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?