?

Thanks,
James

[Chris Bradley]

None of the above.

The ions within the cage will experience their own independent effects (as in A). This is the idea and purpose of a Faraday cage, it replicates a 'universe' free of any other e-fields except that generated within.

The ions outside the cage, on both sides, will affect each other. The e-fields of the outer ions will be effectively 'channeled' around the cage, if you want to think of it like that.

Equally, the ions on the outside are in their own 'e-field universe' and also isolated from the effects of what is within the Faraday cage. So a Faraday cage might be defined as splitting the universe into two parts whose e-fields become independent of each other.

[JamesC]

So more like this ( attached ) - the faraday cage makes 2 universes out of it?

What about grounding the faraday cage, does it make a difference?

Thanks,
James

[Steven Sesselmann]

Chris, or anyone else who cares to give it a shot, I have another question.

Let there be two identical Faraday cages, large enough to accommodate a scientist and some lab equipment, and insulate these two Faraday cages from ground.

On all the lab equipment and measuring equipment that the scientists work with, the ground connection has been firmly connected to the cage.

Now let the scientists Albert and Michael enter their labs, shut the cage door.

Each cage is now charged separately, Alberts cage to -100 KV and Michaels cage +100KV with respect to the outside ground (the cages stand on insulating feet)

Albert and Michael sitting inside each cage of course feel no effect of the charge.

Now instructions go out to both scientists, that they should produce an ion of charge +1

Both say no problem, we know how to do that

Both scientists achive their goal and produce an ion beam, then use a mass selection magnet to prove that the ions have a charge of +1.

Now, is Alberts ion more negative than Michael's ion?

Steven

[Chris Bradley]

More that the e-field lines take a circuitous route around the cage. I dare say there are plenty of ways of describing a Faraday cage in terms of its potential/charge, I would venture a summary as saying that such a cage needs to be able to maintain a uniform charge (that pretty much means its conductive) and that its total net charge doesn't change. The easiest way to do this is to make it out of metal and connect it up to the largest body of charge you can find, say, the earth.

[Chris Trent]

A beautiful twist on relativity Steven.

Well put.

[gabrielArgentina]

Steven inside a Faraday cage you live in a neutral universe...not matter at what MVolts potential are!....any charge will be ,on out side surface of the cage .
Then the atoms for make ions are neutral too, and the ions will be exactly the same for both scientist ....if they remove or add the same electrons amounts..
Gabriel.

[Carl Willis]

Hi James,

The interior of a Faraday cage is only field-free when there is no charge enclosed (Gauss' Law). Typically that is its function--to shield the charge-free interior from exterior charges. But in this case, some units of charge are enclosed. So there will be forces felt by the charges inside toward the cage surface, in addition to forces felt between ions outside the cage and the cage surface. The only thing that can be said with certainty is that the line representing your cage will be an equipotential surface, or alternatively that all electrostatic forces will be normal to its surface on contact.

"Grounding" your equipotential surface only explicitly defines its electrostatic potential, but does nothing to change the magnitude or direction of forces felt between the charges.

Attempting to draw a field map for this configuration by hand is subject to a lot of inaccuracy. In conclusion, Drawing C is the most conceptually accurate (the charges inside and outside the box DO influence each other), though it misses the essential character of the box as an equipotential surface.

-Carl

[JamesC]

Carl,

>The interior of a Faraday cage is only field-free when there is no
>charge enclosed (Gauss' Law). Typically that is its function--to shield
>the charge-free interior from exterior charges.

Thanks! - The lights just went On! - The charge free thing is the missing piece.

I am developing a software simulation based on N-N point charges ( I am not clever enough to implement laplace equations etc ) and have been having good success except when dealing with things like ground planes and faraday cages. For example I have been experimenting with a virtual cone trap and it works but seems overly sensitive. A real cone trap has a grounded drift tube which looks a bit like a faraday cage in the center and I was wondering if I could just partition the simulation or if I need to work something else out.

Partioning ( ie two worlds, ions in and ions out of the drift tube ) is probably a good first pass but.. I think am going to try to simulate a metal by a physical boundary with distributed fixed positive charges and equal number of free to move negative charges. I think the negative charges will distribute within the virtual metal to model to a first pass of accuracy the behaviour of the faraday cage.

