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Faraday cage , current flow.

  1. Jul 31, 2013 #1
    Hi ,imagine a situation , we have a large tesla coil, a faraday cage , a man in it and he cage is grounded.
    Now we switch on the coil high frequency high voltage currents makes a breakdown of air and produces a path so that the current goes and strikes the cage.Now according to faradays law the charge resides on the outside of the cage so does the electric field and the current goes along the outside to ground.
    The man inside the cage is okay, as the electric field strength inside the cage is zero and no current flows there.
    Now we take an good enough insulator for the given tesla coil voltage and we insulate the cage from ground , then we attach a wire to the cage from the inside put a mechanical switch in series with that wire and then we let the wire through the cage so that it doesn't electrically touch it to the ground and ground it.

    Now we switch the coil on , the current goes to the cage once again but it has no path to go to ground this time , so it probably seeks other places with less resistance , now the man stands inside , he flips the switch on the wire so that it forms a path to ground , what happens?
    Is the electric field inside the cage still zero? If yes then how can current flow through that wire to ground and if it does what happens to the man standing inside the cage , both if he touches the wire (considering the wire isn't insulated) and if he doesn't touch the wire.?
     
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  3. Jul 31, 2013 #2

    sophiecentaur

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    The answer to this will depend upon how 'good' the Faraday Cage is. Put a hole in it and it is no longer perfect. The fields where your test wire enters the hole will affect what the effective PD will be between the wire and the man's hand will be. Think of it like a potential divider. You have a source of high emf, a high resistance and then a low resistance in series. The man puts his finger across the low resistance. Safe as long as the low resistance is low enough wrt to the high resistance.
    If you draw yourself a diagram and include all connections (to Earth and to the man etc.) you may get an insight into the problem. I am not exactly sure about your model so I may not be commenting on what's in your head. You could even post it, if you are still in doubt.
     
  4. Jul 31, 2013 #3
    Hello Sophie , and others. :)

    Well the thing is kinda simple at first , a faraday cage , a good one.Now there is a man standing inside it , now the man feels no pain or shock due to the fact that the field lines go outside the cage the current flows on the outside and the voltage also resides on it.
    Now normally taking into account that high voltage is used to electrically strike the cage the current would run down the side of the cage and into ground.
    Now imagine the cage being lifted on a huge insulator so that any direct path to ground from the cage is canceled.
    The potential difference begins to rise on the cage and charge builds up due to the fact that the current has no where to run anymore.
    Now there is a man standing inside the cage he holds a switch in his hands and a wire is attached to the inner side of the cage it runs through the switch and out of the cage and is grounded.
    For the sake of the argument letš consider that the place where the wire exits the cage is small , the opening is not large and it has an insulator so that the high voltage wire doesn't electrically attach to the cage at that point.

    Now when the only way for the current to go to ground is through that wire and the switch what would the current do after the man switches the switch on ?
    Normally I would say there would be a path for the current to flow and it would happily do so , but since this is a faraday cage and all the charge resides on the outer side of it the field strength inside is zero so there is no pushing force inside that could drive the electrons in the wire as long as it goes inside the cage this is the first thing that keeps me wondering and the other would be that if we know that all the charge resides on the outer surface of the cage wouldn't that repel the same charge trying to get inside so that it could go to the path of the wire to ground and form a path for the current to flow ?

    I hope you understand this situation , and could give me some answers as to what happens in case like this?

    If it makes any better you can imagine a perfect conducting sphere instead of a cage as the theory is the same.
     

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  5. Jul 31, 2013 #4

    sophiecentaur

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    Hang on a minute. The fact that the field inside is nominally zero will not stop current flowing through the man from the surface of the cage to the wire. Don't confuse Field and Potential. It's potential that causes current to flow through a conductor from one place to another.

    If things went the way you suggest, you could 'insulate' things from each other by just putting them in bell shaped containers.
     
  6. Jul 31, 2013 #5
    A faraday cage doesn't affect electric circuits. The current will be exactly the same as it would be without the cage. If you take a 9V battery and connect + to the outside of the cage and - to the wire coming out of it and the switch inside is open there will be a voltage drop of 9V across the switch. That means that there will be an electric field across the switch despite the presence of the cage. The hole in the cage may be small but when there is a wire going through it even a very small hole can allow a field inside.
     
  7. Aug 1, 2013 #6
    Wait so so you are saying that if we could somehow just theoretically make the wire go through the cage without a hole in it and + it would be insulated from the place where it exits the cage then there would be no field to drive the electrons on the wire?Just theoretically?

