Displacement Current

Why the chapter opens here

• A current produces a magnetic field (Chapter 4). A magnetic field changing with time produces an electric field (Chapter 6, Faraday). Maxwell asked whether the converse holds: does an electric field changing with time produce a magnetic field?
• He found that it must, on grounds of logical consistency, and the extra source term he introduced is the displacement current. Everything else in the chapter — em waves, the whole spectrum, the electromagnetic nature of light — follows from it.

The inconsistency in Ampere's circuital law (the full 3-mark story)

• Set up: a parallel-plate capacitor $C$ in a circuit carrying a time-dependent current $i(t)$, being charged by a battery through a switch.
• Take a plane circular Amperian loop of radius $r$, centred on the wire and perpendicular to it, and apply Ampere's circuital law $$\oint\vec B\cdot d\vec l = \mu_0\, i(t)$$
• By symmetry $\vec B$ is along the circumference and has the same magnitude everywhere on the loop, so the left side is $B(2\pi r)$ and $$B(2\pi r) = \mu_0\, i(t)$$ giving a non-zero field at a point $P$ on the loop.
• Now keep the same loop but choose a different surface with the same boundary — a pot-shaped surface whose bottom lies between the capacitor plates, or a tiffin-box-shaped surface with a flat circular bottom $S$ between the plates. Neither is pierced by the wire.
• The left side, $\oint\vec B\cdot d\vec l$, is unchanged — it depends only on the loop. But the right side is now zero, because no conduction current crosses these surfaces.
• Contradiction: computed one way there is a magnetic field at $P$, computed the other way there is none. Ampere's law as it stands must be incomplete, and the missing term has to be one that makes the answer the same for every surface bounded by the loop.

Finding the missing term

• Look at what does cross the surface $S$ between the plates: the electric field. For plates of area $A$ carrying charge $Q$, the field between them is $$E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{A\varepsilon_0}$$ uniform over $A$, perpendicular to $S$, and zero outside.
• Electric flux through $S$, by Gauss's law: $$\phi_E = EA = \dfrac{1}{\varepsilon_0}\dfrac{Q}{A}\,A = \dfrac{Q}{\varepsilon_0}$$
• If $Q$ changes with time there is a current $i = dQ/dt$, and $$\dfrac{d\phi_E}{dt} = \dfrac{1}{\varepsilon_0}\dfrac{dQ}{dt} \;\Rightarrow\; \varepsilon_0\dfrac{d\phi_E}{dt} = i$$
• So the term to be added to the enclosed current is $\varepsilon_0\, d\phi_E/dt$. With it, both surfaces give the same total current $i$, and $B$ at $P$ is the same however the surface is chosen. $B$ just inside the gap (point M) equals $B$ just outside (point P), as it should.

Definitions and the generalised law

• Conduction current $i_c$: the current carried by conductors due to the actual flow of charges.
• Displacement current $i_d$: the current due to the changing electric field (electric displacement), $$i_d = \varepsilon_0\,\dfrac{d\phi_E}{dt}$$
• Total current: $$i = i_c + i_d = i_c + \varepsilon_0\dfrac{d\phi_E}{dt}$$
• Ampere-Maxwell law (Maxwell's generalisation of Ampere's circuital law): $$\oint\vec B\cdot d\vec l = \mu_0\, i_c + \mu_0\varepsilon_0\dfrac{d\phi_E}{dt} = \mu_0\!\left(i_c + \varepsilon_0\dfrac{d\phi_E}{dt}\right)$$
• In words: the line integral of $\vec B$ round a closed loop equals $\mu_0$ times the total current — conduction plus displacement — through any surface of which the loop is the perimeter.

Displacement current equals the charging current (the standard 2-mark derivation)

• Let the plates each have area $A$ and separation $d$, with instantaneous charge $Q$ on the positive plate, the capacitor being charged by an ac (or any time-varying) source.
• Field between the plates: $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{A\varepsilon_0}$
• Flux through the area $A$ between the plates: $\phi_E = EA = \dfrac{Q}{\varepsilon_0}$
• Displacement current: $$I_D = \varepsilon_0\dfrac{d\phi_E}{dt} = \varepsilon_0\dfrac{d}{dt}\!\left(\dfrac{Q}{\varepsilon_0}\right) = \dfrac{dQ}{dt}$$
• But $dQ/dt$ is precisely $I_C$, the conduction current feeding the plate. Hence $$I_D = I_C$$
• Consequence: the total current is continuous all the way round the circuit — conduction current in the wires hands over to displacement current in the gap and back again — which is why "current appears to flow through a capacitor" even though no charge crosses the gap. If the charging current is $0.25$ A, the displacement current across the plates is $0.25$ A.

