Part 4 · Chapter 16

Displacement Current and Maxwell's Equations

Ampère's law, as the static chapters left it, was quietly broken — it failed the simplest test of charge conservation. Maxwell's repair was a single extra term, the displacement current, and with it the four equations of electromagnetism finally closed on themselves. Out of that closure fell a prediction no one had asked for: that disturbed fields ride off together at the speed of light.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • Why the static Ampère law contradicts charge conservation.
  • Maxwell's displacement current \(\partial\vec{D}/\partial t\) and the capacitor paradox it resolves.
  • The Ampère–Maxwell law \(\nabla\times\vec{H}=\vec{J}+\partial\vec{D}/\partial t\).
  • The complete set of four Maxwell equations in point and integral form.
  • Boundary conditions for time-varying fields, including the perfect conductor.
  • How the equations predict electromagnetic waves at \(c=1/\sqrt{\mu_0\varepsilon_0}\).
Section 16-1

The Flaw in Ampère's Law

The static Ampère law of Chapter 12 reads \(\nabla\times\vec{H}=\vec{J}\). Take the divergence of both sides. The left side vanishes identically — the divergence of any curl is zero — which forces \(\nabla\cdot\vec{J}=0\). But charge conservation, the continuity equation, says something different whenever charge is piling up:

Continuity equation (charge conservation)
\[ \nabla\cdot\vec{J} = -\frac{\partial\rho_v}{\partial t} \]

The two agree only in statics, where \(\partial\rho_v/\partial t=0\). The cleanest illustration is a charging capacitor: choose an Amperian loop around the lead wire. Cap it with a flat disk and a conduction current \(I\) pierces it; bulge that same surface forward so it passes between the plates and no conduction current crosses it — yet the loop is unchanged. Ampère's law returns two different answers. Something must flow across the gap.

I J_d = ∂D/∂t I I (conduction) = I_d (displacement)
The capacitor paradox: no charge crosses the gap, yet the loop still encircles a "current"
Section 16-2

Displacement Current

Maxwell's insight was that the changing electric field in the gap plays the role of the missing current. As charge builds on the plates, \(\vec{D}\) between them rises, and its time rate \(\partial\vec{D}/\partial t\) is a current density in its own right — the displacement current density:

Displacement current density
\[ \vec{J}_d = \frac{\partial\vec{D}}{\partial t}, \qquad I_d = \int_S \frac{\partial\vec{D}}{\partial t}\cdot d\vec{S} \]

Between the plates \(D=\rho_s=Q/A\), so \(\partial D/\partial t=(1/A)\,dQ/dt=I/A\). Integrating over the plate gives \(I_d=I\) exactly — the displacement current across the gap equals the conduction current in the wire. No charge actually crosses the gap; it is the field that carries the continuity through. This term is tiny in ordinary circuits but dominant at high frequency, and it is the engine of every antenna and waveguide.

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A changing field acts like a current
\[ I_d = \frac{d}{dt}\!\int_S\!\vec{D}\cdot d\vec{S} = \frac{dQ}{dt} = I \]

The displacement current restores charge conservation and makes magnetism respond to a changing \(\vec{E}\), exactly as Faraday made electricity respond to a changing \(\vec{B}\). It is the missing half of the symmetry.

Section 16-3

The Ampère–Maxwell Law

Adding the new term to Ampère's law gives the corrected curl equation. Now take its divergence: the curl term still vanishes, leaving \(\nabla\cdot\vec{J}+\partial(\nabla\cdot\vec{D})/\partial t=\nabla\cdot\vec{J}+\partial\rho_v/\partial t=0\) — continuity is satisfied automatically, for free.

Ampère–Maxwell law (point and integral form)
\[ \nabla\times\vec{H} = \vec{J} + \frac{\partial\vec{D}}{\partial t}, \qquad \oint_C \vec{H}\cdot d\vec{l} = I_{\text{enc}} + \int_S \frac{\partial\vec{D}}{\partial t}\cdot d\vec{S} \]

Read it alongside Faraday's \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\) and the symmetry is unmistakable: a changing \(\vec{B}\) curls up an \(\vec{E}\), and a changing \(\vec{E}\) curls up an \(\vec{H}\). The two fields now feed each other, and that mutual feeding is what lets a disturbance propagate with no charges or wires to carry it.

