As I understand it, a mobile phone and a base station communicate by exchanging photons. The antenna is like one half of a capacitor. Pulsating voltage is applied to the conductive antenna and that makes it loose energy by emitting photons. Similar pulsating voltages also get applied to capacitors, but capacitors are not supposed to conduct direct current. Why do they not loose energy by radiating away photons like an antenna?
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3 Answers
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First of all, the photon picture isn't the best way to visualize and model the operation of a radio-frequency antenna. RF antenna engineering is best performed and understood (IMHO) by means of a model in which high frequency AC power flow through a wire creates electromagnetic waves that propagate away into space.
For this to occur, the antenna impedance must be well-matched to the impedance of free space. As such, every antenna possesses a certain (small) amount of capacitance and inductance- so one plate of a capacitor is not an accurate model of an antenna (although capacitor "plates" affixed to specific points of an antenna are commonly used to add enough capacitance to it so that in conjunction with the antenna's intrinsic inductance, the antenna is resonant at the frequency of the transmitter connected to it).
Now note that in a capacitor with two plates, the electric field gets concentrated into the gap between them- by design, you don't want to "waste" the field by letting it leak out of the gap. Some leakage is inevitable, and at very high frequencies, even a well-designed capacitor will leak enough to radiate small amounts of electromagnetic waves.
answered Apr 16, 2021 at 3:51
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$\begingroup$ Is photon not the exact same thing as an EM wave? Whatever energy an antenna radiates can in theory be counted as an integer number of discrete photons. What is most puzzling to me is the birth of the photon. Does it happen only in atoms, from electrons in excited states or also from free electrons in the metallic lattice? Is there e.g. some difference in design of an antenna and a capacitor that supports or prohibits appearance of electrons in excited states on the metal surface? $\endgroup$– MKaamaCommented Apr 19, 2021 at 16:04
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All elements of a circuit will radiate when there is alternating current involved. Typically though, at the frequencies involved, not very much. It is something that circuit designers must be careful of though.
Capacitors themselves typically are encased in some sort of material which will damp the stray radiation, especially if it is a conductive material and hence makes a little Faraday cage.
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$\begingroup$ Would you say a well-designed capacitor looses the same amount of energy as a well-designed antenna, if connected to the same voltage, current and frequency input signal? Just that it (the energy) does not go very far and is converted to heat on the spot? $\endgroup$– MKaamaCommented Apr 19, 2021 at 16:12
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$\begingroup$ Well you have to define "loses". An antenna is designed to "lose" that energy by radiating it. With an antenna, the only place for energy to go is to either be radiated or converted to heat. A good antenna will do all of the former and none of the latter. For a capacitor, it is designed to store energy but almost all of the time the real goal is for that energy to be released back into the circuit in the right way. So any energy lost to heat is a waste, and this parameter is known as the "parasitic resistance", and lower is better. $\endgroup$– geshelCommented Apr 19, 2021 at 18:00
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I think your question might be valid in some non-technical sense. The truth is, the signal does lose energy. As someone previously stated, it is the power of the signal and the impedance of the atmosphere that matter. As the broadcaster of the signal, you have to push enough power through the air to send the signal. If you want to think of these bodies as the plates of a capacitor, you can. But that’s not really what people mean when they talk about a capacitor. But you bring up a valid point. Capacitance (and inductance) happens all the time and in all directions.
