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UY1: Displacement Current

See why Maxwell added displacement current and how it fixes Ampere's law for charging capacitors.

  • University Physics Year 1
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Learning objectives

  • Use vector and differential or integral calculus to express physical change and accumulation.
  • Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
  • Choose an efficient mathematical method and validate the result physically.
Why this matters + quick links

This page gives the UY1 working model/result for Displacement Current. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

  • Maxwell’s key fix to Ampere’s law: changing electric flux behaves like a current in producing magnetic circulation.
  • Displacement current (definition):
I_D = ε₀dΦ_E/dt
  • Maxwell-Ampere law (integral form):
∮ vector B · d vector l = μ₀(I_(C,enc) + ε₀dΦ_E/dt)
  • Modelling context: this matters whenever E changes in time (charging capacitors, EM waves). In strict magnetostatics, dΦ_E/dt = 0 and you recover the usual Ampere’s law.

Prerequisites: Ampere’s Law, Capacitors And Capacitance, Electric Flux + Gauss’s Law (simple)
Next uses: Mutual Inductance, Electromagnetic Spectrum & Sinusoidal EM Plane Waves

2) Setup

Consider a charging parallel-plate capacitor with circular plates (radius R), neglecting fringing fields.

  • Conduction current in wires: I_C = dq/dt.
  • Field between plates: E = q/(ε₀ A) with A = π R².
  • Electric flux through plate area: Φ_E = EA.
  • We are using a quasi-static model: fields change slowly enough that the capacitor field can be treated as approximately uniform between plates.
Common traps (what displacement current is)
  • Displacement current is not charge physically flowing through the dielectric/vacuum. It is a shorthand for the magnetic effect of a changing electric field.
  • Φ_E is an electric flux (of vector E), not magnetic flux. Don’t swap Φ_E and Φ_B.
  • If you use an Amperian loop of radius r < R, only the enclosed fraction of the changing flux counts.

3) Core derivation/explanation

From Φ_E = EA = q/ε₀,

ε₀dΦ_E/dt = dq/dt = I_C

Define displacement current:

I_D≡ε₀dΦ_E/dt

so during capacitor charging, I_D = I_C in magnitude.

Maxwell-Ampere law (integral form):

∮ vector B · d vector l = μ₀(I_(C,encl) + ε₀dΦ_E/dt)

This removes the surface-choice contradiction in the charging-capacitor geometry.

Checks (sanity)

  • In an ideal charging capacitor, the displacement current between plates matches the wire conduction current in magnitude: I_D = I_C.
  • The computed B(r) between plates scales like B ∝ r for r < R and matches the wire-edge field at r = R.

4) Worked example(s)

For an Amperian circle of radius r < R centered between plates, only a fraction of plate area is enclosed. With uniform E:

I_(D,encl) = I_Cr²/R²

Then

B(2π r) = μ₀ I_Cr²/R² ⇒ B = ((μ₀ I_C)/2π)r/R² (r < R)

At r = R:

B = (μ₀ I_C)/(2π R)

matching the magnetic field magnitude at the wire edge.

5) Practice set (with hints + answers)

  1. A capacitor charges at constant I_C = 0.80 A. What is I_D between plates? Hint: for ideal parallel plates, I_D = I_C. Answer: I_D = 0.80 A.

  2. Plate radius R = 0.040 m, current I_C = 1.2 A. Find B at r = 0.020 m between plates. Hint: use B = ((μ₀ I_C)/2π)r/R². Answer: B = 3.0 × 10⁻⁶ T.

  3. Why can old Ampere’s law fail for a charging capacitor? Hint: try two different surfaces spanning the same loop. Answer: one encloses wire current, another encloses none; displacement-current term restores consistency.

6) Summary + next steps

Displacement current is not charge flow through vacuum; it is a changing electric flux term that produces magnetic effects and completes Ampere’s law.

Next: Mutual Inductance Previous: Induced Electric Field Back To Electromagnetism (UY1)