Background: A Faraday cage is essentially an enclosure of a conducting material. According to electrodynamics, signals cannot pass into or out of this enclosure, because when a signal approaches the cage, in causes charges in the conducting material to accelerate, which creates a signal that cancels out the original signal.

Faraday cages have other interesting properties besides canceling out electromagnetic radiation. An entertaining example was investigated by the BBC, found here, where we learn that cars can effectively shield people inside from lightning strikes. It seems not matter that a car isn't a perfect Faraday cage, what with all the windows and potentially open doors. I hypothesized that cars might also be able to block radio signals, but when I tested some things (below), I got some confusing results. I need to do more tests, but I'm out of ideas.

I decided to do an experiment where I investigated how well a car can insulate against radio waves, in the form of public broadcasting and wifi. First, I took off the antenna from my radio. As expected, the signal got much worse from the car's radio, but it didn't affect a transistor radio inside the car very much. In fact, the transistor radio was loud and clear. This seems to suggest that the car does not do a very good job of canceling out radio waves. Question 1: Why might this be?

It got more interesting when I brought my laptop inside my car. The signal died out completely. I can stand right outside my car and receive wifi signal very well, but when I go inside the car, I am not connected to any network. I did it several times. Confused, I pulled my car close to the wifi transmitter. Same effect. I got out of the car and, holding my laptop, doubled the distance to transmitter. The signal was not strong, but I was still connected.

Although wifi and public radio have different frequencies, still, they are both radio waves. I don't understand how the observed effects of the car could be so different. Can anyone propose further experiments that I might do to try to explain what is happening to the signal?


2 Answers 2


To begin, there is a intimate relationship between the wavelength of the signal, the size of the gaps in the conducting enclosure and the amount of attenuation observed. Take careful notice that the we're not talking about a all or nothing effect.

Secondly lighting isn't just a electomagnetic signal as there is also a substantial motion of charges as well, and the way those currents behave is different from how the pure electromagnetic fields work.

In short, this isn't a phenomena that is easy to address in any depth or completeness with a few home-experiments.

  • $\begingroup$ I recognize that lightning is different from a electromagnetic signal, that example was just there to show that cars do, in fact, have some Faraday cage behavior. The other observed behavior of a Faraday cage is the large attenuation of wifi signals. I think your discussion of the size of gaps and amount of attenuation is constructive, but I also think there is more to be discussed than simply stating the existence of their relationship. $\endgroup$
    – psitae
    Commented Nov 19, 2016 at 22:29
  • $\begingroup$ Check out my question. It relates to your, but I received no comments or answers. physics.stackexchange.com/questions/293125/… $\endgroup$
    – Lambda
    Commented Nov 20, 2016 at 0:40

This is a very good question. One thing that I might add is that WiFi/5G is actually microwave radiation, not radio. Radio signals are much larger, even miles long in wavelength!

Your microwave oven has a wavelength around an inch or so (which you can see on a stick of butter when you microwave it). This is 13.627 GHz. Note that phones and "5G" are 5GHz, which is a wavelength of 2.361 inches.

photon wavelength and electromagnetic frequency are inversely related with the speed of light

Wolfram Alpha is quite useful for this: https://www.wolframalpha.com/input?i=2+cm+electromagnetic+wavelength+to+frequency&assumption=%7B%22FS%22%7D+-%3E+%7B%7B%22PhotonWavelength%22%2C+%22lambda%22%7D%2C+%7B%22PhotonWavelength%22%2C+%22nu%22%7D%7D&assumption=%7B%22F%22%2C+%22PhotonWavelength%22%2C+%22nu%22%7D+-%3E%225+GHz%22


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