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I think the answer is yes. My reasoning is this: Imagine for argument's sake, we could have a charged negative source that has its field blocked by a Faraday cage. We can transport a positive charge outside the cage away from the source at low cost, energetically speaking. We now remove the cage and the positive charge experiences an attractive force due to the exposed field and moves towards the source generating more energy than was used to move it away. If it is possible to shield the static electric field we would have a method to produce free energy which is a violation of the law of conservation of energy. Therefore I conclude it is impossible to shield an electric field or any other static force field (e.g. gravitational. magnetic etc.).

This appears to be a contradiction to the common perception that Faraday cages block electric fields. It may be that they only attenuate rapid changes in the electric field such as RF, rather than completely block a static electric field.

Is my reasoning correct or is there a counter example or a counter argument?

I came to this realisation while constructing a thought experiment that relies on being able to 'switch off' an electric field at will and it would be better for my thought experiment if I am wrong about the impossibility of shielding a static electric field with a Faraday cage, so I will be happy if someone can prove me wrong!

P.S. I think it can also be argued that a negative charge inside the cage would induce a positive charge on the inside surface of the cage and a negative charge on the exterior surface of the cage. This induced negative charge on the exterior surface must surely generate its own field outside the cage, so the exterior electric field would not be much different, to when the cage is not there.

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First of all, a Faraday cage isolates the inside from fields that arise from outside. It does not work the other way around!

In your example you seal a negative charge inside a cage and hope that this will make it easier to move a positive charge away from the cage to an infinite distance. But this doesn't work. You are trying to protect the outside from the inside. The cage/charge system has a net negative charge and will still attract the positive charge.

It is true that, once you have sealed the negative charge inside the cage, you can now easily move the negative charge around the interior of the cage. It is protected from the positive charge outside. You could attempt to create a perpetual motion machine by moving the interior charge around the cage. However, as another answer already indicated, opening and closing the door of the cage is not energetically free.

The door is covered in induced electric charges. Those charges will produce forces that tend to force the door open or prevent it from being closed [1]. It doesn't matter whether the door is heavy or light, because the energy required to move the charges is largely going to be expended on moving the charges themselves.

[1] One way to see this is to imagine cutting out a small patch from a closed Faraday cage. (This is like unlocking the door.) Like a piece of confetti attracted to a charged van de Graaf generator, that small patch of the cage will be attracted to the two charges. It will fly toward whichever one is closest. (If they are exactly equidistant, the slightest disturbance will make it fly toward the closer one.) So energy was released by opening the door. To close it again, you will have to spend energy to drag the piece away from the charge and back to its original location. In a real door, the door is not allowed to fly away. Perhaps it is attached by a chain. This makes things far more complicated, because the chain is conductive, but the overall conclusion -- it costs energy to close the door -- should be the same.

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  • $\begingroup$ However you argument that the charge on the door increases the force to open the door is a good one. $\endgroup$
    – KDP
    Commented May 30 at 17:47
  • $\begingroup$ @KDP the most important part of that answer is "Note - this does require the cage to be grounded". Read the comments on that answer, some of which also point out that the question does not make that assumption, so that answer doesn't apply properly. Grounded means that a real cage, unlike the spherical superconductor in that question, needs to be connected by wires to a source of arbitrary numbers of both positive and negative charges, which get pulled in or pushed out as needed. $\endgroup$
    – Ryan C
    Commented May 30 at 18:13
  • $\begingroup$ From the point of view of an arbitrarily small patch of the cage, how does it know which side is the "inside" and which one the "outside"? $\endgroup$
    – Lawnmower Man
    Commented May 30 at 19:57
  • $\begingroup$ @LawnmowerMan The inside of the patch is nearest the source and charges of the opposite sign on the door try to get as near to the source as possible and gather on the inside surface and like charges are repelled and accumulate on the outside. This is induction at work. The difference is proximity to the source. $\endgroup$
    – KDP
    Commented May 31 at 6:06
  • $\begingroup$ @RyanC Do nor grounded charges still create an excess of charge in the ground and create their own field, albeit spread out and diluted. $\endgroup$
    – KDP
    Commented May 31 at 6:08
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We now remove the cage [...]

