Electromagnetism · Upper Division

Maxwell's Equations

Four equations. Every electromagnetic phenomenon ever observed — radio waves, light, magnetism, the photoelectric effect — follows from them.

PrerequisitesElectricfields(Ch.13)⋅Magneticfields(Ch.15)⋅Induction(Ch.16)⋅VectorsandcalElectric fields (Ch. 13) \cdot Magnetic fields (Ch. 15) \cdot Induction (Ch. 16) \cdot Vectors and calculus (Ch. 21–22)
Learning Goals
  • State Maxwell's four equations in both integral and differential form and identify what each encodes.
  • Explain why Maxwell's displacement current was necessary for consistency of Ampère's law.
  • DerivetheelectromagneticwaveequationfromMaxwell′sequationsandcalculatec=1/(μ0Derive the electromagnetic wave equation from Maxwell's equations and calculate c = 1/\sqrt(\mu_{0}ε0\varepsilon_{0}.
  • Find the B field associated with a given plane wave E field using Faraday's law.
  • Apply Poynting's theorem to relate energy flux to field energy density and Ohmic dissipation.

M.1 The Four Laws in Integral Form

Maxwell's equations unify the laws we have studied individually — Gauss's law, Faraday's law, Ampère's law — into one consistent system, with one crucial addition: the displacement current.

Theorem M.1 — Maxwell's Equations (Integral Form, in vacuum)
I. Gauss's law:   ∮E\cdotdA=Qenc/ε0(electricfieldfromcharges\oint E\cdotdA = Q_{enc}/\varepsilon_{0} \qquad (electric field from charges
II. Gauss's law for magnetism:   ∮B\cdotdA=0(nomagneticmonopoles\oint B\cdotdA = 0 \qquad (no magnetic monopoles
III. Faraday's law:   ∮E\cdotdl=−dΦB/dt(changingBcreatesE\oint E\cdotdl = -d\Phi_B/dt \qquad (changing B creates E
IV. Ampère–Maxwell law:   ∮B\cdotdl=μ0(I+ε0dΦE/dt)(currentandchangingEcreateB\oint B\cdotdl = \mu_{0}(I + \varepsilon_{0} d\Phi_E/dt) \qquad (current and changing E create B

The new term ε₀ dΦ_E/dt is Maxwell's 1865 addition. Without it, Ampère's law was inconsistent — charge conservation was violated at a capacitor plate. With it, the equations are consistent, and they predict electromagnetic waves.

M.2 The Differential Form

Using the divergence theorem and Stokes' theorem, the integral forms convert to differential equations that hold at every point in space:

∇\cdotE=ρ/ε0∇\cdotB=0\nabla\cdotE = \rho/\varepsilon_{0} \qquad \nabla\cdotB = 0(M.1)
∇\timesE=−\partialB/\partialt∇\timesB=μ0J+μ0ε0\partialE/\partialt\nabla\timesE = -\partialB/\partialt \qquad \nabla\timesB = \mu_{0}J + \mu_{0}\varepsilon_{0} \partialE/\partialt(M.2)

Here ρ is charge density (C/m³) and J is current density (A/m²). The differential forms are local — they describe what happens at each point, rather than requiring integration over surfaces and loops. In free space (ρ = 0, J = 0) the equations are perfectly symmetric between E and B.

M.3 Electromagnetic Waves

In free space, take the curl of Faraday's law and substitute the Ampère-Maxwell law:

∇2E=μ0ε0∂2E/\partialt2(waveequationforE)\nabla^{2}E = \mu_{0}\varepsilon_{0} \partial^{2}E/\partialt^{2} \qquad (wave equation for E)(M.3)

This is the wave equation with speed:

c=1/(μ0ε0)=2.998×108m/sc = 1/\sqrt(\mu_{0}\varepsilon_{0}) = 2.998\times10^{8} m/s(M.4)

Maxwell calculated this in 1865 from the measured values of μ₀ and ε₀ — and recognized it as the speed of light. Light is an electromagnetic wave. This was one of the great unifications in the history of physics: electricity, magnetism, and optics were one subject.

Example M.1 — Displacement Current in a Charging Capacitor

A parallel-plate capacitor (plate area A) is being charged by current I. Find the displacement current density between the plates and verify Ampère's law is satisfied.

E between plates:E=σ/ε0=Q/(ε0A).dE/dt=(1/ε0A)dQ/dt=I/(ε0AE = \sigma/\varepsilon_{0} = Q/(\varepsilon_{0}A). \qquad dE/dt = (1/\varepsilon_{0}A) dQ/dt = I/(\varepsilon_{0}A
Displacement current density:Jd=ε0dE/dt=I/A(sameasconductioncurrentdensityinwiresJ_{d} = \varepsilon_{0} dE/dt = I/A (same as conduction current density in wires
Total displacement current:Id=Jd×A=I—exactlytheconductioncurrentIenteringthecapacitorI_{d} = J_{d} \times A = I — exactly the conduction current I entering the capacitor.
Consistency:Ampère's law now works for any surface bounded by the Amperian loop — the result is the same whether the surface passes through the wire (I) or between the plates (I_
Example M.2 — Plane Wave Solution

VerifythatE(z,t)=E0x^sin(kz−\omegat)isasolutiontoMaxwell′sequationsinfreespaceVerify that E(z,t) = E_{0} x̂ sin(kz - \omegat) is a solution to Maxwell's equations in free space, and find the associated B field.