Hopefully that will better model the resulting electric fields without having to figure out something more complicate. Hopefully I shuold then get better modeling of the cone traps etc.

Why do a sim?? Well a) Understand whats going on an b) I want to be able to dynamically control grid voltages etc based on feedback inside the simulation and don't know other software that can do it.

[Chris Bradley]

I was hoping to have covered those points with subtlety by my description. Perhaps to much subtlety! If there isn't a net charge in the cage then some process will likely occur to pull charge off the cage until it is a net zero charge inside. The cage is therefore not holding a constant charge itself (therefore is not a Faraday cage by my definition). Also, if you separate charges within the cage, they may be 'used' to polarise the cage and thus cause e-field externally, which is why there should be suitable clearance between equipment and the cage such that localised charges don't have this effect. Polarisation of the charges on the cage is also possible, of course, with magnetic fields. Magnetic fields ordinarily pass through a conventional metal cage anyway unless there is volume that 'saturates' and 'short circuits' those magnetic fields.

To Steven's question, even if the cage is gigavolts potential above an external ground, if the interior of the cage is just one electron short of net (figuratively speaking) then it will still pull an electron out of the cage despite the cage being well short of electrons(net) itself (or works the other way, exactly the same even if the cage is net positive, it would only give up one electron, so the charges know no difference whatever the cage is at)...at least, I *think* that's what Gauss's law says about this situation!? Can't say I've tried standing in a gigavolt cage myself!

[JamesC]

Sorry if the following comments are obvious to you all but I have just had a few lights turn on and I am excited about it.

Having just implemented my virtual metal in my simulation tool, which I modelled as an evenly distributed fixed positive charges and a equal number of free to move electrons constrained with the boundary of the metals geometry I can see immediately this model is going to go a long way to help me understand what is going on.

I was wondering, yeah why don't two positively charged plates repel? How can two postively charged plates be at equillibrium.

Now I can see there are two things, charge and *charge distribution*.

So two plates can be positively charged , ie have fewer negative charges than positive. But when bought close together first logic thinks, hey these should repel, but then you see the charge distribution works to cancel that out. The electrons act like a counterbalance to the external field so indeed two positively charged plates can have no repulsion. I think the other thing is the metal has a heck-of-alot of electrons so the charge is just really a small bias relative to the total population.

So for the two oppositely charged faraday cages, the charge distributions act to cancel out each others excess/deficit just by shifting around a bit. The real question then becomes one of energy and potential difference not of charge. ie How much energy is needed for albert to give michael is ion and vica versa. Energy of course was put in to charge the two plates in the first place.

Actually I think I can use this technique to create those potential maps you see in SIMION etc from the N:N calculations my sim is making. Whats nice to see is the macro concepts of fields, faraday cages, etc evolving from the fundamental coulomb operation. I will post some pretty pictures when I get the chance

[Richard Hull]

I have always felt that all charges have a return path.

Outside of any totally metal sealed box, (perfect faraday cage), you may apply a charge. This charge is relative to usually local ground if the charging gear is grounded or some other body (isolated generator). The outside world globe/planet is the traditional return as it is local and no need for an infinity.

Likewise if you are inside a sealed sphere of metal and have a generator depositing a charge on the sphere the return path is totally within the sphere as it is its own little universe, electrostatically speaking. The return path for the charge is totally within the sphere.

How did a return path to test the charge created within the sphere get generated outside the sphere? No work was done outside the sphere to separate charge.

If there is spearation outside the sphere by some means and we ground the sphere to discharge it. then will the person inside the sphere see his charging efforts disappear relative to his equipment, (his return path) ???

There should be little difference here regardless of whether you ground your generational gear to the sphere and charge an internal electrode or let the gear's return float and charged the sphere. It is only a matter of where the charge was separated. Polarity or ground issues inside or out should make no difference. The electrostatic separation of charges occured only in the sphere relative to the person in the sphere. A person outside should record nothing whether the sphere outside is grounded or floating as he has done no work in separating charge.

