Maxwell's Equations and Plane Waves: Boundary Conditions, Poynting, Skin Depth and Polarisation

Section 8 of the Electronics and Communication Engineering paper is the one section that cannot be attacked by pattern-matching, because it rests on four equations that are genuinely more compact than the phenomena they describe. Maxwell's equations in differential form are local statements about what a field does at a point; in integral form they are global statements about a surface or a loop, and the two are the same physics related by the divergence and Stokes theorems. Learning both forms is not duplication — a question that gives you a closed surface wants the integral form, and a question that gives you a field expression wants the differential. Combine the two curl equations and you get the wave equation, whose solution in free space is a plane wave with E and H perpendicular to each other and to the direction of travel, in the fixed ratio η₀ = 376.73 Ω. Multiply them and you get the Poynting vector, which says power flows where E × H points — a statement that resolves, among other things, the question of where the energy in a transmission line actually travels. Then the boundary conditions, which are Maxwell's equations applied to a vanishing pillbox and a vanishing loop, and which are the source of the reflection and refraction results: the tangential E and the normal B are always continuous, and the other two components jump by whatever surface charge or current sits there. From those, the whole of plane-wave behaviour follows — reflection coefficients, Brewster and critical angles, polarisation and its handedness, the difference between phase and group velocity, and skin depth, which is the single most useful number in the section because it explains at a stroke why a microwave conductor is plated rather than solid and why a power busbar is hollow.

Maxwell's Equations in Both Forms, and What Each One Says

The four equations, side by side, with the physical statement each makes
LawDifferential formIntegral formWhat it actually asserts
Gauss, electric∇ · D = ρ∮ D · dS = Q_encElectric flux lines begin and end on charge, so D has sources.
Gauss, magnetic∇ · B = 0∮ B · dS = 0There is no magnetic charge, so every B line closes on itself.
Faraday∇ × E = −∂B/∂t∮ E · dL = −dΦ/dtA changing B drives a circulating E; the minus sign is Lenz's law.
Ampère with Maxwell's correction∇ × H = J + ∂D/∂t∮ H · dL = I_enc + dΨ/dtBoth a conduction current and a changing D drive a circulating H.

The term ∂D/∂t in the last row is Maxwell's displacement current, and it is the only one of the four contributions that was added on theoretical rather than experimental grounds. Ampère's original law is inconsistent with charge conservation: take the divergence of ∇ × H = J and the left side is identically zero, forcing ∇ · J = 0, which is false whenever charge accumulates anywhere — a capacitor being charged, for instance. Add the displacement term and the divergence gives 0 = ∇ · J + ∂ρ/∂t, which is exactly the continuity equation. So displacement current is not a patch, it is what makes the set self-consistent, and its physical payoff is the whole of this section: a changing E makes an H, a changing H makes an E, and the pair can therefore sustain each other through empty space with no charge and no current anywhere. That is a radio wave, and it exists because of one term.

🎯 Conductor or dielectric is not a property of the material — it is a property of the frequency
The ratio of conduction to displacement current in a material is the loss tangent σ/(ωε), and it is the quantity that decides which approximation you may use. Copper at 1 GHz: σ = 5.8 × 10⁷ S/m, ε ≈ ε₀ = 8.854 × 10⁻¹² F/m, so σ/(ωε) = 5.8 × 10⁷/(2π × 10⁹ × 8.854 × 10⁻¹²) = 5.8 × 10⁷/0.0556 = 1.04 × 10⁹. Nine orders of magnitude, so copper is a good conductor even at microwave frequencies, and displacement current in it can be ignored entirely. Now take dry earth with σ = 10⁻³ S/m and ε_r = 5. At 1 MHz: σ/(ωε) = 10⁻³/(2π × 10⁶ × 5 × 8.854 × 10⁻¹²) = 10⁻³/2.78 × 10⁻⁴ = 3.60. The two currents are comparable, so this is a lossy dielectric and neither limiting formula applies — you must use the full complex propagation constant. Raise the frequency to 1 GHz and the same soil gives 3.60 × 10⁻³, comfortably a good dielectric. One material, three regimes, decided entirely by frequency. The engineering consequences are immediate and examinable: a low-frequency broadcast signal is attenuated by ground losses in a way a microwave link over the same soil is not, and a radar's ground clutter behaves as a reflection from a dielectric while a long-wave signal partly diffuses into the earth. The working thresholds are σ/(ωε) > 100 for good conductor, < 0.01 for good dielectric, and everything between needs the general solution — and the habit worth forming is to compute the ratio before choosing a formula, not after.

