I understand how a Faraday cage works. The charges within the conductor rearrange to cancel the applied electric field, etc. However, if you look at the following gif from Wikipedia, the field continues after the cage.

enter image description here

Say the Faraday cage is constructed using a perfect electrical conductor (PEC). For a PEC, the reflection coefficient is -1, so the incident wave is completely reflected and there's no transmission. This comes from differential form of Ohm's law, J=sigma*E where sigma is the conductivity. PEC conductivity is infinite so if there's an electric field we'd had infinite current density which is non-physical.

Since all of the electric field will be reflected, there is no field that can be transmitted on the right side of the figure, past the Faraday cage.

Am I missing something? I'd appreciate any help.

  • $\begingroup$ How is the field 'reflected'? You still have a potential from + to -, and the Faraday cage has only suppressed the field within the cage. Those accumulated charges (+/-) on either end keep the field going from + to -. $\endgroup$
    – Jon Custer
    Commented Jun 26, 2017 at 14:42
  • $\begingroup$ The (-) on the far right exists assuming the field exists past the cage, which is what I'm confused about. Since the field is reflected, as is required by a PEC surface, there won't be that (-) on the far right. As for why the field is reflected, I mentioned differential form of Ohm's law, but it can also be seen with boundary conditions for an incident wave. $\endgroup$
    – eichenja
    Commented Jun 26, 2017 at 14:45
  • $\begingroup$ No, it exists on the far right because it establishes the external applied field that the Faraday cage is screening on the interior. $\endgroup$
    – Jon Custer
    Commented Jun 26, 2017 at 14:53
  • $\begingroup$ That's my question, though - why does it exist? Actually, I think I just figured it out. The boundary conditions state that there tangential components of an electric field must be continuous across the boundary, so no tangential field can exist. The normal components may be discontinuous (this case), and the difference between the electric flux densities are equivalent to the surface charge density. Since there can be no field in a conductor, that means the surface charge density on the right side is equal to the electric flux density of the field in the medium to the right of the cage. $\endgroup$
    – eichenja
    Commented Jun 26, 2017 at 14:59

1 Answer 1


One way to think of this is that the charges on the Faraday cage move to produce what is, at first order, an electric dipole, whose field looks like this:

Dipole field

Since electric fields obey superposition, you can see that cancelling a uniform field inside the dipole will enhance the field outside the dipole.

In practice a Faraday cage produces not a simple dipole but a field distribution with all the multipoles needed to cancel the external field inside the cage; the mathematical expression can become quite complicated if the cage has some non-spherical shape like your rectangle. But the same general principle applies: cancellation of the fields inside the cage does not lead to zero field outside the cage.

Incidentally, the wording in your question

Why do electric fields continue? ... there is no field that can be transmitted on the right side of the figure

suggests you're making a common newbie conceptual error, thinking of the electric field as something that "flows" down the arrows from positive charges to negative charges. It's not. The electric field is a property of space around every charge, positive or negative. In particular, if the convention for charge signs had been made differently, you'd be asking how the field gets from the right side of the obstruction to the left. Electromagnetism is completely symmetric if you change the sign of all the charges, so any statements about electromagnetism that depend on who's "negative" versus who's "positive" are unsound.

  • $\begingroup$ Hey Rob. Thanks for the answer, it helped clarify a few things. As a note, when you say "suggests you're making a common newbie conceptual error, thinking of the electric field as something that "flows"...," I don't think that's completely correct. At least according to "Advanced Engineering Electromagnetics" by Balanis, a wave is a traveling entity that does indeed move from one point to another. Similarly, the E and H fields move perpendicularly to that wave (for TEM mode). I ended up resolving the issue by analyzing the behavior with boundary conditions, mentioned in my comment above. $\endgroup$
    – eichenja
    Commented Jun 26, 2017 at 17:39
  • $\begingroup$ @eichenja Hmm. I interpreted your diagram as showing a DC field, which I think is what its author intended. If you're interested in exclusion of electromagnetic waves by a Faraday cage, there are some other issues to consider --- not least whether the cage is large or small compared to the wavelength of the signal, which determines whether the wave can sneak around the cage by diffraction. $\endgroup$
    – rob
    Commented Jun 27, 2017 at 7:00

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