Why does a Faraday cage still work if it's mesh and not a solid conductor? Is it that all you need is enough material for charges to sufficiently reconfigure in order to counterbalance an external electric field?

Also, is there a known mathematical relationship that describes how "thin" you can spread the Faraday cage before it ceases being effective at "stopping" an electric field of a certain magnitude?


1 Answer 1

  1. It doesn't quite eliminate all the electric field. In practice, one often tries to eliminate stay fields from lower frequency fields (i.e. the 60Hz current from the wall) so one only needs a mesh with lattice of some fraction of the wavelength of the components to eliminate.

  2. Real conductors have a finite conductivity $\sigma$, and the electric field decreases exponentially inside the conductor, with amplitude decaying as $e^{-z/d}$ for a distance $z$ inside the conductor. The so-called skin-depth can be found (in the approximation of a "good conductor" where $\sigma\gg \epsilon\omega$) as $d\sim \sqrt{\frac{2}{\mu_0\sigma\omega}}$ where $\omega$ is the frequency of the signal in rad/sec. Thus a thickness of a few skin depths will be enough to practically completely stop the field from penetrating through the conductor.

  • $\begingroup$ forgive my ignorance but you said that faraday cages are used for low frequency fields like 60hz but I often see people using it to block GSM signals which are between 400MHz and 1900MHz $\endgroup$
    – Trey
    Commented Apr 24, 2020 at 15:50
  • 1
    $\begingroup$ @Trey it is possible to do this using a small mesh for then the “cage” is basically a continuous reflecting surface. $\endgroup$ Commented Apr 24, 2020 at 16:29

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