Boundary Conditions: Two Always Continuous, Two That Jump

Apply the integral forms to shapes that shrink onto the interface and the boundary conditions drop out. Shrink a pillbox across the surface: Gauss's electric law gives D_(1n) − D_(2n) = ρ_s, and the magnetic one gives B_(1n) = B_(2n) with no exception because there is no magnetic surface charge. Shrink a loop straddling the surface: Faraday gives E_(1t) = E_(2t), because the enclosed flux goes to zero with the area, and Ampère gives H_(1t) − H_(2t) = J_s. So the pattern is worth stating as a pair of sentences rather than four formulae. The tangential electric field and the normal magnetic flux density are always continuous, at every interface, under all conditions. The normal D and the tangential H jump by exactly the surface charge and surface current that happen to be present, and at an interface between two ordinary dielectrics there are none, so those are continuous too. At a perfect electric conductor the fields inside vanish and the conditions become the ones every waveguide and antenna problem uses: E_t = 0 at the surface, H_t = J_s, B_n = 0, D_n = ρ_s. Read the first of those as the design rule it is — a perfect conductor cannot support a tangential electric field, which is why the walls of a waveguide force the modes they force.

🧠 One field crossing one interface, with a check that closes
Let region 1 be z < 0 with ε_(r1) = 2 and region 2 be z > 0 with ε_(r2) = 5, with no surface charge on the plane z = 0. The field just below the interface is E₁ = 5 x̂ + 10 ŷ + 3 ẑ V/m. Split it before doing anything: the tangential part is 5 x̂ + 10 ŷ, lying in the interface plane, and the normal part is 3 ẑ. Tangential E is continuous, so E_(2t) = 5 x̂ + 10 ŷ, unchanged. Normal D is continuous because ρ_s = 0, so ε₁E_(1z) = ε₂E_(2z), giving 2 × 3 = 5 × E_(2z) and E_(2z) = 1.2 V/m. Hence E₂ = 5 x̂ + 10 ŷ + 1.2 ẑ V/m. Two checks are worth doing because they cost seconds and catch sign errors. First, compute D_n on both sides: D_(1n) = 2 × 8.854 × 10⁻¹² × 3 = 5.312 × 10⁻¹¹ C/m² and D_(2n) = 5 × 8.854 × 10⁻¹² × 1.2 = 5.312 × 10⁻¹¹ C/m² ✓, identical as required. Second, use the refraction law for field lines, tan θ₂/tan θ₁ = ε₂/ε₁, where θ is measured from the normal. Here |E_t| = √(25 + 100) = 11.180, so tan θ₁ = 11.180/3 = 3.7268 giving θ₁ = 74.98°, and tan θ₂ = 11.180/1.2 = 9.3169 giving θ₂ = 83.87°. The ratio 9.3169/3.7268 = 2.5 = ε₂/ε₁ ✓. Read what that means physically: going into the denser dielectric, the field lines bend away from the normal, because the normal component is squeezed while the tangential is untouched. That single asymmetry — one component held fixed and the other scaled — is the mechanism behind every result in the rest of this chapter, from the reflection coefficient to Brewster's angle, and it is worth having seen once on a field you can hold in your hand rather than only inside a formula.

The Wave Equation, Intrinsic Impedance and the Poynting Vector

Take the curl of Faraday's law, substitute Ampère's, and in a source-free region the result is ∇²E = µε ∂²E/∂t², the wave equation, whose solutions travel at v = 1/√(µε). In free space that is c = 1/√(µ₀ε₀) = 3 × 10⁸ m/s, and the fact that this comes out of two electrostatic and magnetostatic constants with no mechanics anywhere in the derivation is the reason Maxwell's theory was believed before it was tested. In a dielectric the same expression gives v = c/√(ε_r µ_r), and for a non-magnetic material v = c/√(ε_r) with √(ε_r) being the refractive index. The uniform plane wave solution has three properties that must be stated as a package: E and H are both transverse to the direction of propagation; they are perpendicular to each other; and their ratio is the intrinsic impedance η = √(µ/ε), which in free space is η₀ = √(µ₀/ε₀) = 376.73 Ω, universally rounded to 377 Ω and often quoted as 120π. Note that η has the dimensions of resistance but dissipates nothing — it is a ratio of field magnitudes, not a component.

The Poynting vector S = E × H has the units of watts per square metre and points in the direction power flows. Its integral form is Poynting's theorem, −∮ S · dA = d/dt ∫(w_e + w_m) dV + ∫ σ|E|² dV, which reads as an energy balance: the power flowing inward through a closed surface either raises the stored electric and magnetic energy inside, or is dissipated as heat, and there is no third destination. For a time-harmonic plane wave the useful quantity is the time-average Poynting vector, S_av = ½ Re{E × H*}, and for a plane wave in a lossless medium this reduces to S_av = E₀²/(2η) with E₀ the peak amplitude. State that convention, because it is where the factor of two lives: with E₀ given as an RMS value the same power is E_rms²/η, and the two differ by exactly a factor of two.

