Maxwell's Displacement Current Explained

Fig 1 — The capacitor paradox: the same Amperian loop C bounds surface S₁ (pierced by the wire, I_enc = I) and surface S₂ (passing between the plates, no conduction current)

Introduction to Displacement Current

By the early 1860s, the laws of electricity and magnetism discovered over the previous century — Coulomb's and Gauss's laws, Ampère's law, Faraday's law — explained every known experiment. Yet when James Clerk Maxwell collected them into a single mathematical system he found that they were inconsistent with each other. Ampère's law, as we saw at the end of Part 5: Ampere's Circuital Law, violates conservation of charge whenever currents vary in time.

Maxwell's fix was to add one term, which he called the displacement current. It says that a changing electric field produces a magnetic field, just as a current does. The term is tiny in most low-frequency circuits, which is why nobody had noticed it. But it changed physics: with it, the equations predict self-sustaining electromagnetic waves travelling at the speed of light. Every radio, radar and antenna depends on this one term.

The Limitations of the Original Ampere's Circuital Law

Ampère's law, ∮𝐇·d𝐥 = ∫𝐉·d𝐀, makes an implicit assumption: the right-hand side must give the same answer for every surface bounded by the loop. Two surfaces sharing the same boundary together form a closed surface, so the assumption is equivalent to requiring zero net current out of every closed surface, ∮𝐉·d𝐀 = 0. In differential form, taking the divergence of ∇ × 𝐇 = 𝐉 gives ∇·𝐉 = 0 (the divergence of a curl is identically zero).

But the continuity equation, which expresses conservation of charge, says:

Continuity equation (conservation of charge)

\nabla \cdot 𝐉 = - \frac{\partial ρ_{v}}{\partial t} \oint_{S} 𝐉 \cdot d 𝐀 = - \frac{d Q_{enc}}{d t}

Current flowing out of a region must reduce the charge inside it. ∇·𝐉 = 0 is only true when charge densities are constant in time.

The original Ampère's law is therefore valid only for steady (DC) currents. Any situation where charge piles up or drains away — a capacitor charging, charge oscillating on an antenna, an RF current in a resonator — lies outside its range.

Understanding the Capacitor Charging Problem

Current Flowing Through Connecting Wires

Connect a parallel-plate capacitor to a source. While it charges, a conduction current I(t) flows in the wires, carrying positive charge onto one plate and removing it from the other. Around the wire there is a magnetic field — measurable with a probe — of magnitude H = I/(2πr), exactly as Ampère's law predicts.

Electric Field Between Capacitor Plates

Between the plates there is no conduction current — the gap is an insulator, or vacuum. What there is, is a growing electric field. With charge Q on plates of area A, Gauss's law gives D = Q/A (so E = Q/(εA)), and the total electric flux through the gap is Ψ_D = DA = Q. As the capacitor charges, Q rises and so does the field:

Rate of change of the electric flux in the gap

\frac{d Ψ_{D}}{d t} = A \frac{d D}{d t} = \frac{d Q}{d t} = I

The rate of change of electric flux in the gap is exactly equal to the conduction current in the wire.

The Problem of Current Continuity

Now apply Ampère's law to a circular loop C around the wire (Fig 1). Using the flat surface S₁ that the wire pierces, I_enc = I. Using the bag-shaped surface S₂ that bulges between the plates, nothing pierces it: I_enc = 0. The magnetic field around C cannot be both I/(2πr) and zero. Experiment settles which is right: the field around the wire is I/(2πr). Something passing through S₂ must be acting as a current — and the only thing there is the changing electric field.

Maxwell's Contribution to Ampere's Law

Maxwell proposed the correction in his 1861–62 paper On Physical Lines of Force and developed it fully in A Dynamical Theory of the Electromagnetic Field (1865). He reasoned from Gauss's law: since ∇·𝐃 = ρ_v, the continuity equation can be rewritten as

Combining continuity with Gauss's law

\nabla \cdot 𝐉 = - \frac{\partial}{\partial t} (\nabla \cdot 𝐃) ⟹ \nabla \cdot (𝐉 + \frac{\partial 𝐃}{\partial t}) = 0

It is not the conduction current 𝐉 alone that has zero divergence (flows in closed loops), but the total 𝐉 + ∂𝐃/∂t. Replacing 𝐉 by this sum in Ampère's law removes the contradiction.

The added term is not an arbitrary patch. It is the unique addition that makes Ampère's law consistent with charge conservation and Gauss's law, reduces to the original law for steady fields (∂𝐃/∂t = 0), and — as Maxwell realised — completes the symmetry with Faraday's law: a changing magnetic field makes an electric field, and now a changing electric field makes a magnetic field.

