Introduction to Electrodynamics · Problem 11.8

Radiation — David J. Griffiths

Physics · radiation-from-discharging-capacitor

Problem

A parallel-plate capacitor CC, with plate separation dd, is given an initial charge (±)Q0(\pm)Q_0. It is then connected to a resistor RR, and discharges, Q(t)=Q0e−t/RCQ(t) = Q_0 e^{-t/RC}. (a) What fraction of its initial energy (Q02/2CQ_0^2/2C) does it radiate away? (b) If C=1C = 1 pF, R=1000 ΩR = 1000\,\Omega, and d=0.1d = 0.1 mm, what is the actual number? In electronics we don't ordinarily worry about radiative losses; does that seem reasonable, in this case?
Intuitive explanation

The key insight is that a discharging capacitor is an accelerating charge distribution—the changing electric field radiates electromagnetic energy according to the Larmor formula. You'll need to integrate the radiated power over the entire discharge process and compare it to the initial stored energy. The result reveals why real circuits usually ignore radiation: the fraction depends sensitively on the ratio of the capacitor's size to the wavelength of radiation it emits, and for typical electronics, this ratio is tiny.

Intuitive explanation

The key insight is that a discharging capacitor is an accelerating charge distribution—the changing electric field radiates electromagnetic energy according to the Larmor formula. You'll need to integrate the radiated power over the entire discharge process and compare it to the initial stored energy. The result reveals why real circuits usually ignore radiation: the fraction depends sensitively on the ratio of the capacitor's size to the wavelength of radiation it emits, and for typical electronics, this ratio is tiny.

Related theory

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