A Faraday cage within a static electric field leaves an internal electric field equal to $\vec{0}$ (as long as there are enough mobile electrons in the conductor to counteract the external $\vec{E}$).

It could be said as "a Faraday cage is fully opaque to static electric field".

On the other hand, it doesn't act on an external magnetic field.

It could be said as "a Faraday cage is fully transparent to static magnetic field".

Wich equations describe transparency (behaviour) of a Faraday cage submitted to different frequencies electro-magnetic fields?

Ideally I would like to find an absorption spectrum of a Faraday cage from 0 Hz to visible light (1 PHz).

  • $\begingroup$ Do you really mean a cage (with holes in) in which case, the frequency dependence amounts to whether the holes are big enough to pass the wavelength of radiation, or are you interested in the combined effects of reflection and attenuation, which are also frequency dependent. I think all of this is answered in the "related questions". $\endgroup$
    – ProfRob
    Dec 30, 2015 at 10:59
  • $\begingroup$ I'm interested in any sort of "Faraday cage" to better understand how they absorb, refract or reflect different wavelengths. Ideally I would like to find a way to draw their absorption spectrum. $\endgroup$
    – dan
    Dec 30, 2015 at 14:03

1 Answer 1


According to What is the relationship between Faraday cage mesh size and attenuation of cell phone reception signals? for a Faraday cage with a characteristic mesh size of $l$, then the cut-off frequency $f_c$ corresponds to a wavelength of $2l$, i.e. $f_c = c/2l$. In these circumstances, the wavevector $$k = \frac{2 \pi}{c} \sqrt{f^2-{f_c}^2}$$

Thus when $f<f_c$, the electric field is exponentially attenuated as $\exp(-\alpha x)$, for a cage of thickness $x$, where (if I've done my sums right) $$ \alpha = \frac{\pi}{c} \sqrt{2(f_c^2 - f^2)}$$

If there are no holes at all (i.e. a metal box) then the frequency dependence is explored in Faraday cage in real life

It is shown that $$\frac{E_t}{E_i} \simeq 4 \frac{\eta_{\rm c}}{\eta_0} \exp(-x/\delta) = 0.47 \omega^{-1/2} \exp(-22 \omega^{1/2} x),$$ where $\eta_c$ is the impedance of the conductive material and $\delta$ is the frequency-dependent "skin depth". The transmitted power fraction would be the square of this. The numbers in the above formula are appropriate for aluminium, but the frequency dependence would be the same for any good conductor. Note that at high frequencies attenuation in the medium is important, but even if the box was very thin, that reflection from the surface is extremely important even at quite modest frequencies.

  • $\begingroup$ Yep. To bring it down to intuition: currents will flow in response to EM radiation but since the induced currents are insufficient to fully cancel the field, some will penetrate. A mesh has less surface conductivity than a sheet, and if waves are small compared to the size of the holes, they get through without ever noticing the conductor. $\endgroup$
    – Floris
    Dec 30, 2015 at 13:42
  • $\begingroup$ You say that $f_c$ for $\lambda=2l$ is the cutoff frequency of the mesh. What is its resonance frequency? $\endgroup$
    – dan
    Dec 30, 2015 at 15:36

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