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Derivation

Displacement Current and Maxwell's Equations

Statement

The magnetostatic Ampère law \(\nabla\times\mathbf{B}=\mu_0\mathbf{J}\) is mathematically inconsistent with local charge conservation whenever the charge density changes in time. Requiring the continuity equation \(\nabla\cdot\mathbf{J}+\partial\rho/\partial t=0\) to hold identically forces the addition of Maxwell's displacement-current term, giving \(\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t\). This single correction closes the four field equations into a self-consistent system.

Why it matters

Without the displacement current the field equations contradict the conservation of charge — the most securely established of all local conservation laws — the instant a current charges a capacitor or a wave propagates. The extra term is not an empirical patch bolted on to fit data; it is uniquely fixed by demanding that the same charge that leaves one region appears in another.

The same term makes the vacuum equations support source-free wave solutions travelling at \(c=1/\sqrt{\mu_0\varepsilon_0}\). Electromagnetic radiation, and with it the identification of light as an electromagnetic phenomenon, exists only because of this correction.

Assumptions
Charge is locally conserved: any change of charge in a volume equals the flux of current through its boundary.If dropped, charge could appear or vanish without a current, continuity fails, and there is no logical lever to force any correction to Ampère's law.
Gauss's law \(\nabla\cdot\mathbf{E}=\rho/\varepsilon_0\) holds instantaneously, including in time-dependent situations.If dropped, we cannot trade \(\partial\rho/\partial t\) for \(\varepsilon_0\,\partial(\nabla\cdot\mathbf{E})/\partial t\), and the correction term could not be written purely in terms of the fields.
The fields are smooth enough that mixed partial derivatives commute and the identity \(\nabla\cdot(\nabla\times\mathbf{B})\equiv 0\) applies pointwise.If dropped (e.g. at idealised surface currents or point sources) the divergence identity must be handled distributionally, and the pointwise argument needs care at the singular support.
Linear response of vacuum: \(\varepsilon_0\) and \(\mu_0\) are constants, not fields.If dropped (nonlinear or dispersive media) the displacement current generalises to \(\partial\mathbf{D}/\partial t\) with a constitutive relation, and the clean vacuum form no longer applies.
Derivation
1
\[ \nabla\times\mathbf{B}=\mu_0\mathbf{J} \]
Start from Ampère's law as obtained magnetostatically from the Biot–Savart law for steady currents. A
2
\[ \nabla\cdot(\nabla\times\mathbf{B})=\mu_0\,\nabla\cdot\mathbf{J} \]
Take the divergence of both sides; divergence is linear so it passes through the constant \(\mu_0\). A
3
\[ \nabla\cdot(\nabla\times\mathbf{B})\equiv 0 \quad\Rightarrow\quad \mu_0\,\nabla\cdot\mathbf{J}=0 \]
The divergence of any curl vanishes identically (a vector-calculus identity, valid for \(C^2\) fields). Hence Ampère's law demands \(\nabla\cdot\mathbf{J}=0\). A
4
\[ \nabla\cdot\mathbf{J}+\frac{\partial\rho}{\partial t}=0 \]
Impose the continuity equation, the local statement of charge conservation. This is the physical input the uncorrected law violates whenever \(\partial\rho/\partial t\neq0\). A
5
\[ \nabla\cdot\mathbf{J}=-\frac{\partial\rho}{\partial t}\neq 0 \quad(\text{time-dependent }\rho) \]
Steps 3 and 4 are contradictory unless \(\partial\rho/\partial t=0\). The magnetostatic law is therefore incomplete; something must be added to restore consistency. B
6
\[ \rho=\varepsilon_0\,\nabla\cdot\mathbf{E} \]
Use Gauss's law to express the charge density through the field, so the offending \(\partial\rho/\partial t\) can be rewritten in field variables. A
7
\[ \frac{\partial\rho}{\partial t}=\varepsilon_0\,\frac{\partial}{\partial t}(\nabla\cdot\mathbf{E})=\varepsilon_0\,\nabla\cdot\frac{\partial\mathbf{E}}{\partial t} \]
Differentiate in time and exchange the order of the space and time derivatives — legitimate for smooth fields (Clairaut/Schwarz). B
8
\[ \nabla\cdot\mathbf{J}+\varepsilon_0\,\nabla\cdot\frac{\partial\mathbf{E}}{\partial t}=0 \quad\Rightarrow\quad \nabla\cdot\!\left(\mathbf{J}+\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}\right)=0 \]
Substitute step 7 into continuity and collect under one divergence. The combination \(\mathbf{J}+\varepsilon_0\,\partial\mathbf{E}/\partial t\) is always divergence-free, exactly the property the curl in Ampère's law requires. B
9
\[ \nabla\times\mathbf{B}=\mu_0\!\left(\mathbf{J}+\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}\right) \]
Replace the divergence-violating source \(\mathbf{J}\) in step 1 by the divergence-free combination from step 8. This is the minimal change consistent with continuity, reducing to the original law when fields are static. C
Result
\[ \nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\,\frac{\partial\mathbf{E}}{\partial t} \]

