Home exercise 5.2.1: Displacement current in a capacitor

Home exercise 5.2.1

Interactive: Displacement Current

Home exercise 5.2.1

Displacement current in a capacitor

A parallel-plate capacitor with plates of area $A$ and separation $d$ (air between the plates, fringing neglected) is connected to the voltage $V (t) = V_{0} cos ω t$ .

Find the electric field $E$ and the displacement current density $J_{d} = \partial D / \partial t$ between the plates.
Find the total displacement current $i_{d}$ through the gap.
Show that $i_{d}$ is equal to the conduction current $i (t)$ in the wire.

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 7-5, and B. M. Notaroš, Electromagnetics, Example 8.1.

Try it yourself first.

Strategy

Section 5.2.2 in a concrete case. The field in the gap is $V / d$ , uniform, so $J_{d} = ε_{0} \partial E / \partial t$ is uniform too and $i_{d} = J_{d} A$ . The conduction current is $d q / d t$ with $q = C V$ the charge on a plate.

Between the plates the field is uniform (the plates are much smaller than the wavelength, so the field follows the voltage without delay), directed from the positive to the negative plate. Let the upper plate be at $z = d$ and carry the voltage $V (t)$ with respect to the lower one. Then

E D J_{d} = - \frac{V ( t )}{d} a_{z} = - \frac{V _{0}}{d} cos ω t a_{z} [V/m], = ε_{0} E, = \frac{\partial D}{\partial t} = \frac{ε _{0} V _{0} ω}{d} sin ω t a_{z} [A/ m^{2}] .

$J_{d}$ is uniform over the plate area, so the displacement current crossing a surface in the gap with normal $+ a_{z}$ (from the lower to the upper plate) is

Answer

i_{d} = \int_{A} J_{d} \cdot d S = \frac{ε _{0} A}{d} ω V_{0} sin ω t = C ω V_{0} sin ω t, C = \frac{ε _{0} A}{d}

The charge on the upper plate is $q = C V$ , and the conduction current flowing in the wire into the upper plate is

i = \frac{d q}{d t} = C \frac{d V}{d t} = - C ω V_{0} sin ω t .

This is the current counted into the upper plate from above. The displacement current of part (b) was counted upward, from the lower plate into the upper plate from below, and it is $- i$ : the total current entering the upper plate, $i + i_{d} = 0$ , vanishes at every instant. In other words, the current $i$ that arrives through the wire leaves the plate through the gap as the displacement current $- i_{d} a_{z} = i (- a_{z})$ . With one consistent reference direction, from the source round the loop and down through the gap, the two are the same current:

Answer

i_{d} = i = C \frac{d V}{d t}, ∣ i_{d} ∣ = ∣ i ∣ = C ω V_{0} ∣ sin ω t ∣

Compare the current in the wire with the displacement current through the gap at every instant.

InteractiveLoading simulation…

Check

This is Maxwell's argument in miniature (Section 5.2): the current that disappears into the plate reappears as $\partial D / \partial t$ in the gap, so the total current $J + \partial D / \partial t$ is continuous round the circuit and $\oint H \cdot d l$ is the same on every surface spanning the loop. The circuit-theory law $i = C d V / d t$ is the displacement current. Units: $(ε_{0} A / d) ω V_{0}$ is $F \cdot s^{- 1} \cdot V = C/s = A$ .