Electronic – Faraday cage mathematics

shielding

We all know the Faraday cage effect: light waves get through the screen on the front of your microwave oven, because their wavelength is much less than the size of the holes, whereas
the microwaves don't get out, because their wavelength is much greater. Yet despite many hours of looking around and dozens of discussions with individuals, I've been unable to
find an analysis of the mathematics of this effect.

Presumably there's a simple argument that shows some kind of exponential attenuation depending on the ratio of wavelength of hole size. Can anyone point me to the literature on this subject?

Best Answer

The following forum post might be interesting for you: Mathematical derivation of the Faraday cage from the Maxwell Equations.

Especially post #4 from Astronuc (please have look in the link above for the complete citation):

Well trivially, it's Gauss's law. Inside the hollow conductor there is not charge, so the enclosed charge is zero, so the electric field is zero everywhere.

Now more directly, consider the most trivial case of the center of a hollow sphere, with 'uniform' charge on the surface. For each charge, there is an equal charge diametrically opposed, and therefore at the center the electrical fields (vectors) are equal and opposite, so they cancel.

Now, consider any point, off-center. One cannot apply the opposite point charge, but rather one must consider opposing surfaces, \$dA\$, which would have charges \$σ_1dA_1\$ and \$σ_2dA_2\$. Now think if two cones with vertices touching (and having same solid angles) and colinear (parallel) axes, with heights \$r_1\$ and \$r_2\$. The \$E\$ from one is just \$\dfrac{σ_1dA_1}{r_1^2}\$ and the other is \$\dfrac{σ_2dA_2}{r_2^2}\$, but realize that \$dA_i\$ is proportional to \$r_i^2dΩ_i\$, where \$dΩ\$ is the solid angle enveloped by cones and subtended by \$dA_i\$.

So \$E_i\$ is proportional to \$\dfrac{1}{r_i^2}\$, and \$dA_i\$ is proportional to \$r_i^2\$, and the term cancel which then leaves equal charges (\$σdΩ\$) opposing each other, and therefore the electric fields cancel, i.e. \$\vec{E}=0\$.