Generally speaking, this could cost a large amount of energy.

A Faraday cage works because the charges inside it are able to minimize their energy, by being able to move freely. If you bring a conductor into existence inside an electric field, first the electric field would permeate the conductor. But then charges would start to move since there exists a lower energy state. After everything has settled and any extra energy has dissipated, the electric field inside the conductor is now zero. The charges are arranged on the boundary of the conductor in a specific way to achieve this.

If I now the Faraday cage a bit, the charges will initially not be in their equilibrium position anymore. The electric field inside the conductor will not necessarily be zero and the energy is not minimized.

I don't know how this works in practice (how much work does it cost to move a Faraday cage?), but I think this easily explains where all the energy comes from.

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  • $\begingroup$ WE only really need to open an arbitrarily lightweight and well balanced door on the cage and we could have an arbitrarily large charge inside the cage so that an energy profit can still be made. No? $\endgroup$
    – KDP
    Commented May 30 at 9:37
  • $\begingroup$ @KDP all these forces/energy happen at the speed of light c. $\endgroup$
    – PhysicsDave
    Commented May 31 at 23:38
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This argument shows that "turning off" the shielding effect without work expenditure is not consistent with the "there is no free lunch" idea (there may be one lunch, but not arbitrary number repeatably, because configuration of the system would have changed).

The Faraday cage shielding is done by free charge particles in the cage, and moving this charge to "turn off" the shielding effect requires that some force does work on these particles (or on the conductor body these particles are attached to). This prevents repeating the process to gain work from nothing.

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A conductor cancels electric fields perpendicular to its surface. It does this because electrons move freely within the conductor.

If there is a component of the electric field parallel to the surface, the electrons are moved by that field, decreasing its magnitude. Equilibrium is reached when the electrons can't shift position to decrease the field, at which point there are no electric field components parallel to the surface.

Electric fields pointing directly away from the surface would try to pull electrons out of the surface (or push them further in), and this nominally gives a field gradient inside the material. Extremely large fields (greater than around a GV/m, depending on the material) can overwhelm the attraction between electrons and their atoms, but I'm going to ignore such extreme cases for the rest of this answer. Extreme fields will arise when the amount of charge inside the cage becomes comparable to the total charge on all freely-moving electrons composing the cage itself... so I'm also ignoring that case.

If the field is changing quickly, the electrons' finite mass becomes an issue - it takes time for the electrons to move and cancel out the field. (The plasma frequency is the relevant frequency scale, and it varies by material.) These frequencies are quite high - metals appear shiny because the electrons are cancelling the oscillating electric fields in photons, bouncing the light rays that hit them. Depending on the material, you're probably looking at finite-electron-mass effects in the far-UV to X-ray regime. For the rest of this answer, I'm going to assume you're not asking about charge distributions changing at such rapid rates.

Some "Faraday cages" are made of wire mesh. In this case, the electrons can't move to the spaces between the wires, so there's a limit on their ability to cancel short-wavelength (high-frequency) EM waves, and this limit is tied to the fine-ness of the mesh. This gives a cutoff in wave transmission around the gap size of the mesh: shorter wavelengths can go through the gaps and longer wavelengths treat the mesh as a continuous surface. Commonly, this lets a Faraday cage block radio waves while being cheaper, lighter, and at least partially transmitting visible light - all of which are convenient. For the rest of this answer, I'm going to assume your wavelengths are significantly longer than the mesh size.

