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Out of curiosity, I estimated the voltage amplitude for typical visible light. Wikipedia says that sunlight hits the earth at about $1000 W / m^2$. The intensity of an electric field is given by

$$I = \frac{\epsilon_0 c}{2} E^2 .$$

If the voltage of a light wave is given by

$$V_0 \cos(x / 2 \pi \lambda)$$ where $\lambda$ is the wavelength of the light, then as $E = - \nabla V$, we have

$$V_0 = 2\pi \lambda \sqrt \frac{2 I}{\epsilon_0 c}.$$ Plugging in $I = 1000 W/m^2$ and $\lambda = 500$nm, the wavelength of blue light, we get $V_0$ is 3 millivolts, which seems reasonable to me.

Then I tried to do this for radio waves. According to Wikipedia, the power of a FM radio transmitter is about $50 kW$. Furthermore, a typical FM wavelength is about 3 meters long. If we are a distance of just 2 kilometers away from the radio tower, using $I = P / 4 \pi r^2$, we can plug in these numbers and get that, for radio waves, $V_0$ is 16 volts, which larger than any typical battery. Why don't the large AC currents induced in electronics near a radio tower ruin the standard functioning of all electrical devices?

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  • $\begingroup$ You might want to take a look at this calculation - check your numbers, they seem high. They come up with 40 mV/m at 5 µW/m$^2$ of power. At 2 km, the power density should be 1 mW/m$^2$. $\endgroup$ – Floris Nov 29 '17 at 4:35
  • $\begingroup$ Light waves do not have "voltage". Even for the simplest electromagnetic wave, a plane wave, we cannot introduce fully descriptive electric potential, because its electric field is not a potential field. Consequently there is no sense to "voltage". One has to work with electric field directly or with vector potential. $\endgroup$ – Ján Lalinský Dec 17 '18 at 3:14
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Look at this plot of Radio Frequency Energy Harvesting - Sources and Techniques maximum received power

power of radio waves

So radio wave fields are of order of microwatts per meter square at a distance of 50 meters from the antenna.

Compare it to the 1000Watts/m^2 from the sun to realize your estimates are wrong.

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If the radio waves are strong enough, they can easily destroy electronics devices. But they need to be very, very strong and able to penetrate the case of the device to do so.

The electromagnetic whiplash (called electromagnetic pulse or EMP) of an atomic or nuclear explosion is sudden enough and strong enough to induce very large currents in metal objects tens to hundreds of km away from the blast. Those currents are sufficient to blow up high-voltage power line switchgear and fry the semiconductor junctions inside all sorts of solid-state electronic devices. To survive, military electronics as used in the battlefield and in aircraft must be carefully "hardened" against EMP with special shielding techniques.

In the case of light: If the box enclosing the electronics is colored plastic or metal, sunlight will not make it inside. Even if it could, the wavelength of the light is far, far shorter than the dimensions of the circuit board with the electronics on it, and so electric currents will not be induced in the board by having sunlight fall on it.

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  • $\begingroup$ For a fascinating discussion of nuclear EMP, see this excellent paper. $\endgroup$ – rob Dec 17 '18 at 1:26
  • $\begingroup$ @rob, most excellent indeed. $\endgroup$ – niels nielsen Dec 17 '18 at 7:23

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