Reading. A magnetic field circulates not only around real (conduction) current \(\mathbf{J}\) but also around a changing electric field. The quantity \(\mathbf{J}_d=\varepsilon_0\,\partial\mathbf{E}/\partial t\), the displacement current density, carries no charge yet sources \(\mathbf{B}\) in exactly the same way as conduction current, so total current \(\mathbf{J}+\mathbf{J}_d\) is conserved and the Amperean loop integral is path-independent whatever surface it caps.

Units check. \([\varepsilon_0\,\partial\mathbf{E}/\partial t]=(\mathrm{C^2\,N^{-1}\,m^{-2}})\cdot(\mathrm{V\,m^{-1}\,s^{-1}})\). With \(\mathrm{V=J\,C^{-1}=N\,m\,C^{-1}}\), this is \(\mathrm{C^2\,N^{-1}\,m^{-2}}\cdot\mathrm{N\,C^{-1}\,s^{-1}}=\mathrm{C\,m^{-2}\,s^{-1}=A\,m^{-2}}\), a current density, matching \([\mathbf{J}]=\mathrm{A\,m^{-2}}\). Both sides of the boxed equation then carry \(\mu_0\times\mathrm{A\,m^{-2}}=\mathrm{T\,m^{-1}}\), the units of \(\nabla\times\mathbf{B}\).

Limiting cases
  • Static / steady currents (\(\partial\mathbf{E}/\partial t=0\)): the term vanishes and the law reduces to magnetostatic Ampère \(\nabla\times\mathbf{B}=\mu_0\mathbf{J}\).
  • Charging capacitor gap (\(\mathbf{J}=0\), \(\partial\mathbf{E}/\partial t\neq0\)): the whole right-hand side is displacement current, giving the same \(\mathbf{B}\) between the plates as the conduction current in the leads.
  • Source-free vacuum (\(\mathbf{J}=0,\ \rho=0\)): pairing with Faraday's law yields \(\nabla^2\mathbf{E}=\mu_0\varepsilon_0\,\partial^2\mathbf{E}/\partial t^2\), a wave with speed \(c=1/\sqrt{\mu_0\varepsilon_0}\).
  • Quasi-static limit (length \(\ell\ll c\,\tau\) for timescale \(\tau\)): \(|\mathbf{J}_d|/|\mathbf{J}|\sim(\ell/c\tau)^2\ll1\), so the displacement current is a negligible correction and circuit theory suffices.
Breaks when
  • Polarisable / magnetisable media. In matter one must use \(\nabla\times\mathbf{H}=\mathbf{J}_f+\partial\mathbf{D}/\partial t\) with \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\); bound and polarisation currents mean the vacuum form \(\varepsilon_0\,\partial\mathbf{E}/\partial t\) is not the correct source and dispersion makes \(\varepsilon\) frequency-dependent.
  • Nonlinear or time-varying constitutive relations. If \(\varepsilon\) depends on field strength or explicitly on time, \(\partial\mathbf{D}/\partial t\neq\varepsilon\,\partial\mathbf{E}/\partial t\) and the neat single-term correction splits into several pieces.
  • Regimes where classical fields fail. At high photon energies or strong fields the classical source term is replaced by quantum-electrodynamic vacuum polarisation and photon–photon effects; Maxwell's linear vacuum law is only an effective low-energy description.
Failure modes
  • "Displacement current is a real flow of charge." No charge crosses the capacitor gap; \(\mathbf{J}_d=\varepsilon_0\,\partial\mathbf{E}/\partial t\) is a field rate, not moving charge.
  • Forgetting \(\varepsilon_0\). Writing \(\nabla\times\mathbf{B}=\mu_0(\mathbf{J}+\partial\mathbf{E}/\partial t)\) is dimensionally wrong — \(\partial\mathbf{E}/\partial t\) is not a current density without the \(\varepsilon_0\).
  • Applying the divergence identity to the wrong term. Students take \(\nabla\cdot\mathbf{J}=0\) as a law rather than as the (false) consequence of the uncorrected equation.
  • Choosing the "wrong" Amperean surface. In the capacitor problem, integrating only over a surface pierced by the wire and ignoring the gap surface gives an apparent contradiction; the displacement current is exactly what restores agreement.
  • Treating the correction as empirical. Believing Maxwell "measured" the term; it is forced by continuity plus Gauss with no new experiment required.
Discussion

