I see lots of discussions about how a Faraday cage can block EM signals, but almost no one addresses the possible difference between incoming signals versus outgoing signals. It seems to me that those are different situations.

For sake of argument, lets say the cage is a hollow superconducting sphere.

I can see how EM fields outside the sphere could re-distribute the charge on the surface of the sphere in a way that there would be no EM field inside the sphere.

However, in the simple case where a point charge is placed in the center of the sphere, the charge on the surface redistributes in such a way that the EM field outside the sphere is as if the sphere was not there.

Similarly, it seems like if the charge inside the sphere were rapidly moved up and down along the z-axis, it would create an EM wave outside the sphere as if the sphere was not there.

Am I missing something here?

**Edit: I used the term "Faraday Cage" because it is in common use and it makes this discussion easier for others to find. But I am asking about a highly idealized case of a superconducting spherical shell with a charge inside but not connected to the shell. For me it provides a good starting point for understanding real Faraday cages.

By the way: I was once told by a physics professor that this is one of the few things Feynman gets wrong in the Feynman lectures. Which shows how easy it is to get this wrong. Either a physics professor at a major university gets this wrong or Feynman.

  • $\begingroup$ "Similarly, it seems like if the charge inside the sphere were rapidly moved up and down along the z-axis, it would create an EM wave outside the sphere as if the sphere was not there." Why does it seem like that to you? $\endgroup$
    – Mike
    Jul 24, 2016 at 14:15
  • $\begingroup$ @Mike to some extent Chuck is right, an incident plane wave from the outside (coming from "infinity") will penetrate into and reflect differently from the way the near and not so far field of a finite sized antenna placed inside the cage gets through the wires forming the cage. $\endgroup$
    – hyportnex
    Jul 24, 2016 at 14:32
  • 3
    $\begingroup$ You can't bring a charge in from infinity to a Faraday cage, as the cage does not allow charge to go in or out. You are correct that bringing a charge in from infinity to a spherically conducting shell with a small hole in it will result on a charge on the exterior of the shell and the E field will look a lot like a point charge. If you force a charge inside a Faraday cage to oscillate, the charges on the inside of the cage will cancel all radiation to the outside world. $\endgroup$ Jul 24, 2016 at 14:43
  • $\begingroup$ @Mike: because in the semi-static case, the coulomb force outside the shell is the same as if the shell was not there. $\endgroup$
    – Chuck
    Jul 25, 2016 at 14:25
  • $\begingroup$ @Chuck That depends on how you assembled the cage — with the charge inside or outside to begin with. We can't just magic a charge into existence inside the cage while retaining the ability to predict what while happen using physics. So the charge was either there to begin with, or was brought in from outside. In that case, see Ross's comment. $\endgroup$
    – Mike
    Jul 25, 2016 at 16:36

3 Answers 3


An "RF cage" is commonly used to keep signals IN as well as OUT (see for example the mesh door of a microwave oven.)

The short answer is - reciprocity says "if it works in one direction, it works in the opposite direction". The induced charges on the sphere (in the case of a charge inside it) are just enough to cancel the field outside exactly - because that's how superconductivity works. The E field must be normal to the surface at every point. The charge redistributes to make it so. Any field due to the charge inside is exactly cancelled by the induced charge on the surface. Note - this does require the cage to be grounded: that is especially important at low frequencies (where the wave length of the radiation is long compared to the size of the cage). Grounding ensures there is no net charge on the sphere-plus-contents, and should eliminate your concern.

So yes, a Faraday cage blocks outgoing signals.

  • 2
    $\begingroup$ This doesn't address the OP's note that a static charge inside a perfectly conducting sphere is not screened. The field outside the conductor (in the case of a charge centered inside a spherical conductor) is exactly the same as if the conductor weren't there. The field due to charges inside are not canceled by induced charges at the surface. I suspect that non-static transverse waves will be screened, but showing that requires an extension to your explanation. $\endgroup$
    – garyp
    Jul 24, 2016 at 14:49
  • 4
    $\begingroup$ @garyp a Faraday cage needs to be grounded - at which point the field is perfectly cancelled. $\endgroup$
    – Floris
    Jul 24, 2016 at 14:59
  • 3
    $\begingroup$ @garyp Another surprisingly important difference is that the cage isn't smooth. The charges inside the conducting sphere respond smoothly to changes due to the internal charge, but as soon as the cage is cut up into wires, it can only respond evenly in a few directions. I've had good luck shielding noisy equipment (as in ~20 dB) in places where I couldn't easily install a grounded cover by wrapping them in aluminum foil that had been balled up before using (while covering them neatly with a carefully shaped and smoothed sheet of fresh foil did very little). $\endgroup$ Jul 24, 2016 at 16:17
  • 1
    $\begingroup$ Don't know. I had put it down to interference between the radiation fields of the many mis-aligned responses to the driving pulse. The original (smooth) attempt was quite a neat job of origami—even if I do say so myself—and fitted very snugly. But I suspect you're right that there was more capacitance in the second attempt. $\endgroup$ Jul 25, 2016 at 0:51
  • 2
    $\begingroup$ @Floris: your comment about the cage needing to be grounded resolves this for me. Grounding it would allow the super conducting shell to obtain a charge that would cancel the charge contained inside the shell. Quite a few websites that discuss Faraday cages leave this out (and ignore the possibility that incoming and outgoing waves might be different cases). One practical significance of this might be Faraday cage phone cases and privacy. Since the case is not grounded, it seems like it would only block incoming signals. $\endgroup$
    – Chuck
    Jul 25, 2016 at 14:16