Wave equation:∂2E/\partialz2=−k2E0sin(kz−\omegat).∂2E/\partialt2=−ω2E0sin(kz−\omegat\partial^{2}E/\partialz^{2} = -k^{2}E_{0} sin(kz-\omegat). \qquad \partial^{2}E/\partialt^{2} = -\omega^{2}E_{0} sin(kz-\omegat.
Condition:k2=μ0ε0ω2→ω/k=c✓k^{2} = \mu_{0}\varepsilon_{0}\omega^{2} \to \omega/k = c ✓
B from Faraday:∇\timesE=−\partialB/\partialt→\partialEx/\partialzy^=kE0cos(kz−\omegat)y^=−\partialB/\partialt\nabla\timesE = -\partialB/\partialt \to \partialE_x/\partialz ŷ = kE_{0}cos(kz-\omegat) ŷ = -\partialB/\partialt
Integrate:B=(E0k/ω)y^sin(kz−\omegat)=(E0/c)y^sin(kz−\omegatB = (E_{0}k/\omega) ŷ sin(kz-\omegat) = (E_{0}/c) ŷ sin(kz-\omegat
Result:B⊥E⊥k^,∣B∣=∣E∣/c.Thefieldsareperpendicularandinphase.✓B ⊥ E ⊥ k̂, |B| = |E|/c. The fields are perpendicular and in phase. ✓

M.4 Energy and the Poynting Vector

Electromagnetic fields carry energy. The energy density stored in the fields is:

u=12ε0E2+B2/(2μ0)u = \frac{1}{2}\varepsilon_{0}E^{2} + B^{2}/(2\mu_{0})(M.5)

The rate of energy flow per unit area is given by the Poynting vector:

S=(1/μ0)E×B[W/m2]S = (1/\mu_{0}) E \times B \qquad [W/m^{2}](M.6)

For a plane wave, S = (E²/μ₀c) ẑ — energy flows in the direction of propagation, as it must. The time-averaged intensity (irradiance) is I = ⟨|S|⟩ = E₀²/(2μ₀c) = cε₀E₀²/2. This connects Maxwell's equations directly to the intensity observed in optics experiments.

Definition M.2 — Common Traps
  • Integral and differential forms are equivalent only with the right calculus theorems: use Gauss for flux and Stokes for circulation.
  • Displacement current is not optional: it is required by charge conservation and predicts EM waves.
  • Free-space waves are transverse: E, B, and propagation direction are mutually perpendicular.
  • The Poynting vector gives energy flux: its direction is the direction of field energy transport, not necessarily wire current direction.
Exercises — M.1–M.4 Maxwell's Equations
1.
Calculatec=1/(μ0ε0)usingμ0=4π×10−7H/mandε0=8.854×10−12F/m.ComparetotheknCalculate c = 1/\sqrt(\mu_{0}\varepsilon_{0}) using \mu_{0} = 4\pi\times10^{-7} H/m and \varepsilon_{0} = 8.854\times10^{-12} F/m. Compare to the known speed of light.
m/s
Straightforward
2.Explainthephysicalmeaningof∇\cdotB=0.Whatwoulditmeanifthiswerenotzero?WhatisExplain the physical meaning of \nabla\cdotB = 0. What would it mean if this were not zero? What is the significance of Dirac's magnetic monopole argument?
Intermediate
3.For a plane wave polarized in the ŷ direction traveling in x, show that the electric and magnetic energy densities are equal. Find the time-averaged total energy density.
Intermediate
4.DerivePoynting′stheorem\partialu/\partialt+∇\cdotS=−J\cdotEfromMaxwell′sequations.InterpreteachtermDerive Poynting's theorem \partialu/\partialt + \nabla\cdotS = -J\cdotE from Maxwell's equations. Interpret each term physically.
Challenging
Key Takeaways
  • ∇\cdotE=ρ/ε0:chargescreatedivergingelectricfieldlines\nabla\cdotE = \rho/\varepsilon_{0}: charges create diverging electric field lines.
  • ∇\cdotB=0:nomagneticmonopoles;Bfieldlinesalwaysformclosedloops\nabla\cdotB = 0: no magnetic monopoles; B field lines always form closed loops.
  • ∇\timesE=−\partialB/\partialt:changingBcreatesacurlingE(Faraday\nabla\timesE = -\partialB/\partialt: changing B creates a curling E (Faraday.
  • ∇\timesB=μ0J+μ0ε0\partialE/\partialt:currentsandchangingEcreateB.ThesecondtermisMaxwell′saddi\nabla\timesB = \mu_{0}J + \mu_{0}\varepsilon_{0}\partialE/\partialt: currents and changing E create B. The second term is Maxwell's addition.
  • Freespace:c=1/(μ0ε0)=3×108m/s—lightisanelectromagneticwaveFree space: c = 1/\sqrt(\mu_{0}\varepsilon_{0}) = 3\times10^{8} m/s — light is an electromagnetic wave.
  • PoyntingvectorS=E\timesB/μ0givespowerflowperarea;energydensityu=12(ε0E2+B2/μ0Poynting vector S = E\timesB/\mu_{0} gives power flow per area; energy density u = \frac{1}{2}(\varepsilon_{0}E^{2}+B^{2}/\mu_{0}.