The logical structure is worth dwelling on: the correction is not derived from data but from an internal inconsistency. The divergence of a curl is identically zero, so any law of the form \(\nabla\times\mathbf{B}=\mu_0(\text{source})\) is only admissible if that source is divergence-free. Conduction current alone is divergence-free only in steady state. Continuity tells us that the shortfall is exactly \(\partial\rho/\partial t\), and Gauss lets us re-express it in the fields, so the unique divergence-free completion is \(\mathbf{J}+\varepsilon_0\,\partial\mathbf{E}/\partial t\). The added term is the minimal object that repairs the mathematics while vanishing in the static limit where the original law was verified.

With this term the four equations become a closed, self-consistent set. Faraday's law had already coupled a changing \(\mathbf{B}\) to a circulating \(\mathbf{E}\); the displacement current supplies the reciprocal coupling of a changing \(\mathbf{E}\) to a circulating \(\mathbf{B}\). This symmetry between the two curl equations is precisely what allows a disturbance to sustain itself: each field's time variation feeds the other's spatial curl, and the pair propagates as a wave. The thread of symmetry is not decorative here — the balanced curl equations are the reason radiation exists.

More deeply, the displacement current is not optional once one adopts the relativistic viewpoint. The four Maxwell equations combine into \(\partial_\mu F^{\mu\nu}=\mu_0 J^\nu\), where \(F^{\mu\nu}\) is the antisymmetric field tensor. Antisymmetry forces \(\partial_\nu\partial_\mu F^{\mu\nu}\equiv 0\), which is exactly the continuity equation \(\partial_\nu J^\nu=0\) in four-vector form. The displacement term is the spatial part of the manifestly covariant source law; it is what makes Ampère's and Gauss's laws two faces of one Lorentz-covariant statement. Charge conservation is then not an extra assumption but a built-in identity of the tensor structure.

Common misconceptions. The name "displacement" is a historical artefact of Maxwell's mechanical-ether picture and should not be read as any physical displacement of a medium; in vacuum there is nothing to displace. Equally, the term is not "small and negligible" in general — it dominates entirely in the capacitor gap and in every radiation field. It is negligible only in the quasi-static regime where feature sizes are small compared with \(c\) times the timescale of variation.

Worked examples
1
\[ \text{Parallel-plate capacitor, area }A=100\ \mathrm{cm^2},\ \text{charging current }I=2.0\ \mathrm{A}. \]
Show the displacement current across the gap equals the conduction current in the leads, and find \(\partial E/\partial t\). A
2
\[ E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0 A}\quad\Rightarrow\quad \frac{\partial E}{\partial t}=\frac{1}{\varepsilon_0 A}\frac{dQ}{dt}=\frac{I}{\varepsilon_0 A} \]
Uniform field of an ideal capacitor from Gauss; differentiate with \(I=dQ/dt\). A
3
\[ I_d=\varepsilon_0\frac{\partial E}{\partial t}\,A=\varepsilon_0\cdot\frac{I}{\varepsilon_0 A}\cdot A=I \]
Integrate the displacement current density over the gap area; the \(\varepsilon_0\) and \(A\) cancel. B
4
\[ \frac{\partial E}{\partial t}=\frac{2.0}{(8.854\times10^{-12})(1.00\times10^{-2})}=2.26\times10^{13}\ \mathrm{V\,m^{-1}\,s^{-1}} \]
Insert \(A=100\ \mathrm{cm^2}=1.00\times10^{-2}\ \mathrm{m^2}\) and \(\varepsilon_0=8.854\times10^{-12}\ \mathrm{F\,m^{-1}}\). A
\[ I_d=2.0\ \mathrm{A}=I,\qquad \frac{\partial E}{\partial t}=2.3\times10^{13}\ \mathrm{V\,m^{-1}\,s^{-1}} \]