Even a "perfect" Faraday cage does not block EM radiation. If a EM radiation hits a cage the incoming photons or get dissipated in the mesh and re-transmitted with longer wavelength to both sides of the mesh (in principle a sort of black body radiation), heating the inside and the surrounding of the cage, or some amount of photons are going through the mesh and it is possible to catch a signal of modulated radiation inside the cage. The same holds for EM radiation produced inside the cage: or full dissipation into longer wavelengths or some amount of EM radiation goes through.

Since the receiver could be tiny in relation to the wavelength of a modulated EM radiation the mesh has to be with tiny holes too are better from massive metals.

BTW Faraday cage was used to show that high voltage current slides on the outer surface of a Faraday hollow sphere made from mesh, meanwhile a human is sitting inside. For blocking EM signals it is better to use Metal sheets.

From Wikipedia:

"To a large degree, though, they shield the interior from external electromagnetic radiation if the conductor is thick enough and any holes are significantly smaller than the wavelength of the radiation." (Faraday cage)

  • 3
    $\begingroup$ Not getting in a lot of down vote cases any comment it has to do with rassistic expression due to my bad English? In any other case it would be nice to get a short explanation about my failure. $\endgroup$ Jul 24, 2016 at 18:02
  • 1
    $\begingroup$ Sorry, I did not ask the question clearly. I was asking about a highly idealized case where the "cage" consisted of a superconducting shell - without holes. $\endgroup$
    – Chuck
    Jul 25, 2016 at 14:49

It is not necessary to specify a hollow superconducting sphere in order to plainly address the the use of a Faraday cage to attenuate the emission of electromagnetic signals originating within the enclosure. Also, we'll just assume a metal foil or panel rather than a metal screen material for the walls (and floor and ceiling) of the enclosure, since the reasons for using screen are unrelated, e.g., simplified ventilation, light weight and visibility advantage over solid panels. I will refer to the cage as a shielded enclosure or shielded room since that is typically where the application lies, i.e., the reduction of EMI/RFI emission for security applications (for example EMSEC in the NSA TEMPEST specifications). I think this model is more applicable than attempting to visualize typically static charge redistribution.

There is an excellent treatment of the problem at Architectural Electromagnetic Shielding Handbook by Leland H. Hemming – 2000.

We will assume the emitting source is placed at a reasonable distance from the wall of the enclosure (inside) such that we are concerned with far field radiation (radiation that has escaped the antenna; 1/6 wavelength is often used as an approximate distance for predominance of radiation field in shielding applications) propagating outwards towards the enclosing shield wall.

Radiation field about a conductor

The attenuation provided by the shield results from three mechanisms: (1) reflection of an electromagnetic wave when it encounters an impedance disontinuity, e.g., the air to metal impedance discontinuity as the wave encounters the shield (2) absorption within the shield material of portions of the wave energy not reflected as it transfers some energy in heating the shield and (3) possible additional reflections within the shield and at the impedance discontinuity as any remnant of the wave encounters another metal (typical shield material) air boundary. The diagram below is for a wave approaching from outside; simply reverse the label, i.e., let "inside of enclosure" in the diagram be "outside of enclosure."

EM wave encountering shield (just reverse to consider inside moving outwards

Electromagnetic radiation is primarily shielded by reflection via mobile charge carriers (electrons or holes) which interact with the radiation. High conductivity of the shield is not required, e.g. on the order of 1 ohm is usually sufficient. Electrical conductivity is not the criterion for shielding (though conductivity enhances shielding) though, since that would require connectivity in the conduction path, e.g., to ground.

Metals, the most common EM shielding material, function mainly by reflection via their free electrons. The secondary mechanism of shielding is absorption via electric and/or magnetic dipoles which interact with the incident EM field. Materials for EMI shielding

The reflection loss depends on the impedance of the shield and the wave impedance. Metals have a much lower surface impedance than the high-impedance electric field and therefore reflect that energy well. The lower impedance magnetic field is comparable in impedance to the metal surface so attenuation of magnetic field is primarily through absorption in the shield. There is typically not an excessive amount of shield thickness required to achieve that end. For example, a Series 81 Shielded Room uses 28-gauge galvanized steel (about 0.0187 inch thick) to exceed NSA 65-694-106/CID 09.12 shielding effectiveness.Series 81 Shielded Room Datasheet (The sum of reflection, absorption or multiple reflections in dB is the shielding effectiveness and the absorption loss is proportional to the thickness of the shield.) Series 81 galvanized shield effectiveness

If you would like to see a more mathematical treatment of the behavior of electromagnetic waves as they encounter a boundary, MIT has some useful work here: Reflection & Transmission of EM Waves


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.