📌 Where the name comes from

Maxwell pictured space filled with an elastic medium (the "aether") whose charges are slightly displaced by an electric field. A changing field would move them, producing a current — a "displacement" current. The aether idea was abandoned after the Michelson–Morley experiment and Einstein's relativity, but the term itself turned out to be exactly right, and the name stuck.

What Is Displacement Current?

Definition of Displacement Current

Displacement current is the rate of change of electric flux (of 𝐃) through a surface. It is not a flow of charge, but it produces a magnetic field in exactly the same way as a conduction current of the same value:

Displacement current

I_{d} = \frac{d Ψ_{D}}{d t} = \frac{d}{d t} \int_{S} 𝐃 \cdot d 𝐀 = ε \frac{d Φ_{E}}{d t}

Measured in amperes. In vacuum ε = ε₀; in a uniform dielectric ε = ε₀ε_r.

Physical Meaning of Displacement Current

A changing electric field is a source of magnetic field. In the capacitor gap, the growing 𝐄 is surrounded by a circulating 𝐇, continuing the field that circles the wires.
Total current is continuous. Conduction current in the wires plus displacement current in the gap form an unbroken loop: current "flows" through a capacitor even though no charge crosses the gap. This is why a capacitor passes AC.
In a dielectric it has two parts. 𝐃 = ε₀𝐄 + 𝐏, so ∂𝐃/∂t = ε₀∂𝐄/∂t + ∂𝐏/∂t. The second part is a genuine motion of bound charges as molecules polarise — a real (if tiny) current. The first part exists even in a perfect vacuum, and that is the part that allows light to cross empty space.

Displacement Current Density

Displacement current density

𝐉_{d} = \frac{\partial 𝐃}{\partial t} = ε \frac{\partial 𝐄}{\partial t} phasor: 𝐉_{d} = j ω ε 𝐄

J_d is in A/m². For sinusoidal fields it is proportional to frequency and leads the electric field by 90° — the same 90° by which a capacitor's current leads its voltage.

Mathematical Expression of Displacement Current

For a parallel-plate capacitor with plate area A and gap d filled with permittivity ε, the field is E = V/d and the capacitance is C = εA/d. The displacement current through the gap is then:

Displacement current in a parallel-plate capacitor

I_{d} = J_{d} A = ε A \frac{d E}{d t} = \frac{ε A}{d} \frac{d V}{d t} = C \frac{d V}{d t}

The familiar capacitor law I = C dV/dt is the displacement current in the gap. For a sinusoid V = V₀ sin ωt, the peak current is ωCV₀ and the capacitor's reactance is X_C = 1/(ωC).

Ampere-Maxwell Law

Modified Integral Form

Ampère–Maxwell law — integral form

\oint_{C} 𝐇 \cdot d 𝐥 = \int_{S} 𝐉 \cdot d 𝐀 + \frac{d}{d t} \int_{S} 𝐃 \cdot d 𝐀 = I_{c} + I_{d}

Now S₁ and S₂ in Fig 1 give the same answer: through S₁ flows conduction current I; through S₂ flows displacement current I_d = dQ/dt = I. The circulation of 𝐇 is the total current — conduction plus displacement.

Modified Differential Form

Ampère–Maxwell law — differential form

\nabla \times 𝐇 = 𝐉 + \frac{\partial 𝐃}{\partial t} phasor (with 𝐉 = σ𝐄): \nabla \times 𝐇 = (σ + j ω ε) 𝐄

Taking the divergence: ∇·(∇ × 𝐇) = 0 = ∇·𝐉 + ∂(∇·𝐃)/∂t = ∇·𝐉 + ∂ρ_v/∂t — exactly the continuity equation. Charge conservation is now built into Maxwell's equations.

Relationship Between Changing Electric Fields and Magnetic Fields

Put the two curl equations side by side, for a region with no conduction current:

The two curl equations in source-free space

\nabla \times 𝐄 = - \frac{\partial 𝐁}{\partial t} \nabla \times 𝐇 = \frac{\partial 𝐃}{\partial t}

A changing 𝐁 creates a circulating 𝐄 (Faraday); a changing 𝐃 creates a circulating 𝐇 (Maxwell). Each field is the source of the other.

This mutual creation is the heart of electromagnetic waves. Suppose a changing current on an antenna produces a changing magnetic field. That field induces a circulating electric field around it, slightly further out; that changing electric field induces a magnetic field further out still; and so on. The disturbance propagates outward and keeps going after the source has stopped. Without the displacement current, the second step would be missing — a changing electric field in empty space would produce nothing — and there would be no radio waves and no light. Part 8 turns this argument into the wave equation and shows that the speed of propagation is 1/√(μ₀ε₀) = c.