Reading. The displacement current in the gap exactly matches the conduction current in the wire, so the Amperean loop integral gives the same \(\mathbf{B}\) whichever capping surface is chosen. No charge crosses the gap.

Units check. \(\varepsilon_0(\partial E/\partial t)A=\mathrm{(F\,m^{-1})(V\,m^{-1}\,s^{-1})(m^2)}=\mathrm{F\,V\,s^{-1}}=\mathrm{C\,s^{-1}=A}.\)

1
\[ E(t)=E_0\cos(\omega t),\quad E_0=1.0\times10^{4}\ \mathrm{V\,m^{-1}},\quad f=1.0\ \mathrm{MHz}. \]
Find the peak displacement current density of an oscillating field (e.g. inside a dielectric-free region of a radio-frequency cavity). A
2
\[ J_d=\varepsilon_0\frac{\partial E}{\partial t}=-\varepsilon_0 E_0\,\omega\sin(\omega t),\qquad J_{d,\text{peak}}=\varepsilon_0 E_0\,\omega \]
Differentiate the field; the amplitude of the sine sets the peak. A
3
\[ \omega=2\pi f=2\pi(1.0\times10^{6})=6.28\times10^{6}\ \mathrm{rad\,s^{-1}} \]
Convert frequency to angular frequency. A
4
\[ J_{d,\text{peak}}=(8.854\times10^{-12})(1.0\times10^{4})(6.28\times10^{6}) \]
Substitute numbers with \(\varepsilon_0=8.854\times10^{-12}\ \mathrm{F\,m^{-1}}\). B
\[ J_{d,\text{peak}}=5.6\times10^{-1}\ \mathrm{A\,m^{-2}} \]

Reading. Even a modest \(10^4\ \mathrm{V\,m^{-1}}\) field oscillating at 1 MHz sources a displacement current density of order \(0.5\ \mathrm{A\,m^{-2}}\) — comparable to conduction currents in real conductors, which is why the term cannot be ignored at radio frequencies.

Units check. \(\mathrm{(F\,m^{-1})(V\,m^{-1})(s^{-1})=(C\,V^{-1}\,m^{-1})(V\,m^{-1})(s^{-1})=C\,m^{-2}\,s^{-1}=A\,m^{-2}}.\)