🧠 A 10 V/m wave carries 133 mW per square metre, and the check is one line
A uniform plane wave in free space has a peak electric field of E₀ = 10 V/m. Then H₀ = E₀/η₀ = 10/376.73 = 0.02654 A/m, that is 26.54 mA/m — and notice how small the magnetic field is in SI numbers, which is a consequence of η₀ being 377 and not of the wave being weak. The time-average power density is S_av = E₀²/(2η₀) = 100/753.46 = 0.1327 W/m², that is 133 mW/m². Check it the other way, from the product of the fields: ½ E₀ H₀ = 0.5 × 10 × 0.02654 = 0.1327 W/m² ✓, and the agreement confirms both the impedance and the factor of two. Now scale it to something recognisable. Solar irradiance at the top of the atmosphere is about 1360 W/m², so this wave is about ten thousand times weaker than sunlight — and yet 10 V/m is a strong field by radio standards, far above what a receiver needs. Run the same arithmetic backwards for a receiver at 1 µV/m: S_av = 10⁻¹²/753.46 = 1.33 × 10⁻¹⁵ W/m², which is femtowatts per square metre and still perfectly usable, because the noise floor computed in the communications chapters is lower still. The lesson is that the field strength and the power density are not two independent facts about a wave; either one fixes the other through η, and a question that gives you one and asks for the other is testing whether you know which convention its amplitude was quoted in.

Propagation Through Lossy Media, and Skin Depth

In a conducting medium the propagation constant becomes complex, γ = α + jβ, so the wave both oscillates and decays: E = E₀ e^(−αz) cos(ωt − βz). In the good-conductor limit σ ≫ ωε, both parts take the same value, α = β = √(πfµσ), and the reciprocal of that is the quantity everything is quoted in terms of. The skin depth δ = 1/α = 1/√(πfµσ) is the depth at which the field amplitude has fallen to 1/e, about 37 per cent, of its surface value. Three properties are worth having by heart because they carry all the engineering. It goes as 1/√f, so a hundredfold rise in frequency shrinks it by ten. It goes as 1/√σ, so a better conductor confines the current more — which is the opposite of most people's first guess. And in the good-conductor limit α = β exactly, which means the wave loses e^(−2π) of its amplitude — a factor of 535 — in a single wavelength inside the metal, so there is no meaningful propagation in a conductor at all, only attenuation.

🎯 Copper from 50 Hz to 1 GHz: 9.3 mm down to 2.1 µm
Copper has σ = 5.8 × 10⁷ S/m and µ = µ₀ = 4π × 10⁻⁷ H/m. At 1 MHz, πfµσ = π × 10⁶ × 1.2566 × 10⁻⁶ × 5.8 × 10⁷ = 2.290 × 10⁸, so α = 1.513 × 10⁴ and δ = 66.1 µm. Now scale by 1/√f and nothing else is needed. At 1 GHz, a thousand times higher, δ = 66.1/√1000 = 2.09 µm. At 50 Hz, twenty thousand times lower, δ = 66.1 × √20000 = 9.35 mm. Read the consequences off those three numbers. At 50 Hz, 9.35 mm is comparable with the radius of a substantial busbar, which is why large power conductors are made hollow or stranded — the metal at the centre carries almost no current and is paying for itself in weight only. At 1 GHz, 2.09 µm is about a fortieth the thickness of household foil, so a microwave component's electrical behaviour is entirely a property of a mirror-thin surface layer; this is exactly why waveguides and connectors are silver-plated over brass, and why surface roughness on the scale of a micron becomes a loss mechanism. That surface layer also has a resistance, the surface resistance R_s = 1/(σδ), which at 1 GHz is 1/(5.8 × 10⁷ × 2.09 × 10⁻⁶) = 8.25 mΩ per square — the standard figure for copper at that frequency, and the number every microwave loss budget starts from. Note that R_s grows as √f, since δ shrinks as 1/√f, so conductor loss in any microwave structure rises with the square root of frequency, never linearly. And for shielding there is a rule of thumb worth its derivation: at a thickness of 5δ the field is down by e⁻⁵, which is 20 log₁₀(e⁻⁵) = −43.4 dB, so five skin depths is about 43 dB of shielding — which at 1 GHz in copper means a 10.5 µm plating is already a good screen, while at 50 Hz it would take 47 mm of solid copper, which is why low-frequency magnetic shielding uses high-permeability steel instead of thicker copper.

