- 2074 Asoj · 3 marks
Conduction and displacement current in capacitor
Short answer
In a parallel-plate capacitor with v(t)=Vm sin ωt, conduction current in wires is i_c = C dv/dt = C Vm ω cos ωt. Between plates, D = ε v/d, displacement current density J_d = ε/d dv/dt, total i_d = J_d S = C dv/dt = i_c. Thus displacement current equals conduction current, ensuring continuity.
Derive conduction and displacement current in a capacitor.
Consider a parallel-plate capacitor with plate area S, separation d and dielectric permittivity ε, connected to a source v(t) = V_m sin ωt.
Conduction current
The current in the connecting wires is the rate of change of charge on the plates:
i_c = dq/dt = C dv/dt, C = εS/d
i_c = C V_m ω cos ωt
No charge crosses the dielectric, so there is no conduction current between the plates (σ ≈ 0). Without any other current, Ampere's law would give a break in current at the capacitor.
Displacement current
Between the plates, E = v/d, so D = εE = εv/d. The displacement current density is
J_d = ∂D/∂t = (ε/d) dv/dt
Total displacement current through the plate area:
i_d = J_d S = (εS/d) dv/dt = C dv/dt = C V_m ω cos ωt
Result
i_d = i_c. The displacement current in the dielectric is equal to the conduction current in the wire, so the total current is continuous through the whole circuit. This is why Maxwell added ∂D/∂t to Ampere's law: ∇×H = J + ∂D/∂t.
✓ Answer checked by Claude Opus 5.5 (high effort)
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