Physics · Electromagnetic Waves · NEET
Start with a charging capacitor. Charge on the plate is Q = CV. The conduction current in the wire is I = dQ/dt. For a parallel plate capacitor, E = Q/(ε0 A), so Q = ε0 A·E = ε0 (E·A) = ε0 ΦE, because flux ΦE = E·A. Now differentiate: I = dQ/dt = ε0 (dΦE/dt). Maxwell defined this right-hand side as the displacement current: id = ε0(dΦE/dt). So id is not a separate mystery — it is just the conduction current rewritten in terms of the electric field between the plates.
Ampere's law needs a current passing through ANY surface bounded by the loop. If the surface passes through the wire, it sees the conduction current I. If the surface bulges between the plates (through the gap), no charge crosses it, so it sees zero conduction current — a contradiction. Maxwell fixed this: the changing electric flux in the gap gives id = ε0(dΦE/dt), and this exactly equals I. So both surfaces now give the same answer, and current stays continuous through the circuit.
ΦE is the electric flux = E × A (field strength times plate area), measured in volt·metre. As the capacitor charges, E grows, so ΦE grows with time. dΦE/dt is the rate at which this flux changes each second. A fast-changing field gives a large displacement current; a steady (DC, fully charged) field gives dΦE/dt = 0, so id = 0. That is why displacement current only exists WHILE the field is changing.
No. In a vacuum or air gap, no electrons cross the space between the plates. 'Displacement current' is a name Maxwell chose, but nothing is actually carried across. What flows is the effect of a changing electric field, which produces a magnetic field exactly as a real current would. So id has the units and the magnetic effect of a current, without any moving charge.
For a capacitor, Q = CV, so I = dQ/dt = C(dV/dt). Since id = I, we get id = C(dV/dt). This is the most useful form for NEET numericals: given capacitance C and the rate of change of voltage dV/dt, the displacement current is simply their product. You rarely need to compute flux directly in exam problems — use id = C(dV/dt).
A parallel plate capacitor of capacitance 20 µF is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires and the displacement current through the plates of the capacitor would be, respectively,
To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 µF, the rate of change of applied variable potential difference (dV/dt) must be
A capacitor of capacitance C is connected across an ac source of voltage V given by V = V₀ sin ωt. The displacement current between the plates of the capacitor would then be given by
Try the real previous-year questions from this chapter — each with the answer and a full solution.
They are equal in magnitude but different in nature. Conduction current is real moving charge in the wire; displacement current is due to a changing electric field in the gap. Their magnitudes match so current stays continuous: id = I = C(dV/dt).
The ampere (A), same as any current. Even though ε0(dΦE/dt) is built from permittivity and flux, the combination has the units of current.
Yes. Once fully charged, V and E are constant, so dΦE/dt = 0 and dV/dt = 0. Therefore id = 0. Displacement current exists only while the field is changing.
To fix a contradiction in Ampere's circuital law for a charging capacitor, where a surface through the gap gave zero current while a surface through the wire gave I. Adding id = ε0(dΦE/dt) removed the inconsistency and led to the prediction of electromagnetic waves.
ε0 = 8.85 × 10⁻¹² C²/(N·m²) or F/m. In most NEET problems you use the shortcut id = C(dV/dt), so ε0 rarely needs to be plugged in directly.
