5.2.2 The displacement current equals the conduction current

Let us make the argument quantitative for a parallel-plate capacitor of plate area $A$ and spacing $d$ , filled with a medium of permittivity $ε$ (Figure 5.2.3). The plate voltage is $v (t)$ . Between the plates, with fringing neglected, the field is uniform (Section 3.4),

E Ψ = \frac{v}{d}, D = ε \frac{v}{d}, = \int_{S} D \cdot d S = D A = \frac{ε A}{d} v = C v, (5.2.6)

where $C = ε A / d$ is the capacitance of Section 3.4. The displacement current through any surface $S$ across the gap is therefore

i_{d} = \int_{S} \frac{\partial D}{\partial t} \cdot d S = \frac{d Ψ}{d t} = C \frac{d v}{d t} = i, (5.2.7)

which is exactly the circuit-theory law of the capacitor. The relation $i = C d v / d t$ , used since the first year, is the second Maxwell equation applied to the gap. Home exercise 5.2.1 carries this out for a sinusoidal voltage.

(a) The capacitor and an Amperian loop in the gap.

(b) H_\phi in the gap (solid) and around the feeding wire (dashed). The displacement current of a parallel-plate capacitor. (a) The plate voltage v(t) sets D=\varepsilon v/d in the gap, and d\Psi/dt=C\,dv/dt is the current i in the wire. The generalised Ampère law on the dashed loop C_\rho gives the magnetic field inside the gap. (b) For circular plates of radius R the field grows linearly with \rho inside the gap and joins the 1/\rho field of the wire at the rim: the magnetic field does not notice the gap. — **Figure 5.2.3.** (b) $H_{ϕ}$ in the gap (solid) and around the feeding wire (dashed). The displacement current of a parallel-plate capacitor. (a) The plate voltage $v (t)$ sets $D = ε v / d$ in the gap, and $d Ψ/ d t = C d v / d t$ is the current $i$ in the wire. The generalised Ampère law on the dashed loop $C_{ρ}$ gives the magnetic field inside the gap. (b) For circular plates of radius $R$ the field grows linearly with $ρ$ inside the gap and joins the $1/ ρ$ field of the wire at the rim: the magnetic field does not notice the gap.

The displacement current also has a magnetic field, as a current should. Apply Eq. (5.2.5) to the dashed circle $C_{ρ}$ of radius $ρ$ centred on the axis inside the gap, where $J = 0$ . By symmetry $H = H_{ϕ} (ρ) a_{ϕ}$ , and $\partial D / \partial t$ is uniform over the disc spanning the circle, so

⟹ H_{ϕ} 2 π ρ = \frac{\partial D}{\partial t} π ρ^{2} H_{ϕ} = \frac{ρ}{2} \frac{\partial D}{\partial t} = \frac{ρ}{2} \frac{i}{A} (5.2.8)

for circular plates with $\partial D / \partial t = i / A$ . At the rim of plates of radius $R$ , $A = π R^{2}$ and $H_{ϕ} = i / (2 π R)$ , which is the field of the wire outside: the magnetic field joins continuously, and the field of the current "through" the capacitor is the field of a current spread uniformly over the plate (Figure 5.2.3b, Exercise 5.2.1).

The picture

Think of the lines of $D$ between the plates as the continuation of the current. When $q$ grows, new lines appear in the gap at the rate $d q / d t$ , and the magnetic field circulates around the growing bundle of lines exactly as it circulates around the wire. The flux lines of $D$ carry the current across the gap.

Quick check

A $1 μ F$ capacitor is charged so that its voltage rises at $10 V/ms$ . How large is the displacement current through the gap?

$i_{d} = C d v / d t = 1 0^{- 6} \times 1 0^{4} = 10 mA$ , equal to the conduction current in the leads at the same instant. The size and shape of the plates do not matter: only $C d v / d t$ does.