Problems
  1. Starting only from \(\nabla\times\mathbf{B}=\mu_0\mathbf{J}\), show explicitly that this law is incompatible with a spherically symmetric radial current \(\mathbf{J}=J(r,t)\hat{\mathbf{r}}\) feeding a growing point charge. Which equation must be added?
    Solution Take the divergence: \(\nabla\cdot(\nabla\times\mathbf{B})=0=\mu_0\nabla\cdot\mathbf{J}\), so the law requires \(\nabla\cdot\mathbf{J}=0\). But a current feeding a growing charge has \(\nabla\cdot\mathbf{J}=-\partial\rho/\partial t\neq0\) (by continuity, since \(dQ/dt=\oint\mathbf{J}\cdot d\mathbf{A}\neq0\)). The two are contradictory, so the magnetostatic law fails. Adding the displacement current \(\mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t\) restores consistency: with Gauss, \(\varepsilon_0\partial(\nabla\cdot\mathbf{E})/\partial t=\partial\rho/\partial t\) exactly cancels the shortfall.
  2. A capacitor with circular plates of radius \(R=5.0\ \mathrm{cm}\) is charged by \(I=3.0\ \mathrm{A}\). Find the magnetic field magnitude at radius \(r=2.0\ \mathrm{cm}\) from the axis, midway between the plates.
    Solution Displacement current is uniform over the plate: \(J_d=I/(\pi R^2)\). Enclosed by radius \(r\): \(I_{d,\text{enc}}=I\,(r^2/R^2)\). Ampère–Maxwell on a circle: \(B(2\pi r)=\mu_0 I_{d,\text{enc}}=\mu_0 I r^2/R^2\), so \(B=\mu_0 I r/(2\pi R^2)\). Numerically \(B=(4\pi\times10^{-7})(3.0)(0.020)/[2\pi(0.050)^2]=(2\times10^{-7})(3.0)(0.020)/(0.0025)=4.8\times10^{-6}\ \mathrm{T}\). So \(B\approx4.8\ \mu\mathrm{T}\).
  3. Show that in a source-free vacuum the Maxwell curl equations combine to give a wave equation, and identify the propagation speed.
    Solution With \(\mathbf{J}=0,\rho=0\): take the curl of Faraday \(\nabla\times\mathbf{E}=-\partial\mathbf{B}/\partial t\): \(\nabla\times(\nabla\times\mathbf{E})=-\partial(\nabla\times\mathbf{B})/\partial t\). Use \(\nabla\times(\nabla\times\mathbf{E})=\nabla(\nabla\cdot\mathbf{E})-\nabla^2\mathbf{E}=-\nabla^2\mathbf{E}\) (since \(\nabla\cdot\mathbf{E}=0\)), and \(\nabla\times\mathbf{B}=\mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t\). Then \(\nabla^2\mathbf{E}=\mu_0\varepsilon_0\,\partial^2\mathbf{E}/\partial t^2\), a wave equation with speed \(c=1/\sqrt{\mu_0\varepsilon_0}=2.998\times10^{8}\ \mathrm{m\,s^{-1}}\). The displacement term is essential; without it the right-hand side vanishes and no wave exists.
  4. Estimate the ratio of displacement to conduction current density in copper (\(\sigma=5.96\times10^{7}\ \mathrm{S\,m^{-1}}\)) at \(f=50\ \mathrm{Hz}\), for an ohmic field \(J=\sigma E\). Comment on why displacement current is negligible in power engineering.
    Solution Conduction: \(J_c=\sigma E\). Displacement peak: \(J_d=\varepsilon_0\omega E\). Ratio \(J_d/J_c=\varepsilon_0\omega/\sigma=(8.854\times10^{-12})(2\pi\cdot50)/(5.96\times10^{7})=(8.854\times10^{-12})(314)/(5.96\times10^{7})=2.78\times10^{-9}/5.96\times10^{7}=4.7\times10^{-17}\). Utterly negligible, so at mains frequency in a good conductor the displacement current plays no role — quasi-static circuit theory is exact to extraordinary precision. It becomes comparable only when \(\omega\sim\sigma/\varepsilon_0\), the plasma/relaxation frequency, far into the optical/UV for metals.
  5. A uniform electric field between capacitor plates increases at \(\partial E/\partial t=1.0\times10^{12}\ \mathrm{V\,m^{-1}\,s^{-1}}\). Compute the magnetic field at radius \(r=3.0\ \mathrm{cm}\) from the symmetry axis, treating the field region as extending well beyond \(r\).
    Solution With no conduction current in the gap, Ampère–Maxwell over a circle of radius \(r\): \(B(2\pi r)=\mu_0\varepsilon_0(\partial E/\partial t)(\pi r^2)\), so \(B=\tfrac12\mu_0\varepsilon_0 r\,(\partial E/\partial t)\). Numerically \(B=\tfrac12(4\pi\times10^{-7})(8.854\times10^{-12})(0.030)(1.0\times10^{12})\). Compute \(\mu_0\varepsilon_0=1/c^2=1.113\times10^{-17}\ \mathrm{s^2\,m^{-2}}\); then \(B=\tfrac12(1.113\times10^{-17})(0.030)(1.0\times10^{12})=\tfrac12(3.34\times10^{-7})=1.7\times10^{-7}\ \mathrm{T}\). So \(B\approx0.17\ \mu\mathrm{T}\), induced purely by the changing electric field.