That leaves us with two separate cases of concern for low-frequency, low-amplitude electromagnetic waves interacting with contiguous:

  1. Equal and opposite charges are available: Inside the Faraday cage, you've got positive and negative charges which try to form a dipole electric field. The electrons inside the cage's material move around to cancel out the portions of the electric field intersecting the interior surface of the cage / metal shell. This distorts the dipole in the interior space, but no fields reach the exterior. Your energy considerations don't really apply since the additional energy necessary to get +q into place is cancelled by the energy to get -q into place. The cancellation is approximate, but differences are going to relate to the exact geometry of the cage and placement of the charges within it, and roughly relate to one or more of the caveats above.
  2. There is a net charge inside the cage: The exact distribution inside the cage will cause electrons to shift around on the interior surface of the metal Faraday cage, and within the cage's metal. The metal will cancel out electric fields within the metal, but there will still be a net charge. If the "shielded" interior charge is +1 Coulomb, this will translate to there being -1 Coulomb shifted toward the interior surface, and a corresponding +1 Coulomb left behind and distributed over the exterior surface. The electric field on the exterior will be perpendicular to the surface at all points, and so only the external geometry will matter outside the Faraday cage - not the distribution of the charges in the interior space or the geometry of the cage's internal surface.

Note also we can exchange "interior" and "exterior" in all the above without loss of generality. The only thing that matters is that the electrons are able to interact with all the electric field lines that cross from one region to the other.

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  • $\begingroup$ If the cage was a sphere of thickness 1 mm then there would be a field in the metal. $\endgroup$
    – PhysicsDave
    Commented May 30 at 19:33
  • $\begingroup$ What is the conclusion, for example, for the question in the title? $\endgroup$
    – Peter Mortensen
    Commented May 31 at 20:06
  • $\begingroup$ @PeterMortensen that a static field might or might not be blocked, depending on the distribution of the charges, i.e. whether you're in case 1 or 2. $\endgroup$
    – Sarah Messer
    Commented Jun 3 at 13:22
  • $\begingroup$ @PhysicsDave Yes, that "field in the metal" is covered by the phrase "nominally gives a field gradient inside the metal" in paragraph 3. $\endgroup$
    – Sarah Messer
    Commented Jun 3 at 13:23
  • $\begingroup$ "The metal will cancel out electric fields within the metal...." is a bit confusing $\endgroup$
    – PhysicsDave
    Commented Jun 3 at 15:18
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When the charge $-q$ is inside the Faraday cage, the cage must have a charge $+q$ to shield it (in addition this $+q$ will be distributed in the right pattern over the outside cage walls, that's how the Faraday cage works).

We must assume that the charge has flown to it through some earth connection (just the earth in the grid connection would suffice) or we must assume (if there is no ground connection) that the charge $+q$ was already present from the beginning. Since you do not specify it, let's assume the latter case: To move the $-q$ charge and the $+q$ charged cage away from each other requires energy, you then might gain some energy by bringing the positive charge outside closer to the $-q$ charge (staying away from the $+q$ cage!) but those two energies can obviously cancel.

If, on the other hand, we assume the former case (grounded cage) then the work done for moving the $-q$ away from the cage is less, since the $+q$ charge of the cage will gradually flow away during the process, but still some work is required. The interesting question remaining is if this again will cancel the energy gained by your subsequent manipulation of the positive charge outside, there must be something changing that amount of energy as well, in this scenario. You could still see this as a kind of paradox! (But clearly there are opposite energies that can cancel, and since your question doesn't ask for the computational details we cannot address this here...)

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Faraday cages are by definition isolated, i.e., not connected to ground. In electronics, many circuits use a cage that is also grounded, (called shielding), so that is a source of confusion for many.

In your question (ungrounded) there will be a field, from the positive shell to the negative charge, all inside the cage. The point you are missing which is the purpose of the cage is to shield the inside from external charges. So in your case, the negative charge is still detectable outside the cage … but no matter how hard you try, you will not be able to affect it.

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Well, first of all, it is clear that Faraday cages work, as they have been experimentally tested many times. You just have to search for a video of them working in YouTube, or watch this. So clearly your reasoning must be wrong.

Now, where is the "energy expenditure" hidden? My guess is that, while you move the charge inside the cage, you also change how the electric field outside the cage is. Remember that the charge inside the cage create a field outside the cage. This is modified when moving the charge around inside the cage, which should impact the field outside the cage and whatever is creating the exterior field in the first place. I believe that the energy expenditure is moved to here (probably the cage heats up because of the internal movement of charges). Still, there could be a clearer/nicer way of explaining why you are not generating free energy!

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