Reflection and Refraction: the Coefficients, and the Power Bookkeeping

At normal incidence on the plane boundary between two media of intrinsic impedances η₁ and η₂, matching the tangential E and H at the interface gives the two coefficients directly: the reflection coefficient Γ = (η₂ − η₁)/(η₂ + η₁) and the transmission coefficient τ = 2η₂/(η₂ + η₁), with the identity 1 + Γ = τ always holding. The limits fix the signs. Onto a perfect conductor, η₂ = 0 gives Γ = −1: total reflection with a phase reversal, so E is forced to zero at the surface and a standing wave forms with its first electric-field maximum a quarter wavelength out. At 3 GHz, where λ = 100 mm, that maximum sits at 25 mm from the wall, and knowing this locates every field probe in a slotted line. Onto a denser dielectric, η₂ < η₁ gives a negative Γ, again a phase reversal but only partial reflection. Onto a rarer one, Γ is positive and there is no phase reversal. The standing-wave ratio in medium 1 is SWR = (1 + |Γ|)/(1 − |Γ|), and it is worth noting that SWR cannot distinguish Γ = +1/3 from Γ = −1/3 because it discards the phase.

⚠️ The reflected power is 11 per cent and the transmitted is 89, but τ² is 44 — the identity is not 1 − Γ² = τ²
A wave in free space strikes a lossless dielectric with ε_r = 4 at normal incidence. Then η₁ = 376.73 Ω and η₂ = 376.73/√4 = 188.37 Ω, so Γ = (188.37 − 376.73)/(188.37 + 376.73) = −188.37/565.10 = −1/3 exactly, and τ = 2(188.37)/565.10 = 2/3. Check the identity: 1 + Γ = 1 − 1/3 = 2/3 = τ ✓. Now the power. The reflected power fraction is |Γ|² = 1/9 = 11.11 per cent, and by conservation the transmitted fraction must be 8/9 = 88.89 per cent. But |τ|² = 4/9 = 44.44 per cent, which is neither of those, and a candidate who writes 1 − |Γ|² = |τ|² gets 0.889 = 0.444 and has to choose which of two correct-looking numbers to submit. The identity 1 − |Γ|² = |τ|² is false at a dielectric interface, and the reason is that τ is a ratio of field amplitudes while power depends on both the field and the impedance. The correct bookkeeping carries the impedance ratio: transmitted power fraction = |τ|² × η₁/η₂ = (4/9)(376.73/188.37) = (4/9)(2) = 8/9 = 88.89 per cent ✓, and now 11.11 + 88.89 = 100 ✓. The same field, moving into a medium of half the impedance, carries twice the power for a given amplitude — which is exactly the missing factor. The special case where the naive identity does hold is a transmission line with the same Z₀ on both sides of a lumped discontinuity, and that is where most people first meet it, which is why the mistake survives. The standing-wave ratio in medium 1 here is SWR = (1 + 1/3)/(1 − 1/3) = 2, and if you instead sent the wave the other way, from the dielectric into air, Γ would be +1/3, the phase reversal would vanish, and SWR would still read 2 — the same mismatch, opposite phase.

At oblique incidence two new phenomena appear, and both are angles that can be computed exactly. Snell's law n₁ sin θ₁ = n₂ sin θ₂ governs the direction, and when the wave travels from dense to rare there is a critical angle beyond which no transmitted direction exists: θ_c = arcsin(n₂/n₁). From ε_r = 4, so n₁ = 2, into air, θ_c = arcsin(0.5) = 30°, and past that angle the reflection is total. That is total internal reflection and it is the mechanism of the optical fibre in the next chapter. Separately, for parallel polarisation — the electric field in the plane of incidence — there is an angle at which the reflection coefficient is exactly zero, the Brewster angle tan θ_B = n₂/n₁. From air into ε_r = 4, θ_B = arctan 2 = 63.435°, and at that incidence the refracted ray comes off at arcsin(sin 63.435°/2) = 26.565°, so θ_B + θ_t = 90° exactly — the reflected and refracted directions are perpendicular, which is the geometric statement of the same fact. There is no Brewster angle for perpendicular polarisation, and that asymmetry is precisely why light reflected off water at a shallow angle is strongly polarised, and why polarising sunglasses work.

Polarisation, and Phase Against Group Velocity

Polarisation describes the path traced by the tip of the electric field vector at a fixed point as time passes. Write the field of a wave travelling in +z as the sum of two transverse components and everything follows from their relative amplitude and relative phase. If the two components are in phase, the sum stays on a fixed line and the wave is linearly polarised — E = x̂ 3 cos(ωt − βz) + ŷ 4 cos(ωt − βz) is linear at arctan(4/3) = 53.13° from the x axis. If the amplitudes are equal and the phases differ by 90°, the tip traces a circle and the wave is circularly polarised. Anything else traces an ellipse. The axial ratio of an elliptical wave is the ratio of major to minor axis: E = x̂ 3 cos(ωt − βz) + ŷ 4 sin(ωt − βz) traces an ellipse with semi-axes 4 and 3, so the axial ratio is 4/3 = 1.333, that is 20 log₁₀(4/3) = 2.50 dB. A perfectly circular wave has an axial ratio of 0 dB and a linear one of infinity, so the axial ratio in decibels is the standard way of quoting how nearly circular a real antenna's output is.