Where each current lives

• In a steady electric field inside a conducting wire, $d\phi_E/dt = 0$ so $i_d = 0$.
• In most real regions both are present, since no medium is a perfect conductor or a perfect insulator.
• There can be large regions with no conduction current at all but a displacement current due to a time-varying field — and there a magnetic field exists with no current source nearby. This prediction is verified: the field measured at M between the plates equals the field at P just outside.

Why no steady current flows through a capacitor

• Once the capacitor is fully charged on a dc supply, the potential difference across it equals the emf, the field between the plates is constant, so $\phi_E$ is constant and $I_D = \varepsilon_0\, d\phi_E/dt = 0$.
• The gap is an insulator, so no conduction current can cross it either. Hence no steady-state current in the capacitor branch.
• While charging, $Q$ (and hence $E$ and $\phi_E$) increases with time, so a momentary displacement current $I_D = dQ/dt$ flows, equal to the conduction current in the wire. While discharging, $Q$ decreases, so a momentary current again flows, in the opposite sense, until the capacitor is fully discharged.

Maxwell's equations in vacuum

• Together with the Lorentz force $\vec F = q(\vec E + \vec v\times\vec B)$ they express all the basic laws of electromagnetism.
• Significance of the last two: a magnetic field changing with time gives rise to an electric field (Faraday), and an electric field changing with time gives rise to a magnetic field (displacement current). Time-dependent electric and magnetic fields therefore generate each other — and that mutual regeneration is what makes a self-propagating electromagnetic wave possible.
• The symmetry is not perfect: there is no known magnetic charge (monopole) to appear on the right of the second equation.

Diagrams described in words

• Ampere's-law inconsistency figure: draw a battery, switch and wire feeding a parallel-plate capacitor. Around the wire draw a circular loop of radius $r$ with its plane perpendicular to the wire, and mark a point $P$ on it. Then draw (i) a flat disc bounded by the loop, pierced by the wire, and (ii) a pot-shaped or tiffin-shaped surface with the same rim whose flat bottom $S$ sits between the plates and is pierced by no wire. Show uniform field arrows between the plates crossing $S$.
• Field between the plates: draw the two plates with parallel arrows for the uniform $\vec E$ pointing from the positive to the negative plate; at a point M between them show $\vec B$ as a circle in a plane parallel to the plates, i.e. along the perimeter of a circle centred on the axis — so $\vec E$ and $\vec B$ are perpendicular there. A cross-sectional view shows concentric circles of $\vec B$ around the axis.

Standard results for this subtopic

• $I_D = \varepsilon_0\dfrac{d\phi_E}{dt}$; $\phi_E = \dfrac{Q}{\varepsilon_0}$ for a parallel-plate capacitor; $I_D = I_C = \dfrac{dQ}{dt}$.
• $\oint\vec B\cdot d\vec l = \mu_0(i_c + i_d)$.
• $C = \dfrac{\varepsilon_0 A}{d}$ and $I = C\dfrac{dV}{dt}$: for circular plates of radius $12$ cm separated by $5.0$ cm charged at a constant $0.15$ A, $C \approx 80$ pF, $dV/dt \approx 1.9\times10^{10}$ V s$^{-1}$, and the displacement current across the plates is $0.15$ A.
• Kirchhoff's junction rule remains valid at each plate provided the displacement current is included along with the conduction current.
• For a capacitor across an ac source, $I_{rms} = V_{rms}\,\omega C$ and $I_0 = \sqrt2 I_{rms}$; the displacement current equals the conduction current at every instant.
• Magnetic field at radius $r$ from the axis between circular plates of radius $R$: $$B = \dfrac{\mu_0 r}{2\pi R^2}\, I_D$$ For $R = 6.0$ cm, $r = 3.0$ cm, $C = 100$ pF, $V_{rms} = 230$ V, $\omega = 300$ rad s$^{-1}$: $I_{rms} = 6.9\ \mu$A, $I_0 = 9.76\ \mu$A, and the amplitude of $B$ is about $1.63\times10^{-11}$ T.