The Displacement Current from Charge Conservation

Derivation

The Displacement Current from Charge Conservation

D-161 Home PU-204 Threads fields · symmetry Depends on amperes-law-from-biot-savart, charge-continuity-equation

Statement

The magnetostatic Ampère law ∇×B = μ₀J forces ∇·J = 0, which contradicts the continuity equation ∇·J + ∂ρ/∂t = 0 whenever the charge density is time-dependent. Requiring consistency with continuity and with Gauss's law ∇·E = ρ/ε₀ forces the addition of exactly one extra term, Maxwell's displacement current ε₀ ∂E/∂t, giving the corrected law ∇×B = μ₀(J + ε₀ ∂E/∂t).

Why it matters

This is the single amendment that closes Maxwell's equations into a self-consistent set and makes electromagnetic waves possible. Without the displacement current, a charging capacitor, an antenna, or any circuit with a time-varying field would violate charge conservation at the level of the field equations.

The argument is also a template for how a conservation law can dictate the form of a field equation: continuity is not an optional add-on but a mathematical constraint that Ampère's law, in its static form, simply cannot satisfy. The repair is essentially unique, which is why it was a genuine prediction rather than a fit.

Assumptions

The fields are smooth enough that ∇·(∇×B) = 0 holds identically.If B is not twice differentiable (e.g. at an idealised surface current sheet or a point source), the divergence-of-curl identity must be interpreted distributionally, and the pointwise contradiction below need not appear in that form. Gauss's law ∇·E = ρ/ε₀ is taken as independently established.Without it the extra term cannot be tied to ρ, and the divergence of the correction could not be evaluated; the uniqueness argument then collapses. The continuity equation ∇·J + ∂ρ/∂t = 0 is exact (charge is locally conserved).If charge could appear or vanish nonlocally, no correction to Ampère's law could restore consistency, and the whole construction would be void. The correction added to μ₀J is a genuine vector field determined locally by the electromagnetic fields, not an arbitrary divergence-free field.Uniqueness is only up to the curl of an arbitrary vector field (a divergence-free ambiguity). Insisting the extra term be built locally from E and ρ via Gauss's law fixes it to ε₀ ∂E/∂t; dropping this locality assumption reopens that gauge-like freedom.

Derivation

∇×B = μ₀J

Start from Ampère's law in its magnetostatic form, taken as a prior result. A

∇·(∇×B) = μ₀ ∇·J

Apply the divergence to both sides; divergence is linear so μ₀ passes through. A

∇·(∇×B) ≡ 0 ⟹ μ₀ ∇·J = 0 ⟹ ∇·J = 0

The divergence of any curl vanishes identically (vector calculus identity), so the left side is zero for all smooth B; μ₀ ≠ 0. B

∇·J = −∂ρ/∂t

Independently, the continuity equation (prior result) expresses local charge conservation. A

−∂ρ/∂t = 0 (forced by steps 3 and 4)

Equating the two expressions for ∇·J. This is the contradiction: static Ampère's law demands the charge density be constant in time everywhere, which fails for any charging or discharging system. B

∇×B = μ₀J + C

Postulate a correction: add an unknown vector field C to the right-hand side, to be determined by demanding consistency. B

0 = ∇·(∇×B) = μ₀ ∇·J + ∇·C ⟹ ∇·C = −μ₀ ∇·J = μ₀ ∂ρ/∂t

Take the divergence again and substitute continuity (step 4). The correction must have exactly this divergence to remove the contradiction. B

ρ = ε₀ ∇·E ⟹ μ₀ ∂ρ/∂t = μ₀ε₀ ∂(∇·E)/∂t = ∇·(μ₀ε₀ ∂E/∂t)

Use Gauss's law to eliminate ρ, then commute the divergence with the time derivative (permissible for smooth fields on a fixed spatial domain). The right-hand side is now manifestly a divergence. C

∇·C = ∇·(μ₀ε₀ ∂E/∂t) ⟹ C = μ₀ε₀ ∂E/∂t + ∇×W

Matching divergences fixes C up to the curl of an arbitrary field W (a divergence-free ambiguity). Locality — the correction is built from the fields already in play through Gauss's law — selects W = 0. This is the uniqueness step. C

∇×B = μ₀J + μ₀ε₀ ∂E/∂t = μ₀(J + ε₀ ∂E/∂t)