ℹ️ Handedness, stated by the convention that decides it
Handedness has two rival conventions in the literature, so a question about it is only answerable once you name the one you are using. This chapter uses the IEEE convention, which is the one engineering and antenna practice follows: point the thumb of the appropriate hand along the direction of propagation, and if the fingers curl in the sense in which the field rotates with increasing time, that is the handedness. Apply it to E = x̂ E₀ cos(ωt − βz) + ŷ E₀ sin(ωt − βz), propagating in +z. At z = 0 and t = 0 the field points along +x; a quarter period later, at ωt = 90°, it points along +y. So the rotation is x → y → −x → −y. Now put the right thumb along +z: the right-hand fingers curl exactly that way, so this is right-hand circular polarisation, and its phasor form is the standard E₀(x̂ − j ŷ)e^(−jβz). Reverse the sign of the ŷ term, E = x̂ E₀ cos(ωt − βz) − ŷ E₀ sin(ωt − βz), and you have LHCP. The optics convention looks back along the ray instead of along it and therefore calls the same wave left-handed, which is why a bare 'right-hand circular' with no convention stated is genuinely ambiguous. Two consequences make handedness worth getting right. A receiving antenna of one handedness is in principle completely blind to a wave of the other, which is the basis of polarisation-diversity links and of the fact that a circularly polarised signal reflecting off a flat conductor comes back with its handedness reversed and is therefore rejected. And a linear antenna receiving a circular wave picks up exactly half the power, a fixed 3 dB polarisation loss, whichever way the linear element is oriented — which is precisely why satellite navigation and broadcast links use circular polarisation: it removes the receiver's orientation from the link budget at a known, constant cost.

Phase velocity v_p = ω/β is the speed at which a point of constant phase moves, and group velocity v_g = dω/dβ is the speed at which a wave packet, and therefore information, moves. In a non-dispersive medium β is proportional to ω and the two are equal. In a dispersive medium they are not, and the difference is a real and measurable effect rather than a bookkeeping artefact: v_p can exceed c without any physical principle being violated, because no energy and no signal travels at v_p. What may never exceed c is v_g. The cleanest exact example in the syllabus is propagation through an ionised medium such as the ionosphere, where the relative permittivity is ε_r = 1 − (f_p/f)² with f_p the plasma frequency. Below f_p the permittivity is negative, β is imaginary, and the wave does not propagate at all — it is reflected, which is exactly how long-distance short-wave broadcasting works and why the usable band shifts between day and night.

🎯 A phase velocity of 1.25c, a group velocity of 0.8c, and a product that is exactly c²
Take an ionised layer with plasma frequency f_p = 9 MHz and send a signal through it at f = 15 MHz. Then ε_r = 1 − (9/15)² = 1 − 0.36 = 0.64, so the refractive index is n = √0.64 = 0.8. Phase velocity: v_p = c/n = 3 × 10⁸/0.8 = 3.75 × 10⁸ m/s, which is 1.25 times the speed of light, and nothing is wrong. Group velocity: derive it rather than quoting it. From β = (ω/c)√(1 − ω_p²/ω²) you get ω² = ω_p² + c²β², and differentiating, 2ω dω = 2c²β dβ, so v_g = dω/dβ = c²β/ω = c²/v_p. Numerically v_g = 9 × 10¹⁶/3.75 × 10⁸ = 2.4 × 10⁸ m/s = 0.8c, and note that this equals cn, so in this medium v_p = c/n and v_g = cn — reciprocal factors either side of c. Multiply: 3.75 × 10⁸ × 2.4 × 10⁸ = 9.0 × 10¹⁶ = c² ✓, exactly. The relation v_p v_g = c² is not a coincidence of this example but a property of any medium whose dispersion has the form ω² = ω₀² + c²β², and it recurs identically in the hollow waveguide of the next chapter, where the cutoff frequency plays the role of f_p. Two readings worth carrying. Because v_g < c always here, information is late, not early — a 15 MHz signal crossing this layer is delayed by a factor of 1/0.8 = 1.25 relative to free space, and that delay is measurable and is how ionospheric sounding works. And because v_g depends on frequency, a wide signal spreads: the components at different frequencies arrive at different times, which is dispersion, and it is the same phenomenon that limits an optical fibre's bit rate in the next chapter. Push f down towards f_p = 9 MHz and n → 0, so v_g → 0 and the group is halted altogether — the wave is turned back, which is the reflection that makes short-wave propagation possible.

