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Maxwell's equations and the displacement current

The displacement current $\varepsilon_0\partial\vec{E}/\partial t$ completes Ampère's law; Maxwell's four equations in vacuum and in matter predict waves at $1/\sqrt{\mu_0\varepsilon_0}$.

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1. What you will learn

By the end of this lesson you will be able to compute displacement currents and the magnetic fields they produce, state Maxwell's equations in vacuum and in matter, and find the speed of electromagnetic waves in media.

2. What you already have

You have Gauss's law, $\nabla \cdot \vec{B} = 0$, Faraday's law, and Ampère's law for steady currents, along with the continuity equation for charge. You know that the divergence of a curl is always zero. This lesson finds a contradiction among these laws and resolves it, completing the equations of electromagnetism.

3. Words for this lesson

4. A changing electric field is a source of magnetism

Take the divergence of Ampère's law, $\nabla \times \vec{B} = \mu_0\vec{J}$. The left side is zero for any field, so the law demands $\nabla \cdot \vec{J} = 0$. But charge conservation says $\nabla \cdot \vec{J} = -\partial\rho/\partial t$, which is not zero while charge accumulates, as on a charging capacitor. Using Gauss's law, $\partial\rho/\partial t = \varepsilon_0\nabla \cdot (\partial\vec{E}/\partial t)$, so the combination $\vec{J} + \varepsilon_0\,\partial\vec{E}/\partial t$ always has zero divergence. Maxwell replaced $\vec{J}$ by it:

$$\nabla \times \vec{B} = \mu_0\vec{J} + \mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t}.$$

The new term, the displacement current, says a changing electric field produces a magnetic field just as a current does. Between the plates of a charging capacitor it equals the current in the wires, so the magnetic field is continuous around the circuit. With it, Maxwell's equations are complete:

$$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}, \quad \nabla \cdot \vec{B} = 0, \quad \nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}, \quad \nabla \times \vec{B} = \mu_0\vec{J} + \mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t}.$$

In empty space, each changing field generates the other, and together they propagate as waves with speed $c = 1/\sqrt{\mu_0\varepsilon_0}$ — the measured speed of light.

5. The method, step by step, and how to check it

1. Find the changing field. Between parallel plates, $E = Q/\varepsilon_0A$, so $\varepsilon_0\,dE/dt\cdot A = dQ/dt = I$.
2. Choose an Amperian loop and a surface. The law holds for any surface bounded by the loop; the total of conduction and displacement current through it is the same for every surface.
3. Solve with symmetry. Circular plates give a field circling the axis, growing linearly between the plates and falling as $1/s$ outside them.

Checks. The total displacement current in a capacitor gap must equal the conduction current in its wires. In steady situations the displacement current vanishes and magnetostatics returns. The units: $\varepsilon_0\,\partial E/\partial t$ is in A/m². And $1/\sqrt{\mu_0\varepsilon_0}$ must come out to $3.00 \times 10^8$ m/s.

6. Why the correction is needed

Draw a loop around the wire leading to a charging capacitor. Ampère's law says $\oint\vec{B} \cdot d\vec{l}$ equals $\mu_0$ times the current through any surface bounded by that loop. A flat disk pierced by the wire gives $\mu_0I$. A surface bulging out to pass between the plates, where no charge flows, gives zero. The same loop cannot have two circulations, so the original law must be incomplete.

Maxwell's term resolves it: the bulging surface is crossed by the growing electric field, whose displacement current equals $I$. Both surfaces now give $\mu_0I$. Maxwell proposed the term in 1861 largely from this kind of consistency argument, before any experiment had detected it. It is one of the great examples of theory, pursued with care, predicting new physics.

7. The prediction of light

In empty space, with no charges or currents, the curl equations become $\nabla \times \vec{E} = -\partial\vec{B}/\partial t$ and $\nabla \times \vec{B} = \mu_0\varepsilon_0\partial\vec{E}/\partial t$. Taking the curl of one and substituting the other gives the wave equation, $\nabla^2\vec{E} = \mu_0\varepsilon_0\,\partial^2\vec{E}/\partial t^2$, for waves traveling at $1/\sqrt{\mu_0\varepsilon_0}$.

