Displacement current
A changing electric field acts like a current, the displacement current i_d = ε₀ dΦ_E/dt. Adding it makes Ampere's circuital law true everywhere.
The puzzle of the charging capacitor
Ampere's circuital law says ∮ B·dl = μ₀i. The line integral of B round a closed loop equals μ₀ times the current through any surface bounded by that loop. Maxwell found a problem with this. Take a capacitor being charged, with a current i in its wires. Draw a circular loop round the wire just outside one plate. For a flat surface bounded by the loop, the current i passes through it, so B is not zero. Now stretch the surface into a pot shape that passes between the plates. No charge crosses the gap, so the law gives B = 0. The same loop gives two different answers, so the law is incomplete. Maxwell noticed that between the plates the electric field is changing.
Finding the displacement current
- 1Let the plates have area A and charge Q at time t. The field between them is E = Q/(ε₀A).
- 2The electric flux through the pot-shaped surface is Φ_E = EA = Q/ε₀.
- 3As Q changes, dΦ_E/dt = (1/ε₀) dQ/dt.
- 4dQ/dt is the current i in the wires, so ε₀ dΦ_E/dt = i.
- 5★So, ε₀ dΦ_E/dt acts as a current between the plates. Maxwell called it the displacement current, i_d = ε₀ dΦ_E/dt, and it equals the conduction current.
★ key step
The Ampere-Maxwell law
Maxwell added the displacement current to Ampere's law: ∮ B·dl = μ₀(i_c + i_d) = μ₀i_c + μ₀ε₀ dΦ_E/dt. Here i_c is the conduction current, carried by moving charges. Outside the capacitor only i_c exists; between the plates only i_d exists. The total, i_c + i_d, has the same value all along the circuit, so the law now gives one answer for every surface. The big idea is this: just as a changing magnetic field produces an electric field (Faraday's law), a changing electric field produces a magnetic field. This is what makes electromagnetic waves possible.
Conduction current vs displacement current
In a charging capacitor, i_d in the gap equals i_c in the wires, so the current is continuous round the whole circuit.
Worked example: displacement current
While a parallel plate capacitor is charging, the electric field between its plates rises at 10¹² V m⁻¹ s⁻¹. The plate area is 0.02 m². Find the displacement current between the plates and the conduction current in the wires. (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
- Electric flux Φ_E = EA, so dΦ_E/dt = A dE/dt = 0.02 m² × 10¹² V m⁻¹ s⁻¹ = 2 × 10¹⁰ V m s⁻¹.
- i_d = ε₀ dΦ_E/dt = 8.85 × 10⁻¹² × 2 × 10¹⁰.
- i_d = 17.7 × 10⁻² A ≈ 0.18 A.
- The conduction current in the wires equals the displacement current, so it is also about 0.18 A.
i_d ≈ 0.18 A; the conduction current is also about 0.18 A
Every mobile phone call depends on this idea. A changing electric field creates a magnetic field, and a changing magnetic field creates an electric field, so the two keep producing each other and travel from the tower to your phone as radio waves.
2 marks: “What is displacement current? Write its expression and say why Maxwell introduced it.” The current due to a changing electric field, i_d = ε₀ dΦ_E/dt. It removes the contradiction in Ampere's law for a charging capacitor and keeps the current continuous.
3 marks: “For a parallel plate capacitor being charged, show that the displacement current between the plates equals the conduction current in the wires. Write the Ampere-Maxwell law.” E = Q/(ε₀A); Φ_E = Q/ε₀; ε₀ dΦ_E/dt = dQ/dt = i; ∮ B·dl = μ₀(i_c + ε₀ dΦ_E/dt).
1 marks: “When is the displacement current between the plates of a capacitor zero?” When the electric field between the plates is not changing, for example when the capacitor is fully charged by a DC source.
A changing E makes B (Maxwell), just as a changing B makes E (Faraday). i_d = ε₀ dΦ_E/dt.