Between circular capacitor plates of radius R, symmetry and the Ampère–Maxwell law give the magnetic field produced by the displacement current at radius r:

Magnetic field between circular capacitor plates

B (r) = {\begin{cases} \frac{μ_{0} I_{d} r}{2 π R^{2}} & r < R \\ \frac{μ_{0} I_{d}}{2 π r} & r ≥ R \end{cases}

Identical in form to the field inside and outside a solid wire carrying a uniform current (Part 5). The displacement current "fills" the gap just as a conduction current fills a wire.

Applications of Displacement Current

Capacitors

Every capacitor in an AC or RF circuit conducts by displacement current. The reactance X_C = 1/(2πfC) falls with frequency because the displacement current for a given voltage, ωCV, rises with frequency. DC-blocking capacitors, decoupling capacitors and the coupling capacitors in RF matching networks all rely on this. At high enough frequency the displacement current also flows through unintended capacitances — between PCB traces, across an inductor's windings, from a transistor's collector to its base — producing parasitic coupling and self-resonance.

Electromagnetic Wave Propagation

In empty space there are no charges and no conduction currents, so the only source of magnetic field is the displacement current. A radio wave travelling from a transmitter to a receiver is sustained entirely by Faraday's law and the displacement current alternately regenerating each other. Maxwell computed the speed of such waves from the measured values of ε₀ and μ₀, found it equal to the measured speed of light, and concluded in 1865 that light is an electromagnetic wave. Heinrich Hertz generated and detected radio waves in 1887–88, confirming the prediction.

RF and Microwave Circuits

In a material with conductivity σ and permittivity ε, the total current density is (σ + jωε)E. The ratio of conduction to displacement current classifies the material at a given frequency:

Conduction-to-displacement current ratio and loss tangent

\frac{| J_{c} |}{| J_{d} |} = \frac{σ}{ω ε} = \tan δ \begin{array}{l} σ / ω ε ≫ 1 : good conductor \\ σ / ω ε ≪ 1 : good (low-loss) dielectric \end{array}

The loss tangent tan δ quoted on every PCB laminate datasheet is exactly this ratio (in low-loss dielectrics the "σ" also includes polarisation losses). The crossover frequency where conduction and displacement currents are equal is f = σ/(2πε).

PCB substrates: FR-4 (tan δ ≈ 0.02) versus Rogers RO4350B (≈ 0.0037) or PTFE laminates (≈ 0.001). The lower the ratio, the lower the dielectric loss in microstrip lines and patch antennas.
Seawater, soil and the human body behave as conductors at low frequencies and as lossy dielectrics at microwave frequencies — which determines how far radio waves penetrate them.
Field solvers include the displacement current in every cell; without it they could not model resonance, radiation or wave propagation.

Numerical Examples

Displacement Current in a Parallel-Plate Capacitor

Problem: An air-gap parallel-plate capacitor has plates of 10 cm × 10 cm separated by 1 mm. The voltage across it is rising at 10⁶ V/s. Find the capacitance and the displacement current in the gap.

C = ε₀A/d; I_d = C dV/dt

C = \frac{8.854 \times 10^{- 12} \times 0.01}{0.001} = 88.54 pF I_{d} = 88.54 \times 10^{- 12} \times 10^{6} = 88.5 μA

Answer: C = 88.5 pF | I_d = 88.5 μA — equal to the conduction current in the leads

Total current is continuous around the circuit: 88.5 μA of conduction current in the wires becomes 88.5 μA of displacement current in the gap.

Displacement Current Density of a 1 GHz Field

Problem: A 1 GHz electric field E = 100 sin(ωt) V/m exists in free space. Find the peak displacement current density. What is it for the same field at 50 Hz?

J_d,peak = ε₀ ω E₀

J_{d} = 8.854 \times 10^{- 12} \times 2 π \times 10^{9} \times 100 = 5.56 A/m² J_{d, 50 Hz} = 2.78 \times 10^{- 7} A/m²

Answer: 5.56 A/m² at 1 GHz; 0.28 μA/m² at 50 Hz

The same field strength gives a displacement current 20 million times larger at 1 GHz than at 50 Hz. This is why the displacement current went unnoticed in nineteenth-century low-frequency experiments, and why it dominates at radio frequencies.

Magnetic Field Between Charging Capacitor Plates

Problem: A capacitor with circular plates of radius R = 5 cm is charged by a current of 2 A. Find the magnetic field between the plates at r = 3 cm and at r = 8 cm from the axis.