Maxwell computed this speed from electrical measurements of $\varepsilon_0$ and $\mu_0$ made with capacitors and coils, and found it matched the measured speed of light. He concluded that light is an electromagnetic wave, uniting optics with electricity and magnetism. Heinrich Hertz generated and detected such waves at radio frequencies in 1887, and within two decades they carried messages across the Atlantic.

8. Displacement current is small but essential

In ordinary circuits the displacement current is hard to notice. In a capacitor it equals the conduction current, but its magnetic field is spread over the plates' area and is weak: microtesla for amperes of current. Inside good conductors the displacement current is utterly negligible compared with the conduction current, because the fields there are tiny.

Its importance is structural. Without it there are no electromagnetic waves: no light, no radio, no radar. With it, a changing electric field anywhere in space produces a magnetic field, which in turn produces an electric field, and the disturbance spreads. At high frequencies it dominates, which is why capacitors pass alternating current: the displacement current in the gap grows with frequency.

9. Maxwell's equations in matter

Inside materials, polarization and magnetization add bound charges and currents. Moving them into $\vec{D}$ and $\vec{H}$ leaves equations with only free sources: $\nabla \cdot \vec{D} = \rho_f$, $\nabla \cdot \vec{B} = 0$, $\nabla \times \vec{E} = -\partial\vec{B}/\partial t$, and $\nabla \times \vec{H} = \vec{J}_f + \partial\vec{D}/\partial t$. A time-varying polarization also contributes a polarization current, $\partial\vec{P}/\partial t$, the motion of bound charge as dipoles change.

For linear materials, $\vec{D} = \varepsilon\vec{E}$ and $\vec{B} = \mu\vec{H}$, and waves travel at $1/\sqrt{\mu\varepsilon}$, slower than in vacuum by the refractive index $n = \sqrt{\varepsilon_r\mu_r}$. Boundary conditions follow from the equations at interfaces and determine how light reflects and refracts, the subject of the next unit.

10. Symmetry between electric and magnetic

In empty space Maxwell's equations are nearly symmetric: swap $\vec{E}$ for $c\vec{B}$ and $c\vec{B}$ for $-\vec{E}$ and they map into each other. The asymmetry lies in the sources: there are electric charges and currents but no magnetic ones, as far as anyone has found. If magnetic monopoles existed, adding their charge and current density would make the equations fully symmetric.

Paul Dirac showed in 1931 that even one monopole anywhere in the universe would explain why electric charge comes in multiples of a basic unit. The equations' near symmetry, and the question of whether it is exact, remain a live topic in particle physics and in materials called spin ices, whose excitations behave like emergent magnetic monopoles.

11. Charge conservation is built in

Taking the divergence of the Ampère–Maxwell law and using Gauss's law gives the continuity equation, $\nabla \cdot \vec{J} + \partial\rho/\partial t = 0$, automatically. Maxwell's equations cannot be solved with sources that violate charge conservation; conservation is not an extra assumption but a consequence.

Charge conservation holds to extraordinary precision. Experiments searching for the decay of an electron into a neutrino and a photon, which would destroy charge, find its lifetime exceeds $6.6 \times 10^{28}$ years, set by the Borexino detector in Italy. The deep reason is a symmetry: electromagnetism's gauge invariance, the freedom to change the potentials without changing the fields, implies charge conservation through Noether's theorem.

12. Signals in cables and circuit boards

Electrical signals travel along wires as electromagnetic waves guided by the conductors, at $1/\sqrt{\mu\varepsilon}$ of the insulating material, not of the metal. In a coaxial cable with solid polyethylene, $\varepsilon_r = 2.25$, signals travel at $2.0 \times 10^8$ m/s, two thirds of $c$; foamed insulation raises this to $80$ percent. On a circuit board made of FR-4, $\varepsilon_r \approx 4.4$, signals travel at about half the speed of light.

At gigahertz clock rates this matters enormously. A signal takes about $7$ ns to cross a meter of circuit board, and designers of computer motherboards and data-center networks match trace lengths to within millimeters so that bits sent in parallel arrive together. The speed follows directly from Maxwell's equations in matter.

13. In the world: signal timing in circuit boards

In a computer, data move between processor and memory as electromagnetic waves guided by copper traces on circuit boards. Their speed is $c/\sqrt{\varepsilon_r}$ of the board material: for the common FR-4 laminate, $\varepsilon_r \approx 4.4$, about half the speed of light, or roughly $7$ ns per meter. At the data rates of modern memory, several billion transfers per second, a bit occupies only a few centimeters of trace.

