The missing piece Maxwell needed to complete his equations, and why light, radio, and gamma rays are all fundamentally the same phenomenon at different frequencies.
Ampere's Circuital Law (Chapter 4) works perfectly for steady currents, but breaks down in one specific situation: a charging capacitor. Current flows in the wires up to the plates, but no actual charge crosses the gap between them — so which current do you use in Ampere's Law for a Gaussian surface that bulges between the plates?
Maxwell resolved this by proposing that a changing electric field is, in every meaningful sense, equivalent to a current — even where no charge is physically moving.
With displacement current added, Ampere's Law becomes the Ampere–Maxwell Law:
Any accelerating electric charge radiates an electromagnetic wave. A charge oscillating back and forth (as in an antenna) produces a continuous EM wave at the frequency of oscillation — this is exactly how radio and TV transmitters work: an oscillating current in the antenna generates a matching electromagnetic wave that propagates outward.
EM waves carry energy and momentum, distributed between the electric and magnetic fields:
Visible light is just one narrow slice of a vast continuum of electromagnetic waves, all travelling at exactly the same speed c, distinguished only by frequency (and correspondingly, wavelength).
A parallel plate capacitor is being charged, and a conduction current of 2 A flows through the connecting wires. What is the displacement current between the plates?
Solution: For the Ampere–Maxwell Law to be consistent regardless of which surface is chosen, displacement current must exactly equal conduction current: Id = Ic = 2 A.
Find the wavelength of an electromagnetic wave with frequency 5×10¹⁴ Hz.
Solution: λ = c/f = (3×10⁸) / (5×10¹⁴) = 6×10⁻⁷ m = 600 nm — squarely in the visible light range (orange light).
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