Substitute the minimal choice C = μ₀ε₀ ∂E/∂t back into step 6. A

∇·(J + ε₀ ∂E/∂t) = ∇·J + ∂ρ/∂t = 0 ✓

Verify: the divergence of the total current density now vanishes identically by continuity plus Gauss, so the corrected law is consistent for all time-dependent ρ. B

Result

∇×B = μ₀J + μ₀ε₀ ∂E/∂t, with displacement current density J_d = ε₀ ∂E/∂t

Reading. A time-varying electric field sources a magnetic field exactly as a real current does. The added term is the unique local correction that makes the total current density J + ε₀ ∂E/∂t divergence-free, so that the field equation never contradicts charge conservation. In a charging capacitor, the conduction current in the wire is handed off to the displacement current in the gap without interruption.

Units check. ε₀ has units C²·N⁻¹·m⁻² = F·m⁻¹; ∂E/∂t has units (V·m⁻¹)·s⁻¹. Their product: (F·m⁻¹)(V·m⁻¹·s⁻¹) = (C·V⁻¹·m⁻¹)(V·m⁻¹·s⁻¹) = C·s⁻¹·m⁻² = A·m⁻², a current density, matching J. Both sides of the corrected law then carry units of μ₀·(A·m⁻²) = T·m⁻¹, the units of ∇×B. ✓

Limiting cases

Statics / steady state (∂E/∂t = 0): the correction vanishes and the law reduces to magnetostatic Ampère's law ∇×B = μ₀J. Consistent because steady currents have ∇·J = 0 already.
Source-free region (J = 0): ∇×B = μ₀ε₀ ∂E/∂t. Combined with Faraday's law this yields the wave equation with speed c = 1/√(μ₀ε₀).
Low frequency / quasi-static: the displacement term is smaller than the conduction term by roughly ωε₀/σ in a conductor, so it is negligible in ordinary circuits but dominant in vacuum and dielectrics.
Uniform static ρ, no motion: both correction and conduction current vanish; no magnetic field is produced, as expected of a stationary electrostatic charge.

Breaks when

Media with polarization and magnetization. In matter the correct free-current form is ∇×H = J_free + ∂D/∂t; naively using ε₀ ∂E/∂t ignores the bound polarization current ∂P/∂t and gives wrong fields inside dielectrics.
Singular or lower-dimensional sources. At an idealised surface current, line current, or point charge the fields are not smooth, the identity ∇·(∇×B) = 0 must be read distributionally, and the pointwise manipulation of steps 2–3 no longer applies without care.
Relativistically, at the three-vector level. The clean statement is the covariant ∂_μF^μν = μ₀J^ν whose consistency is ∂_νJ^ν = 0; splitting into E and B obscures the frame dependence and can mislead if the observer is accelerating or the geometry is curved.
Non-conservative charge models. If one (incorrectly) uses a source theory that violates continuity, no correction exists that can rescue Ampère's law — the whole uniqueness argument presupposes exact local charge conservation.

Failure modes

"Displacement current is a real flow of charge." It is not; ε₀ ∂E/∂t is a field term with the units of current density, not moving charge. Nothing physically drifts across a vacuum capacitor gap.
Forgetting μ₀ vs μ₀ε₀. Writing the added term as μ₀ ∂E/∂t instead of μ₀ε₀ ∂E/∂t drops a factor of ε₀ and breaks the units.
Using ∮B·dl = μ₀I_enc for a capacitor and getting different answers for different surfaces. The resolution is that the flat and bulging surfaces enclose the same total (conduction + displacement) current, not the same conduction current.
Confusing ∂E/∂t with ∂D/∂t in dielectrics. Inside matter one must use D = ε₀E + P; the polarization current is often the dominant part.
Claiming Ampère's law "is wrong." It is not wrong; it is incomplete for time-dependent sources. The static form remains exact whenever ∂E/∂t = 0.

Discussion

The deepest content of this derivation is that a field equation is constrained by the conservation law of its source. Because the divergence of a curl is identically zero, any equation of the form ∇×(field) = μ₀(source current) silently asserts that the source current is divergence-free. For magnetostatics that is fine — steady currents form closed loops. But charge conservation only forces ∇·J = 0 when ∂ρ/∂t = 0. The displacement current is precisely the object whose divergence supplies the missing ∂ρ/∂t, and Gauss's law is what lets us write that time derivative as the divergence of something built from E.