∂B/∂t → E ∇×E = −∂B/∂t ∂D/∂t → H ∇×H = J + ∂D/∂t
The closed loop: each changing field is the source of the other
Section 16-4

Maxwell's Equations

With the displacement current in place, the four laws of the whole course stand complete. Each says one thing about the sources of divergence and curl of the fields:

LawPoint formIntegral form
Gauss (electric)\(\nabla\cdot\vec{D}=\rho_v\)\(\oint\vec{D}\cdot d\vec{S}=Q_{\text{enc}}\)
Gauss (magnetic)\(\nabla\cdot\vec{B}=0\)\(\oint\vec{B}\cdot d\vec{S}=0\)
Faraday\(\nabla\times\vec{E}=-\dfrac{\partial\vec{B}}{\partial t}\)\(\oint\vec{E}\cdot d\vec{l}=-\dfrac{d\Phi}{dt}\)
Ampère–Maxwell\(\nabla\times\vec{H}=\vec{J}+\dfrac{\partial\vec{D}}{\partial t}\)\(\oint\vec{H}\cdot d\vec{l}=I+\dfrac{d\Psi_e}{dt}\)

They are completed by the constitutive relations that describe the medium — \(\vec{D}=\varepsilon\vec{E}\), \(\vec{B}=\mu\vec{H}\), \(\vec{J}=\sigma\vec{E}\) — from Chapters 7 and 13. Two divergence laws fix the sources of the fields; two curl laws fix how they change in time. Everything in Parts 5 through 7 — waves, transmission lines, waveguides, antennas — is just these four equations applied to a particular geometry.

Section 16-5

Boundary Conditions

At an interface between two media the fields must obey matching rules, obtained by shrinking Gaussian pillboxes and Amperian loops onto the surface. For time-varying fields the relations keep the very same form as in statics, because the fields stay finite:

Electromagnetic boundary conditions
\[ E_{1t}=E_{2t}, \quad H_{1t}-H_{2t}=K, \quad D_{1n}-D_{2n}=\rho_s, \quad B_{1n}=B_{2n} \]

Tangential \(\vec{E}\) and normal \(\vec{B}\) are always continuous; tangential \(\vec{H}\) jumps by any free surface current \(K\), and normal \(\vec{D}\) jumps by any free surface charge \(\rho_s\). At the surface of a perfect conductor the interior fields vanish, so outside it \(E_t=0\) and \(B_n=0\): the field must meet the metal head-on, supported by induced surface charge and current. These conditions are exactly what shape the modes of a waveguide and the reflection of a wave in Parts 6 and 7.

Section 16-6

The Prediction of Waves

In a source-free linear medium (\(\vec{J}=0\), \(\rho_v=0\)), take the curl of Faraday's law and substitute Ampère–Maxwell. Using \(\nabla\times(\nabla\times\vec{E})=\nabla(\nabla\cdot\vec{E})-\nabla^2\vec{E}\) with \(\nabla\cdot\vec{E}=0\), the two curl equations fold into a single second-order equation:

The wave equation
\[ \nabla^2\vec{E} = \mu\varepsilon\,\frac{\partial^2\vec{E}}{\partial t^2}, \qquad v=\frac{1}{\sqrt{\mu\varepsilon}} \]
z, c E B E ⊥ B ⊥ direction of travel
The prediction: self-sustaining E and B ride off together at v = 1/√(με)
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Light is an electromagnetic wave
\[ c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3\times10^{8}\ \text{m/s} \]

Plug in the measured \(\mu_0\) and \(\varepsilon_0\) — two constants from electrostatics and magnetostatics — and out comes the speed of light. Maxwell concluded that light is an electromagnetic wave, the deepest single result in classical physics, and the doorway to Part 5.

Section 16-7

Worked Examples

1 Displacement current in a capacitor

Problem. A parallel-plate capacitor (area \(0.01\ \text{m}^2\), gap \(1\ \text{mm}\), air) has \(v(t)=50\sin(2\pi\cdot5000\,t)\ \text{V}\). Find the peak displacement current.

Solution. \(I_d=C\,dv/dt\) with \(C=\varepsilon_0 A/d\):

Working
\[ C=\frac{\varepsilon_0(0.01)}{10^{-3}}=88.5\ \text{pF}, \quad I_{d,\max}=C\,(50)(2\pi\cdot5000)\approx 0.14\ \text{mA} \]
2 Conduction vs displacement

Problem. A material has \(\sigma=10^{-4}\ \text{S/m}\), \(\varepsilon_r=5\), at \(f=1\ \text{MHz}\). Find the ratio \(J_c/J_d\) (the loss tangent).

Solution. Use \(J_c/J_d=\sigma/\omega\varepsilon\):

Working
\[ \frac{\sigma}{\omega\varepsilon}=\frac{10^{-4}}{(2\pi\cdot10^{6})(5\varepsilon_0)}\approx 0.36 \]
3 Speed of light

Problem. Compute the wave speed in vacuum from \(\mu_0\) and \(\varepsilon_0\).

Solution. Use \(c=1/\sqrt{\mu_0\varepsilon_0}\):

Working
\[ c=\frac{1}{\sqrt{(4\pi\times10^{-7})(8.854\times10^{-12})}}\approx 3.00\times10^{8}\ \text{m/s} \]
4 Magnetic field in the gap

Problem. A circular capacitor carries a uniform displacement current density \(J_d=0.1\ \text{A/m}^2\). Find \(H\) at radius \(r=2\ \text{cm}\) from the axis.