    Well normally yes i understand that a potential difference would cause current to flow.
    By the resistances sophie did you mean something like if the wire which is in the cage would be a low ohmic resistance one the chances of the man holding it being shocked would be less than if it was a higher resistance one ?
     
  8. Aug 1, 2013 #7

    Baluncore

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    An arc will immediately form across the switch or between the wire and the cage.
    I doubt he will have time to flip the switch.
    Complex circuits made from conductors become irrelevant at high voltages and frequencies in an ionisable atmosphere.
     
  9. Aug 1, 2013 #8

    sophiecentaur

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    Your model is a mixture of ideal and practical elements. You have an 'almost ideal' cage and an isolated wire going through a hole in it. You connect them together. If they started with different potentials then a current will flow to bring the PD to zero. The value of that current will depend upon the PD and the series resistance in the circuit. The total charge that will pass will also depend upon the capacitances of those two charged objects and their initial Potentials.
    Bottom line is that the problem is best considered, not by dealing with the fields (in the same way that normal circuit problems are not suited to the 'field' approach) but by considering PD, Capacity and Resistance. Try to re-draw your thought model as a circuit, replacing physical pictures with equivalent circuit elements and it may be easier to see what's going on. Of course, you need to be using the right circuit but the topology of the system is fairly simple so you should manage it. It's what they do in antenna design and analysis, very often and its a handy tool.
     
  10. Aug 1, 2013 #9
    If a current is flowing through a resistor there is always a field. The general form of ohms law is J=σE. Current density, conductivity and e. field are always dependent on each other. So if there is a current inside the cage there has to be a field there too.
    But of course if all you want to do is calculate the current you should look at the PD and not the e. field.
     
  11. Aug 1, 2013 #10
    Ok I see , thanks so far for the answers.
    Now I draw a simple schematic or something that looks alike :D
    Now I have a further question to which I again feel like knowing the answer but just to make sure.

    Now to the left of the picture I have my primary setup the one which I was describing here with the faraday cage or shell and the switch and a large tesla coil putting an high voltage arc on it.
    Now clearly there is a current going through the arc and along the sides of the cage or shell so even if the shell would be insulated the wire inside it that goes to ground would cause a PD and current would flow.so far so good.

    Now to the right of my picture you can see a structure which is like a sphere in a sphere in a sphere each of those layers are pretty close to one another with some small dielectric between them, in other words I have formed a spherical capacitor with a given capacitance in it.
    Now as normally I apply a given potential to the outer layer , then again there is another wire attached to the innermost sphere or as in the capacitor analogy to the other plate.
    Now the question is (assuming the spheres are isolated from ground) what would happen if the switch inside is left open at first and only the +ve potential attached? Would some charge be set on or between the spheres ? I think it shouldn't because for a ordinary capacitor to charge you need both of the plates attached to different potentials right?
    Now when I would turn the switch on what would happen then ? Assuming the voltage on the +ve is not high enough to cause dielectric breakdown through the spheres and layers between them ?
     

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  12. Aug 1, 2013 #11

    sophiecentaur

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    You would just have a capacitor, corresponding to the inner and outer spheres. The intermediate sphere would have no effect. When you connect the switch, you will get a charge on that capacitor corresponding to Q = CV
    There will be a field between the spheres of V/spacing V/m (approx).
    You are now going to say "But there's no field inside the sphere", I suspect. But the situation is not an empty sphere, where the shell theorem applies. There's a sphere at Earth Potential in there and that alters everything.
    Tell me my prediction was wrong and I'll be very pleased. :smile:
     
  13. Aug 2, 2013 #12
    There is also a capacitance between the wire and the spheres. You can use an equivalent circuit diagram to see what's going on. In that diagram you have to put capacitors in between all the conductors, not just between the spheres.
     
  14. Aug 2, 2013 #13
    Looking at my last drawing , can you please tell me what would happen if I would attach a switch and a wire in series to the inner sphere?
    As you correctly said Sophie , that the inner sphere has no actual effect as the two outer one already form a capacitor but is there is a third middle layer between those two what happens to it , does it get charged? And if so then could I discharge it after given intervals and the whole sphere capacitor would discharge with it or no?
     
  15. Aug 3, 2013 #14

    sophiecentaur

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    It cannot get charged because it's not connected to anything. It will get polarised, though, with + on one face and - on the other.
    Instead of coming up with this long list of possible scenarios, why not get the basics sorted out? The answers to all these supplementary questions would be more obvious to you, I think. You can get most answers by replacing these spheres with simple capacitors.
     
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