Physically, the term completes the symmetry between Faraday's law and Ampère's law: a changing B makes an E (Faraday), and a changing E makes a B (Maxwell). This mutual induction in vacuum is self-sustaining and propagates as light. The measured value 1/√(μ₀ε₀) ≈ 3.00×10⁸ m·s⁻¹ matching the speed of light was Maxwell's decisive evidence that light is an electromagnetic phenomenon, and it all follows from the single term forced here.

The construction also illustrates the interplay of the two "threads" of this unit. The symmetry thread appears in the E–B duality above and in the fact that the covariant source equation ∂_μF^μν = μ₀J^ν makes continuity automatic through the antisymmetry of F^μν. The fields thread appears in treating ε₀ ∂E/∂t as a genuine local source of B on equal footing with matter current.

The uniqueness claimed in the statement is exact only modulo a divergence-free field: any ∇×W could be added to the correction without spoiling consistency, since its divergence vanishes. What removes this freedom is the demand that the extra term be a local functional of the electromagnetic field expressible through the field equations already assumed — Gauss's law then delivers ε₀ ∂E/∂t and nothing else. In the fully relativistic formulation this ambiguity is absent from the start: the antisymmetric field tensor makes ∂_νJ^ν = 0 an identity, so the displacement current is not added but is already contained in ∂_μF^μν. Maxwell's "correction" is, from the covariant viewpoint, simply the space part of an equation that was consistent all along.

Common misconceptions. The displacement current does not require a medium (Maxwell's original elastic-ether picture is obsolete); it exists in perfect vacuum. It is also not merely a bookkeeping trick to make Ampère's loop integral surface-independent — that surface-independence is a consequence of the deeper requirement of consistency with continuity, which is the real driver.

Worked examples

Magnetic field inside a charging capacitor. Circular parallel plates, radius R = 5.0 cm, charged by a steady conduction current I = 2.0 A. Find B at the plate edge (r = R) in the gap.

Set up: by symmetry B is azimuthal and uniform on the Amperian circle of radius r in the gap. A

∮B·dl = μ₀ε₀ dΦ_E/dt = μ₀ I_d,enc, with total displacement current I_d = I

In the gap there is no conduction current; the displacement current equals the conduction current feeding the plates, I_d = ε₀(dΦ_E/dt) = I. B

B·(2πr) = μ₀ I_d,enc = μ₀ I (r²/R²) ⟹ B = μ₀ I r / (2πR²)

Uniform field between plates ⟹ displacement current enclosed scales as area, (πr²)/(πR²). Rearranged symbolically before inserting numbers. B

At r = R: B = μ₀ I / (2πR) = (4π×10⁻⁷ T·m·A⁻¹)(2.0 A) / (2π × 0.050 m)

Evaluate at the edge; the R² and one factor of R cancel. A

B = (4×10⁻⁷ × 2.0)/(2 × 0.050) T = 8.0×10⁻⁶ T = 8.0 μT

Reading. The displacement current in the gap produces exactly the magnetic field a real 2.0 A wire would at the same radius — the field lines close seamlessly across the gap.

Rate of change of the field in a capacitor. Square plates of area A = 1.0×10⁻² m², conduction current I = 1.0 A charging the gap (vacuum). Find dE/dt and the displacement current density J_d.

Set up: uniform field between plates, E = σ/ε₀ = Q/(ε₀A). A

dE/dt = (1/(ε₀A)) dQ/dt = I/(ε₀A)

Differentiate E in time; dQ/dt = I is the conduction current. Symbolic result first. B

dE/dt = (1.0 A) / [(8.85×10⁻¹² F·m⁻¹)(1.0×10⁻² m²)] = 1.0 / (8.85×10⁻¹⁴) V·m⁻¹·s⁻¹

Insert numbers. A

J_d = ε₀ dE/dt = I/A = (1.0 A)/(1.0×10⁻² m²) = 1.0×10² A·m⁻²

The displacement current density equals the conduction current density feeding the plate, as required by continuity across the plate surface. B

dE/dt = 1.1×10¹³ V·m⁻¹·s⁻¹, J_d = 1.0×10² A·m⁻²

Reading. A very large field rate corresponds to a modest current density because ε₀ is tiny. Multiplying J_d by the plate area recovers exactly the 1.0 A conduction current — the hand-off is exact.

