Electromagnetic Waves - Displacement Current

A parallel plate capacitor has circular plates of radius $r = 0.1\text{ m}$. It is being charged by an external source such that the electric field between the plates is changing at a rate of $\frac{dE}{dt} = 5 \times 10^{12} \text{ V/m}\cdot\text{s}$. Calculate the displacement current $I_d$ between the plates.

$I_d = \epsilon_0 A \frac{dE}{dt}$ Given: $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2)$ $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi \text{ m}^2$ $\frac{dE}{dt} = 5 \times 10^{12} \text{ V/m}\cdot\text{s}$ Substituting the values: $I_d = (8.854 \times 10^{-12}) \times (0.01 \times 3.1415) \times (5 \times 10^{12})$ $I_d = 8.854 \times 0.031415 \times 5 \approx 1.39 \text{ A}$

The displacement current is calculated using the rate of change of the electric field multiplied by the area of the plates and the permittivity of free space. In this case, the changing electric field induces a current even though no physical charges are moving across the gap.

Show that the displacement current between the plates of a capacitor is equal to the conduction current $I_c$ when the charge $Q$ on the capacitor is changing.

Electric flux $\Phi_E = \frac{Q}{\epsilon_0}$ (by Gauss's Law). Displacement current is defined as $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$. Substituting $\Phi_E$: $I_d = \epsilon_0 \frac{d}{dt} \left( \frac{Q}{\epsilon_0} \right)$ Since $\epsilon_0$ is constant: $I_d = \epsilon_0 \frac{1}{\epsilon_0} \frac{dQ}{dt}$ $I_d = \frac{dQ}{dt}$ Since $I_c = \frac{dQ}{dt}$, it follows that $I_d = I_c$.

This proof demonstrates that the current is continuous across the circuit. The conduction current in the wires becomes the displacement current in the dielectric/gap between the plates.