Case Study
Passage with linked questions
Case Set 1
Case AnalysisPassage
A physics teacher demonstrates the charging of a parallel plate capacitor connected to a time-varying current source. She draws two different Amperian surfaces sharing the same circular boundary of radius r outside the capacitor: Surface S1 is a flat disc that cuts through the connecting wire, while Surface S2 is a pot-shaped surface that passes between the capacitor plates without touching any conductor. When students apply Ampere's circuital law ∮B·dl = µ₀i to both surfaces, they get different answers for the magnetic field at the same point P — a contradiction. The teacher explains that Maxwell resolved this by introducing a new concept. She writes on the board: 'The electric field between the plates is E = Q/(Aε₀), and as the capacitor charges, this field changes with time.'
Question 1: What additional current did Maxwell introduce to remove the inconsistency in Ampere's circuital law, and what causes it?
- Maxwell introduced the displacement current iₐ = ε₀(dΦ_E/dt), which arises due to the time-varying electric flux between the capacitor plates, not due to any physical flow of charges.
- This displacement current has the same magnitude as the conduction current in the wire and acts as a source of magnetic field, ensuring that both Surface S1 and S2 give the same value of B at point P.
Question 2: Write the generalised Ampere-Maxwell law and explain what 'total current' means in this context.
- The Ampere-Maxwell law is: ∮B·dl = µ₀iₓ + µ₀ε₀(dΦ_E/dt). The total current is the sum of conduction current (iₓ, due to flow of charges in conductors) and displacement current (iₐ = ε₀ dΦ_E/dt, due to changing electric flux).
- Outside the capacitor plates, only iₓ = i exists and iₐ = 0; between the plates, iₓ = 0 and only iₐ = i exists. In both regions, the total current equals i, giving a consistent magnetic field at P.
Question 3: Using the electric flux between the capacitor plates, derive the expression for displacement current and show it equals the conduction current i charging the capacitor.
- The electric field between plates of area A with charge Q is E = Q/(Aε₀). The electric flux through Surface S2 is Φ_E = EA = Q/ε₀.
- Differentiating: dΦ_E/dt = (1/ε₀)(dQ/dt). Since dQ/dt = i (the conduction current), we get dΦ_E/dt = i/ε₀.
- Therefore displacement current iₐ = ε₀(dΦ_E/dt) = ε₀ × (i/ε₀) = i. This shows that the displacement current between the plates exactly equals the conduction current in the wire, resolving the contradiction.