Exam Craft: the Order of Operations for a Plane-Wave Question

The EC paper carries 65 questions for 100 marks in 180 minutes, with General Aptitude taking 10 questions for 15 marks. In this section the time pressure bites in a specific way: almost every question is a chain of three or four substitutions, and a slip anywhere in the chain gives a plausible wrong number rather than an obviously absurd one. The marking policy should therefore shape your habits. A wrong MCQ costs one third of the question's value, so an electromagnetics MCQ you can narrow to two options is worth answering while a blind four-way guess is not; MSQ and NAT carry no negative marking, so every numerical answer is worth submitting even from a partial calculation. The single most useful habit is to compute intermediate quantities in a fixed order and check each one as you go, since almost every result in this chapter has a natural cross-check that costs a few seconds.

  1. Classify the medium first, with σ/(ωε). Above 100 use the good-conductor formulae and δ = 1/√(πfµσ); below 0.01 use the lossless ones; in between neither applies and you need the general γ. Choosing the formula before the ratio is the commonest route to a confidently wrong answer.
  2. Write η for both media before touching Γ. η = √(µ/ε), so for a non-magnetic dielectric η = η₀/√ε_r. Once η₁ and η₂ are down, Γ, τ, SWR and both power fractions follow mechanically.
  3. Check 1 + Γ = τ, then check the powers sum to one using |τ|² η₁/η₂. Both checks cost seconds. If |Γ|² + |τ|² comes to something other than 1 that is not an error — the impedance factor is missing.
  4. Say which amplitude convention you are in before writing a power. S_av = E₀²/2η for peak E₀, E_rms²/η for RMS. The two differ by exactly two, which is 3 dB, and that is the size of most wrong answers in this section.
  5. For handedness and for group velocity, name the convention in the answer. IEEE handedness looks along the direction of propagation; the optics convention looks back along it and reverses every label. And v_p > c is never an error to be corrected — check v_g instead, and v_p v_g = c² where the dispersion allows it.

Key takeaways

  • Displacement current is what makes Maxwell's set self-consistent, and it is what makes radio possible. Take the divergence of Ampère's law without it and you are forced into ∇ · J = 0, which contradicts a charging capacitor; with it you recover the continuity equation exactly, and a changing E and H can then sustain each other in empty space.
  • Conductor or dielectric is a statement about frequency, not about the material. σ/(ωε) for copper at 1 GHz is 1.04 × 10⁹; for dry earth it is 3.60 at 1 MHz and 3.60 × 10⁻³ at 1 GHz — the same soil is a lossy medium on medium wave and a good dielectric at microwave. Compute the ratio before choosing a formula.
  • Two boundary components are always continuous and two jump. Tangential E and normal B never jump; normal D and tangential H jump by ρ_s and J_s. Hence at ε_(r1) = 2 into ε_(r2) = 5, E₁ = 5x̂ + 10ŷ + 3ẑ becomes 5x̂ + 10ŷ + 1.2ẑ, and tan θ₂/tan θ₁ = 9.317/3.727 = 2.5 = ε₂/ε₁ ✓.
  • η₀ = 376.73 Ω ties the field strength to the power density, so either one fixes the other. A peak E₀ = 10 V/m in free space has H₀ = 26.54 mA/m and carries E₀²/2η₀ = 133 mW/m², confirmed by ½E₀H₀. Say whether your amplitude is peak or RMS — the two differ by exactly 3 dB.
  • Skin depth goes as 1/√(fσ), and it decides how metal is used. Copper gives 9.35 mm at 50 Hz, 66.1 µm at 1 MHz and 2.09 µm at 1 GHz, with R_s = 1/(σδ) = 8.25 mΩ per square at 1 GHz. Five skin depths is about 43 dB of shielding, which is why microwave parts are plated and power busbars are hollow.
  • 1 − |Γ|² = |τ|² is false at a dielectric interface. Free space onto ε_r = 4 gives Γ = −1/3 and τ = 2/3, so |Γ|² = 11.11 per cent reflected but |τ|² = 44.44 per cent — the transmitted power is |τ|² η₁/η₂ = 88.89 per cent, and only then does the bookkeeping close.
  • Critical and Brewster angles are exact and easy to confuse. From n = 2 into air, θ_c = arcsin(0.5) = 30°; from air into n = 2, θ_B = arctan 2 = 63.435°, at which the refracted ray leaves at 26.565° so that θ_B + θ_t = 90°. Brewster exists only for parallel polarisation.
  • v_p > c is not an error, and v_p v_g = c² when the dispersion allows it. In a plasma with f_p = 9 MHz at f = 15 MHz, n = 0.8, so v_p = 3.75 × 10⁸ and v_g = 2.4 × 10⁸ m/s, product exactly 9 × 10¹⁶. Information travels at v_g, which never exceeds c.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Maxwell added the displacement current term, the rate of change of D, to the right-hand side of Ampere's circuital law. Taking the divergence of the corrected law shows that it is equivalent to which statement?