Designers therefore lay out the parallel traces of a memory bus with matched lengths, adding serpentine detours to the shorter ones, so that all the bits of a word arrive within a few picoseconds of each other. High-speed networks in data centers use specialized low-loss laminates with lower $\varepsilon_r$, both for speed and for lower loss. Every one of these choices applies Maxwell's equations in matter.

14. In the world: coaxial cable

Cable television, radio antennas, laboratory instruments and the connections inside cell towers use coaxial cable: a center conductor, an insulating layer, and a surrounding shield. The signal travels as an electromagnetic wave in the insulator, at $c/\sqrt{\varepsilon_r}$: about $66$ percent of $c$ with solid polyethylene, over $80$ percent with foamed polyethylene or air-spaced designs.

Engineers call this fraction the velocity factor and use it constantly. Cutting a cable to a quarter wavelength for an antenna matching section, or timing signals from GPS antennas, requires the speed in the cable, not in air. The same equations explain why coax shields so well: the fields are confined between the conductors, and outside, where the currents on the two conductors cancel, there is essentially no field at all.

15. A capacitor gap is not magnetically empty

Because no charge crosses the gap of a capacitor, it seems there should be no magnetic field around it. But the electric field between the plates is changing, and a changing electric field is a source of magnetic field. Around the gap the field is exactly what the wire's current would produce at the same distance; measurements confirm it.

A second misconception is that displacement current is a flow of something. Nothing is displaced in vacuum; the name comes from Maxwell's mechanical model of the ether, since abandoned. It is simply the rate of change of the electric field, entering the equations where a current would.

16. The displacement current in a capacitor

1. A parallel-plate capacitor with plate area $A$ carries charge $Q(t)$. Write the field between the plates.

   $E = \dfrac{Q}{\varepsilon_0A}$

   Uniform between large plates.

2. Find the electric flux.

   $\Phi_E = EA = \dfrac{Q}{\varepsilon_0}$

   Through a surface between the plates.

3. Find the displacement current.

   $I_d = \varepsilon_0\dfrac{d\Phi_E}{dt} = \dfrac{dQ}{dt}$

   Maxwell's term.

4. Compare with the conduction current.

   $I_d = I$

   The current charging the plates.

5. Evaluate for $I = 2.0$ A.

   $I_d = 2.0\ \text{A}, \quad \dfrac{dE}{dt} = \dfrac{I}{\varepsilon_0A}$

   For $A = 0.010$ m², $dE/dt = 2.3 \times 10^{13}$ V/(m s).

17. The magnetic field between circular plates

1. Circular plates of radius $R = 5.0$ cm charge with $I = 2.0$ A. Choose a circle of radius $s < R$ between them.

   $\oint\vec{B} \cdot d\vec{l} = B \cdot 2\pi s$

   Symmetry about the axis.

2. Find the enclosed displacement current.

   $I_{d,\text{enc}} = I\dfrac{\pi s^2}{\pi R^2}$

   Proportional to the enclosed area.

3. Solve for the field.

   $B = \dfrac{\mu_0Is}{2\pi R^2}$

   Linear in $s$.

4. Evaluate at $s = 2.5$ cm.

   $B = \dfrac{2 \times 10^{-7} \times 2.0 \times 0.025}{(0.050)^2} = 4.0\ \mu\text{T}$

   Small but real.

5. Evaluate at the edge.

   $B(R) = \dfrac{2 \times 10^{-7} \times 2.0}{0.050} = 8.0\ \mu\text{T}$

   Same as next to the wire at that distance.

6. Evaluate beyond the plates.

   $B(2R) = 4.0\ \mu\text{T}$

   Like a wire, falling as $1/s$.

18. From Maxwell's equations to the speed of light

1. Take the curl of Faraday's law in empty space.

   $\nabla \times (\nabla \times \vec{E}) = -\dfrac{\partial}{\partial t}(\nabla \times \vec{B})$

   Curl both sides.

2. Use the Ampère–Maxwell law.

   $\nabla \times \vec{B} = \mu_0\varepsilon_0\dfrac{\partial\vec{E}}{\partial t}$

   No currents in empty space.

