0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$
  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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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