I_d = 2 A, spread uniformly over the plate area

B (3 cm) = \frac{2 \times 10^{- 7} \times 2 \times 0.03}{{0.05}^{2}} = 4.8 μT B (8 cm) = \frac{2 \times 10^{- 7} \times 2}{0.08} = 5.0 μT

Answer: 4.8 μT at 3 cm (inside) | 5.0 μT at 8 cm (outside the plate edge)

Here μ₀/(2π) = 2 × 10⁻⁷ is used, so B = μ₀I_d r/(2πR²) inside. Outside the plates the field is exactly what the 2 A in the wire produces at 8 cm. Measurements of this field (first convincingly by Carver and Rajhel in 1974) confirm that displacement current generates a real magnetic field.

Seawater — Conductor or Dielectric?

Problem: Seawater has σ ≈ 4 S/m and ε_r ≈ 81. Find the ratio of conduction to displacement current at 1 MHz and 10 GHz, and the crossover frequency where the two are equal.

ε = 81 × 8.854 × 10⁻¹² = 7.172 × 10⁻¹⁰ F/m; ratio = σ/(ωε)

{(\frac{σ}{ω ε})}_{1 MHz} = \frac{4}{2 π \times 10^{6} \times 7.172 \times 10^{- 10}} = 888 {(\frac{σ}{ω ε})}_{10 GHz} = 0.089

Crossover: f = σ/(2πε)

f_{c} = \frac{4}{2 π \times 7.172 \times 10^{- 10}} = 888 MHz

Answer: ratio ≈ 888 at 1 MHz (conductor) | ≈ 0.09 at 10 GHz (lossy dielectric) | f_c ≈ 0.89 GHz

(In reality the permittivity of water also falls and its dielectric loss rises at microwave frequencies, but the classification holds.) Either way, seawater is extremely lossy: radio waves penetrate only centimetres to metres, which is why submarine communication uses VLF/ELF frequencies where the skin depth is several metres. Part 8 computes the attenuation.

Conduction and Displacement Current in FR-4

Problem: FR-4 has ε_r = 4.4 and tan δ = 0.02 at 1 GHz. Find the effective conductivity that represents its dielectric loss at 1 GHz, and the ratio of displacement to loss current.

σ_eff = ω ε tan δ

σ_{eff} = 2 π \times 10^{9} \times 4.4 \times 8.854 \times 10^{- 12} \times 0.02 = 4.9 \times 10^{- 3} S/m \frac{J_{d}}{J_{loss}} = \frac{1}{0.02} = 50

Answer: σ_eff ≈ 4.9 mS/m at 1 GHz; displacement current is 50× the loss current

FR-4 is a good dielectric, but its effective conductivity grows in proportion to frequency (at constant tan δ). At 10 GHz it is ten times larger, which is why high-frequency boards move to low-loss laminates. Dielectric loss in a microstrip line grows roughly as f·tan δ.

Summary

Concept	Expression	Key point
Continuity equation	∇·𝐉 = −∂ρ_v/∂t	Conservation of charge
Displacement current	I_d = dΨ_D/dt = ε dΦ_E/dt	Changing electric flux acts as a current
Displacement current density	𝐉_d = ∂𝐃/∂t (phasor jωε𝐄)	Grows with frequency
Ampère–Maxwell (integral)	∮ 𝐇·d𝐥 = I_c + d/dt ∫ 𝐃·d𝐀	Resolves the capacitor paradox
Ampère–Maxwell (differential)	∇ × 𝐇 = 𝐉 + ∂𝐃/∂t	Maxwell's fourth equation
Capacitor	I_d = C dV/dt	How capacitors pass AC
Material classification	σ/(ωε) = tan δ	≫1 conductor, ≪1 dielectric

The original Ampère's law conflicts with conservation of charge whenever charge densities change.
Maxwell added the displacement current ∂𝐃/∂t, making the total current 𝐉 + ∂𝐃/∂t continuous everywhere.
A changing electric field produces a magnetic field; together with Faraday's law this allows fields to regenerate each other and propagate as waves.
At RF the displacement current dominates in dielectrics and empty space; the ratio σ/(ωε) decides whether a material behaves as a conductor or an insulator.

We now have all four laws. Part 7: The Four Maxwell's Equations Explained assembles them into Maxwell's equations and shows how they work together.

Put It Into Practice

Use the NEXT ANTENNAS calculators to apply the ideas from this article to real designs.

Antenna Design Patch Antenna Calculator Width, length, effective permittivity and inset feed of a rectangular microstrip patch for your frequency and substrate. Open calculator → Transmission Lines Microstrip Calculator Trace width for a target impedance, effective permittivity and guided wavelength on any PCB substrate. Open calculator → Impedance Matching Smith Chart Visualiser Plot impedances, reflection coefficient and VSWR, and build matching networks step by step. Open calculator → All Tools All RF Calculators Browse every NEXT ANTENNAS calculator for RF, microwave and antenna design in one place. Open calculator →

Maxwell's Displacement Current