I hope this is clear. What say the other gurus?.

Richard Hull

[gabrielArgentina]

Richard wrote <It is only a matter of where the charge was separated.>
This is the most important statement in the understanding of any electrostatic system!!.

When I see the typical example of located charge like in a punctual nonexistent body or some sphere , and put some charges in them ...the first questions is..where come this charges that is putting there?. in the initial message ,we have this kind of system...and we see all type of interaction between this +charges ...but where is the electrons that you remove for make this charges .? why is not in the drawing ?..why is not take in account the attractions between this electrons and this protons?.

Then the second most important is ..reduce the size of infinity to a handled size..like a Faraday cage normal room size!!.
Now we are ready to make the right questions and obtain the right answers.
Gabriel.
Ps:this is not a criticize on this thread ,I only wish make a remark in a common misunderstand.

[ScottC]

I would say that you have to consider the shield gap size of the cage also. If it's not a solid, then it will only shield frequencies with wavelengths longer than the cage "gaps", otherwise it's a semi-transparent just like a fusor grid.

A single ground rod could be considered a (very bad) Faraday Cage.

Scott

[Richard Hull]

I am aware of the leakage to HF rf of even a half way decent screen room with door. Some cell phones work fine in a cage of haphazard construction.

In my example I was considering an ideal, perfectly conducting, metallic sphere, (sealed with no real door), and speaking of electrostatic charges at effectively DC.

I was not speaking of a typical RF screen room.

Richard Hull

[gabrielArgentina]

I have an other Dummies questions ...that i like help to resolve.
In the very ugly hand drawing you can see 6 electroscope positive charged.
At left a force questions F1,F2,F3...Which one is right?..means "L,R" ,left and right attraction force ,"C" central repulsion force.
At right a charge questions CH,1,2,3,...Which one is right?.
After see how kindly the forum answer this simple kind of question ...I put my own ..and I hope get similar kind of answer ...

Take my thanks in advance!.
Gabriel.

[Chris Bradley]

You're drifting off the topic from another thread here, Gabriel. This isn't a faraday cage. What will happen here is any charge will shoot up the line that is sticking out of the top so as to distribute the charge. The electroscope works by pulling one set of charges up out of the lower leaves so that there is a balance between charges low and high, but still is net overall for the leaf and stem structures. IN this case you need to pull charge out of the leaves within the earthed box you are showing to create those local charges, hence is not a Faraday Cage.

[Carl Willis]

Chris is right. A dead horse is about to get beaten here. Gabriel's question is a transparent return to the attraction-repulsion debate that Gabriel already has two threads going on. He should put this in his own thread.

-Carl

[DaveC]

It's gotten a little foggy as to what the questions are that belong to this particular thread. It seems they were something along the line of what effect a "Faraday" cage...or shield has with respect to charged particles on the inside.

We should first note that the Faraday cage, is assymmetric (so to speak) in its effects.

It shields an internal charged particle from the influences of any external charge distribution, and/or potential. Yet the same internal charge, within the Faraday cage, is perfectly detectable from outside the cage! The cage does not "hide" and internal charge.

This can easily be seen and understood, as the charge within is screened from any external field, potential, or charge distribution by the assumed infinite conductivity of the Faraday cage. For a cage made with other than infinite conductivity, some effect CAN and usually will be detected, if instruments of sufficient sensitivity are used.
(Those who use electrometers in the sub-picoamp ranges will know this from firsthand experience.)

The infinite conductivity, which is approximated rather well by the common conductor materials like silver, copper and aluminum almost completely eliminates any non-uniform charge distribution on the inside. There IS a charge distribution present, since any metal is "transparent" to charge...but because the conductivity is so great, there is effectively no tangential (circumferential) potential or charge difference and thus all internal effects are in nominally perfect balance.

With a charge inside the Faraday Cage, its presence rearranges the previously balanced charge distribution in the cage metal, to create a charge distribution on the surface, which is in exact proportion to the geometric position of the charge on the inside. (We speak here of what are sometimes called "image charges" ) A negative image charge distribution sets up on the inner surface of the cage, while the "image" of that image charge sets up on the exterior of the cage.