    1. The divergence of J plus the rate of change of the charge density is zero, which is the equation of continuity.
    2. The divergence of J is identically zero, so charge can never accumulate anywhere.
    3. The divergence of B is zero, so there are no isolated magnetic charges.
    4. The curl of E equals minus the rate of change of B, which is Faraday's law.
    Show answer

    Answer: A — The divergence of J plus the rate of change of the charge density is zero, which is the equation of continuity.

    The divergence of any curl is identically zero, so taking the divergence of the corrected law gives zero equal to the divergence of J plus the rate of change of the divergence of D, and Gauss's law replaces the second term by the rate of change of the charge density. That is exactly the continuity equation, so the displacement term is what makes the set consistent with charge conservation rather than an optional refinement. The tempting answer that the divergence of J is zero is precisely what the uncorrected law forces, and it is false whenever charge accumulates, as it does on the plates of a capacitor being charged. The magnetic Gauss law and Faraday's law are separate equations of the set and are untouched by this manipulation.

  2. A soil has a conductivity of 2 x 10^(-3) S/m and a relative permittivity of 10, with the permittivity of free space taken as 8.854 x 10^(-12) F/m. Above what frequency does its loss tangent fall below 0.01, the threshold at which it may be treated as a good dielectric?

    1. 36.0 MHz
    2. 359.5 MHz
    3. 2.26 GHz
    4. 3.60 GHz
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    Answer: B — 359.5 MHz

    The loss tangent is the ratio of conduction to displacement current, sigma divided by omega times epsilon. First form the absolute permittivity, 10 x 8.854 x 10^(-12) = 8.854 x 10^(-11) F/m. Setting the ratio to 0.01 gives omega = sigma/(0.01 epsilon) = 2 x 10^(-3)/(8.854 x 10^(-13)) = 2.259 x 10^9 rad/s, and the frequency is that divided by 2 pi, which is 3.595 x 10^8 Hz, that is 359.5 MHz. The tempting 2.26 GHz is the angular frequency quoted as though it were a frequency, so it is larger by exactly 2 pi and is the single most common slip in loss-tangent arithmetic. The value 3.60 GHz uses a threshold of 0.001 instead of 0.01, and 36.0 MHz slips a decade in the other direction. The point of computing the ratio at all is that the same soil is a lossy medium on medium wave and a good dielectric at microwave.

  3. A uniform plane wave in air at 1.5 GHz strikes a large perfectly conducting sheet at normal incidence, with c taken as 3 x 10^8 m/s. How far in front of the sheet does the first maximum of the total electric field lie?

    1. 25 mm
    2. 200 mm
    3. 50 mm
    4. 100 mm
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    Answer: C — 50 mm

    A perfect conductor has zero intrinsic impedance, so the reflection coefficient is exactly minus one: the reflection is total and carries a phase reversal, which forces the total tangential electric field to vanish at the surface. The incident and reflected waves then form a standing pattern whose nulls sit at the surface and at every half wavelength out from it, with the maxima exactly midway between, so the first maximum is a quarter wavelength out. At 1.5 GHz the free-space wavelength is 3 x 10^8/1.5 x 10^9 = 200 mm, so the answer is 50 mm. The tempting 100 mm is a half wavelength, which is the spacing between consecutive maxima and also the position of the second null, not of the first maximum; 200 mm is the wavelength itself and 25 mm would be the answer at twice this frequency.

  4. The plane z = 0 carries no free surface charge and separates a dielectric of relative permittivity 2 filling z < 0 from a dielectric of relative permittivity 8 filling z > 0. In the first region the electric field makes an angle of 30 degrees with the normal to the interface. What angle does it make in the second region?

    1. 49.1 degrees
    2. 8.2 degrees
    3. 60.0 degrees
    4. 66.6 degrees
    Show answer

    Answer: D — 66.6 degrees

    Tangential E is continuous across the interface and, with no surface charge, normal D is continuous too, which means the normal component of E is scaled down by the permittivity ratio. Taking the tangent of the angle from the normal as the tangential component over the normal one, the two conditions combine into tan(theta_2)/tan(theta_1) = epsilon_2/epsilon_1 = 4. So tan(theta_2) = 4 x tan(30) = 4 x 0.5774 = 2.309 and theta_2 = 66.6 degrees, bent away from the normal on entering the denser dielectric. The tempting 49.1 degrees uses the square root of the permittivity ratio, borrowing the refractive-index scaling that belongs to Snell's law for the direction of a propagating ray rather than to a static field crossing a boundary. The value 8.2 degrees applies the ratio the wrong way round, and 60.0 degrees is the complement of the given angle.

  5. Copper has a conductivity of 5.8 x 10^7 S/m and the permeability of free space, 4 pi x 10^(-7) H/m. A copper sheet 30 micrometres thick is used as a shield at 4 MHz. By how many decibels is the field amplitude attenuated in passing through the sheet?