3. Expand the double curl.

   $\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \nabla^2\vec{E} = -\nabla^2\vec{E}$

   $\nabla \cdot \vec{E} = 0$ with no charge.

4. Combine into a wave equation.

   $\nabla^2\vec{E} = \mu_0\varepsilon_0\dfrac{\partial^2\vec{E}}{\partial t^2}$

   The form $\nabla^2f = \tfrac{1}{v^2}\partial_t^2f$.

5. Read off the speed.

   $v = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$

   Built from electrical constants.

6. Evaluate the expression.

   $v = \dfrac{1}{\sqrt{4\pi \times 10^{-7} \times 8.854 \times 10^{-12}}} = 2.998 \times 10^{8}\ \text{m/s}$

   The speed of light.

7. Repeat for a medium.

   $v = \dfrac{1}{\sqrt{\mu\varepsilon}} = \dfrac{c}{n}, \quad n = \sqrt{\varepsilon_r\mu_r}$

   The refractive index from material constants.

8. Evaluate for polyethylene, $\varepsilon_r = 2.25$.

   $v = \dfrac{2.998 \times 10^8}{1.5} = 2.0 \times 10^{8}\ \text{m/s}$

   Signals in coaxial cable.

19. Your turn: what is the displacement current between plates of area $0.02$ m² if the field increases at $10^{12}$ V/(m s)?

1. Write the displacement current.

   $I_d = \varepsilon_0A\dfrac{dE}{dt}$

   Maxwell's term.

2. Substitute the values.

   $I_d = 8.85 \times 10^{-12} \times 0.02 \times 10^{12}$

   SI units.

3. Your turn: work this step out. Its working is at the end of the packet.

   Evaluate the current.

20. Guided practice

A capacitor is charging. A few centimeters from the wire leading to it, the magnetic field is $40$ μT. What is the field the same distance from the axis, level with the gap between the plates, if the plates are smaller than that distance?

• $40$ μT, the same
• zero, since no charge crosses the gap
• $20$ μT, half, since only one plate is charging
• $80$ μT, doubled by both plates

21. Guided practice

Complete the worked solution: a capacitor with circular plates of radius $10$ cm charges with current $7$ A. At $s = 3$ cm from the axis between the plates, find the enclosed displacement current in A, the magnetic field in μT, and the field at the plates' edge in μT.

1. Scale the current by the area fraction.

   $I_{d,\text{enc}} = 7 \times \left(\dfrac{3}{10}\right)^2 =$ _p

   Uniform displacement current.

2. Apply Ampère–Maxwell at radius $s$.

   $B = \dfrac{2 \times 10^{-7} \times I_{d,\text{enc}}}{3 \times 10^{-2}} =$ _q

   In μT.

3. Apply it at the edge.

   $B(R) = \dfrac{2 \times 10^{-7} \times 7}{0.10} =$ _r

   The largest field between the plates.

22. Guided practice

Match each of Maxwell's equations to its name.

23. Practice

Between the circular plates of radius $R$ of a charging capacitor, $\mu_0I/2\pi R = 864$ μT. Fill in the magnetic field level with the gap at $s = R/2$, $R$ and $2R$.

24. Practice

The electric field between two plates of area $40$ cm² is increasing at $19 \times 10^{12}$ V/(m s). What is the displacement current, in mA? Use $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m.

Answer: mA

25. Practice

A capacitor with circular plates of radius $50$ cm is being charged by a current of $3$ A. What is the magnetic field between the plates, $2$ cm from the axis, in μT?

Answer: μT

26. Somewhere new

A coaxial cable is insulated with foamed polyethylene, relative permittivity $\varepsilon_r = 1.44$ and $\mu_r = 1$. How fast do signals travel along it, in units of $10^8$ m/s? Use $c = 2.998 \times 10^8$ m/s.

Answer: × 10⁸ m/s

27. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

28. Test question

Between the circular plates of radius $R$ of a charging capacitor, $\mu_0I/2\pi R = 821$ μT. Fill in the magnetic field level with the gap at $s = R/2$, $R$ and $2R$.

29. What you can do now

You can use the complete Maxwell equations. Explain to someone why there is a magnetic field around the gap of a charging capacitor.

Working for the steps left to you

19. Your turn: what is the displacement current between plates of area $0.02$ m² if the field increases at $10^{12}$ V/(m s)?, step 3