Problems

Starting from ∇×B = μ₀J, show explicitly that it is inconsistent for a spherically symmetric radial current J = J(r,t) r̂ feeding a growing point charge.
Solution
For a point charge Q(t) at the origin the current is radial, J = (I/4πr²) r̂ with I = dQ/dt. This field is purely radial and curl-free-sourced, but continuity gives ∇·J = −∂ρ/∂t ≠ 0 since the enclosed charge grows. Yet static Ampère's law requires ∇·J = 0. Contradiction. Adding ε₀∂E/∂t with E = (Q/4πε₀r²) r̂ gives total current density J + ε₀∂E/∂t = (I/4πr²)r̂ + (İ /4πr²)·... — evaluating, the total current density is divergence-free and, being purely radial and spherically symmetric, sources zero B (as symmetry demands there is no preferred azimuthal direction). Consistency is restored.
A capacitor with circular plates (R = 4.0 cm) has a uniform field increasing at dE/dt = 5.0×10¹² V·m⁻¹·s⁻¹. Find (a) the total displacement current and (b) B at r = 2.0 cm.
Solution
(a) I_d = ε₀(dE/dt)(πR²) = (8.85×10⁻¹²)(5.0×10¹²)(π×0.040²) = (8.85×10⁻¹²)(5.0×10¹²)(5.03×10⁻³) = 0.223 A ≈ 0.22 A. (b) Inside, B = μ₀ε₀(dE/dt)(r/2) = (4π×10⁻⁷)(8.85×10⁻¹²)(5.0×10¹²)(0.010) = (1.257×10⁻⁶)(8.85×10⁻¹²)(5.0×10¹²)(0.010) = 5.6×10⁻⁷ T ≈ 0.56 μT.
Show that in a good conductor of conductivity σ at angular frequency ω, the displacement current is negligible compared with the conduction current when ω ≪ σ/ε₀. Evaluate σ/ε₀ for copper (σ = 5.96×10⁷ S·m⁻¹).
Solution
With J = σE and E ∝ e^−iωt, the displacement current is ε₀∂E/∂t = −iωε₀E. The ratio of magnitudes is |ε₀∂E/∂t| / |σE| = ωε₀/σ, which is ≪ 1 when ω ≪ σ/ε₀. For copper, σ/ε₀ = (5.96×10⁷)/(8.85×10⁻¹²) = 6.7×10¹⁸ s⁻¹, so displacement current is negligible in copper up to optical/X-ray frequencies.
Using the corrected Ampère law with J = 0 together with Faraday's law ∇×E = −∂B/∂t, derive the vacuum wave equation for E and identify the wave speed.
Solution
Take the curl of Faraday: ∇×(∇×E) = −∂(∇×B)/∂t. Left side: ∇(∇·E) − ∇²E = −∇²E (since ∇·E = 0 with no charge). Right side, using ∇×B = μ₀ε₀∂E/∂t: −μ₀ε₀∂²E/∂t². Thus ∇²E = μ₀ε₀∂²E/∂t², a wave equation with speed c = 1/√(μ₀ε₀) = 3.00×10⁸ m·s⁻¹. Without the displacement current the right side would vanish and no wave would exist.
For the covariant source equation ∂_μF^μν = μ₀J^ν, prove that charge conservation ∂_νJ^ν = 0 is automatic, and explain how this relates to the displacement current.
Solution
Take ∂_ν of both sides: ∂_ν∂_μF^μν = μ₀∂_νJ^ν. The left side is a contraction of the symmetric operator ∂_ν∂_μ with the antisymmetric tensor F^μν = −F^νμ; a symmetric–antisymmetric contraction is identically zero. Hence ∂_νJ^ν = 0 for any consistent field configuration. The spatial (ν = i) components of ∂_μF^μν = μ₀J^ν expand to ∇×B − μ₀ε₀∂E/∂t = μ₀J — the displacement current appears as the ∂₀F⁰ⁱ piece. So in the covariant form the term is not added by hand; it is the time-derivative-of-E part of the divergence of the field tensor, and continuity holds identically.