Thus, seen from the outside, and at some distance away... it is as though the Faraday cage is not there.

The transparency of a metal to the influence of an external field. can be seen in the process by which the Electrophorus is "charged" - by grounding the plate in the presence of an electrostatic field, and then ungrounding and removing it from the field.

Charge displaced from the plate by the external field, is conducted away to the essentially infinite earth ground, leaving the plate neutral in the presence of the external field, but clearly "charged" when it is removed from the field's influence.

While this is not a closed surface, what happens is the same. The field reaches through the metal, so to speak, and displaces the nominally uniform charge distribution

Whatever view one takes of "charge" itself, doesn't really matter, here.

Don't know if that helps.


Dave Cooper

[Chris Bradley]

I'm not sure exactly what you are trying to say here (I wasn't quite sure about Richard's point either).

As far as I think I understand it, a perfect conductor will polarise perfectly inside-to-outside through the thickness of the conductor, the distribution of charge accumulating around the regions *inside* the cage that have an equal-but-opposite non-net charge, in a minimise-the-energy-attempt to 'neutralise' that region of space. The corresponding outer surface of that conductor will have a uniform distribution (or whatever matches the charge distributions external to it).

In reality the external surface will have some degree of polarisation on it because the conductor will be able to hold up some electric gradient by virtue of its finite conductance.

For AC (/RF) electric fields, if the charges do not re-distribute themselves quickly enough around the cage conductor then, again, you can end up with charge gradients being transiently held whilst the fields fluctuate. Hence, the relation of the conduction speed (~speed of light) to dimensions is germaine, plus if the conductor is a 'bit too thin' (relative to its conductance, thus more or less likely to hold a small field gradient) then such fields may appear to transit through the cage.

Does this match, or oppose, your comments on the 'transparency of metal'? I'm sorry but I'm not sure I fully understood your points.

best regards,

Chris MB.

[JamesC]

>As far as I think I understand it, a perfect conductor will polarise perfectly
>inside-to-outside through the thickness of the conductor, the distribution
>of charge accumulating around the regions *inside* the cage that have
> an equal-but-opposite non-net charge,

Yes, I am much happier with the world now that I am beginning to appreciate the low level processes of charge distributions involved

I think accepting statements like, "a faraday cage cancels all external fields", and "metals are transparent", as absolute truth can easily lead to the building up of ideas that can easily lead you down a wrong path.

I am not yet 100% clear but the 'truth' is slowly emerging - at least in my mind.

A metal is not transparent, in that a field doesn't just go through it or something, instead a metal has properties that allow it's charge to redistribute in such a way as it appears transparent. This is a different thing because its actual transparancy depend on those properties, like thinness, type of metal, charge, frequency of electric field etc.

Equally a faraday cage doesn't really just cancel all external fields, instead it is an arrangment of metals that have the properties that allow charge to redistribute to implement the effect of cancellation. Again this is a different thing because it depends on the geometry of the cage, metal type etc and so on.

I know this is obvious after a while but to the newbie taking "Laws" as absolute truth seems completely reasonable and then ideas build up on those shaky truths and then suddenly you have invented some new physics.

I like to know that macro laws are the result of micro behaviours because it helps understand when and where the macro laws can apply.

[DaveC]

Chris -

I held back responding, to decide if you were responding to my post. But.. no disagreement with your explanation, either. I think you are considering the Faraday "cage" in the same manner as I was.

My calling the metal "transparent" is probably an oversimplification, and strictly speaking not exactly correct, since the metal's conductivity ( finite or otherwise) removes the information about charge distribution on the inside, giving only a gross quantitative effect. The excess charge outside the isolated faraday cage equals the enclosed charge, and appears more or less isotropic. On the inside, the charge could have a quite arbitrary arrangement, giving the same external result.

I think the essential point with regard to James' questions is that Faraday cage makes an almost perfect equipotential surface - DC conditions assumed here. If the cage is connected to an earth ground, the cage potential is that of the local earth potential. If left "floating" the cage potential will represent the average potential in space for the amount of charge enclosed.

Dave Cooper