Solution. Ampère–Maxwell with only \(J_d\): \(H(2\pi r)=J_d(\pi r^2)\), so \(H=\tfrac12 r J_d\):

Working
\[ H=\tfrac12(0.02)(0.1)=1.0\times10^{-3}\ \text{A/m} \]
5 Charge from Gauss's law

Problem. A region has \(\vec{D}=(3x\,\hat{a}_x+4y\,\hat{a}_y-2z\,\hat{a}_z)\ \text{nC/m}^2\). Find the volume charge density.

Solution. Apply \(\rho_v=\nabla\cdot\vec{D}\):

Working
\[ \rho_v=\frac{\partial}{\partial x}(3x)+\frac{\partial}{\partial y}(4y)+\frac{\partial}{\partial z}(-2z)=3+4-2=5\ \text{nC/m}^3 \]
6 Displacement current density

Problem. In a dielectric (\(\varepsilon_r=2.5\)), \(\vec{E}=5\sin(2\pi\times10^{8}t)\,\hat{a}_x\ \text{V/m}\). Find the peak \(J_d\).

Solution. Use \(J_d=\varepsilon\,\partial E/\partial t\), peak \(=\varepsilon_0\varepsilon_r E_0\omega\):

Working
\[ J_{d,\max}=(2.5\varepsilon_0)(5)(2\pi\times10^{8})\approx 0.070\ \text{A/m}^2 \]
Review

Chapter Summary

The flaw

Static \(\nabla\times\vec{H}=\vec{J}\) forces \(\nabla\cdot\vec{J}=0\), contradicting continuity.

Displacement current

\(\vec{J}_d=\partial\vec{D}/\partial t\); equals the conduction current across a capacitor gap.

Ampère–Maxwell

\(\nabla\times\vec{H}=\vec{J}+\partial\vec{D}/\partial t\); satisfies continuity automatically.

The four equations

Two divergence (sources), two curl (time change); plus \(\vec{D}=\varepsilon\vec{E}\), \(\vec{B}=\mu\vec{H}\), \(\vec{J}=\sigma\vec{E}\).

Boundary conditions

\(E_{1t}=E_{2t}\), \(B_{1n}=B_{2n}\); \(H\) and \(D\) jump by \(K\) and \(\rho_s\); conductor: \(E_t=0\).

Waves

\(\nabla^2\vec{E}=\mu\varepsilon\,\partial^2\vec{E}/\partial t^2\); \(c=1/\sqrt{\mu_0\varepsilon_0}\); light is electromagnetic.

Practice

Problems

For each item, identify which Maxwell equation or relation applies, then substitute carefully. Difficulty rises down the list.

  1. A capacitor (area \(20\ \text{cm}^2\), gap \(0.5\ \text{mm}\), air) has \(dv/dt=10^{6}\ \text{V/s}\). Find the displacement current.
  2. Show, by taking the divergence of the Ampère–Maxwell law, that it implies the continuity equation.
  3. A field \(\vec{D}=D_0\sin(\omega t)\,\hat{a}_z\) fills a region. Write the displacement current density and its peak value.
  4. At what frequency are the conduction and displacement currents equal in a medium with \(\sigma=5\times10^{-3}\ \text{S/m}\), \(\varepsilon_r=80\)?
  5. Given \(\vec{D}=(2x\,\hat{a}_x-3y\,\hat{a}_y+z\,\hat{a}_z)\ \text{nC/m}^2\), find \(\rho_v\).
  6. Verify that \(\vec{B}=(2y\,\hat{a}_x-3x\,\hat{a}_y)\ \text{T}\) is a legitimate magnetic field by testing \(\nabla\cdot\vec{B}=0\).
  7. Find the wave speed and refractive index in a non-magnetic dielectric with \(\varepsilon_r=4\).
  8. A \(1\ \text{GHz}\) field has \(E_0=100\ \text{V/m}\) in free space. Find the peak displacement current density.
  9. For a circular parallel-plate capacitor of radius \(a\), show that the induced \(H\) at the plate edge is \(H=\tfrac12 a\,\partial D/\partial t\).
  10. State the boundary conditions on \(\vec{E}\) and \(\vec{H}\) at the surface of a perfect conductor, and explain why \(E_t=0\).
  11. Starting from the four equations in a source-free medium, derive the wave equation for \(\vec{H}\).
  12. Explain physically why the displacement current is negligible at \(50\ \text{Hz}\) in a copper wire but dominant in the gap of a microwave capacitor.
Tip: the whole chapter turns on one repair — adding \(\partial\vec{D}/\partial t\) to Ampère's law so that charge is always conserved. Once that term is in, the four equations are symmetric: changing \(\vec{B}\) makes \(\vec{E}\), changing \(\vec{E}\) makes \(\vec{H}\), and the feedback supports a wave. When solving, first ask which equation governs your quantity — divergence laws for charge and flux, curl laws for time variation — then reach for the constitutive relation to swap \(\vec{D}\leftrightarrow\vec{E}\) or \(\vec{B}\leftrightarrow\vec{H}\). Everything from Chapter 17 onward is these four equations in a new costume.