    1. 7.89 dB
    2. 3.94 dB
    3. 8.69 dB
    4. 0.91 dB
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    Answer: A — 7.89 dB

    Copper at these frequencies is a good conductor, so the skin depth is one over the square root of pi f mu sigma. That product is pi x 4 x 10^6 x 1.2566 x 10^(-6) x 5.8 x 10^7 = 9.159 x 10^8, whose square root is 3.026 x 10^4, so the skin depth is 33.0 micrometres. The sheet is therefore 30/33.0 = 0.908 skin depths thick, and the amplitude falls by that many nepers. Converting, 0.908 nepers x 8.686 dB per neper = 7.89 dB. The tempting 3.94 dB applies 10 log to what is an amplitude ratio rather than a power ratio, halving every answer of this kind. The value 8.69 dB is the attenuation over exactly one skin depth, obtained by assuming the sheet is one skin depth thick rather than computing the thickness in skin depths, and 0.91 dB quotes the neper count as though nepers and decibels were the same unit.

  6. A uniform plane wave travels in free space and carries a time-average power density of 5 W/m^2, with the intrinsic impedance of free space taken as 376.73 ohm. What is the peak amplitude of its magnetic field?

    1. 0.0133 A/m
    2. 0.163 A/m
    3. 0.115 A/m
    4. 61.4 A/m
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    Answer: B — 0.163 A/m

    For a plane wave in a lossless medium the time-average Poynting magnitude is half the peak electric field times the peak magnetic field, and since the two are related by the intrinsic impedance this can be written entirely in H as eta times H_0 squared over two. Solving, H_0 = square root of (2 x 5/376.73) = square root of 0.02654 = 0.163 A/m. The tempting 0.115 A/m is the root-mean-square value, which is the peak divided by root two, and it is the classic three-decibel error of this section: the factor of two in the power formula belongs to a peak amplitude and disappears if the amplitude quoted is already an rms one. The value 61.4 A/m is numerically the peak electric field in volts per metre, obtained by multiplying by the impedance where one should divide, and 0.0133 A/m divides the power density by the impedance without taking any square root at all.

  7. A uniform plane wave travelling in free space, where the intrinsic impedance is 376.73 ohm, strikes at normal incidence the flat face of a lossless non-magnetic dielectric whose relative permittivity is 9. Which of the following statements about this interface are correct?

    1. The reflection coefficient is -0.5, so a quarter of the incident power is reflected.
    2. Three quarters of the incident power is transmitted, even though the square of the transmission coefficient is only 0.25.
    3. The transmission coefficient is 1.5, since 2 eta_2/(eta_2 + eta_1) exceeds unity here.
    4. Because one minus the squared reflection coefficient must equal the squared transmission coefficient, only a quarter of the incident power reaches the second medium.
    Show answer

    Answer: A — The reflection coefficient is -0.5, so a quarter of the incident power is reflected.; B — Three quarters of the incident power is transmitted, even though the square of the transmission coefficient is only 0.25.

    The second medium is non-magnetic, so its intrinsic impedance is 376.73 divided by the square root of 9, that is 125.58 ohm. Then the reflection coefficient is (125.58 - 376.73)/(125.58 + 376.73) = -0.5 exactly, and the transmission coefficient is 1 + that, namely 0.5, which is below unity here because the wave is entering a lower-impedance medium. The reflected power fraction is the squared magnitude, 0.25, so by conservation three quarters is transmitted. The identity that one minus the squared reflection coefficient equals the squared transmission coefficient is false at a dielectric interface and is the trap the question is built on: the transmitted power is the squared transmission coefficient multiplied by the impedance ratio eta_1/eta_2, which is 0.25 x 3 = 0.75, and only then does the bookkeeping close.

  8. A wave travelling in the plus z direction has an electric field with an x component of amplitude 8 V/m and a y component of amplitude 6 V/m, the two components being exactly in phase. The wave is linearly polarised; at what angle to the x axis does the field vector lie?

    1. 53.13 degrees
    2. 45.00 degrees
    3. 26.57 degrees
    4. 36.87 degrees
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    Answer: D — 36.87 degrees

    When two transverse components are exactly in phase their sum stays on a fixed line for all time, so the wave is linear and the line's direction is fixed by the ratio of the amplitudes alone. Measured from the x axis, that angle is the arctangent of the y amplitude over the x amplitude, arctan(6/8) = arctan(0.75) = 36.87 degrees. The tempting 53.13 degrees is arctan(8/6), the angle measured from the y axis instead, and the two are complements of each other so both look plausible unless the reference axis named in the question is checked. The value 45.00 degrees would require equal amplitudes, and 26.57 degrees is arctan(0.5). Note that a phase difference of ninety degrees with these unequal amplitudes would have given an ellipse instead, and only equal amplitudes with that phase difference give a circle.