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Displacement current is one of those topics where students memorise i_d = ε₀ dΦ_E/dt without knowing why it exists. The reason is a short and clever argument about a capacitor, and once you see it, the formula and the exam questions follow. This post walks through the argument, two worked examples and the usual mistakes.
What goes wrong with Ampere's law?
Ampere's circuital law says that the line integral of B around a closed loop equals μ₀ times the current passing through any surface bounded by that loop. The NCERT Class 12 chapter Electromagnetic Waves applies it to a wire charging a parallel-plate capacitor.
Draw a circular loop around the wire, just outside the capacitor.
Choose a flat surface across the loop. The wire passes through it, so B(2πr) = μ₀i and there is a magnetic field.
Now choose a pot-shaped surface with the same rim, whose bottom sits between the plates. No wire passes through it, so the right-hand side is zero and B = 0.
Same loop, two answers. NCERT concludes that the law "must be missing something".
What is displacement current?
Something does pass through the surface between the plates: the electric field. With charge Q on plates of area A, the field is E = Q/(ε₀A), so the electric flux is Φ_E = EA = Q/ε₀. As the capacitor charges, dΦ_E/dt = (1/ε₀) dQ/dt = i/ε₀. So
ε₀ dΦ_E/dt = i
NCERT calls this term the displacement current, i_d. Adding it to Ampere's law gives the same answer for every surface: i = i_c + i_d. Outside the plates only conduction current flows; between them only displacement current. The source of a magnetic field is not just moving charge but also a changing electric field.
Maxwell's 1865 paper A Dynamical Theory of the Electromagnetic Field builds this term in. He writes that the variations of the electrical displacement "must be added to the currents ... to get the total motion of electricity". Our profile of James Clerk Maxwell follows where that led.
Why does it matter beyond the capacitor?
A changing magnetic field makes an electric field (Faraday's law), and now a changing electric field makes a magnetic field. Each can sustain the other through empty space. That is an electromagnetic wave. NCERT says this effect "explains the existence of radio waves, gamma rays and visible light". In the same 1865 paper, Maxwell concluded "that light is an electromagnetic disturbance propagated through the field according to electromagnetic laws".
The JEE (Main) 2026 syllabus opens Unit 15 with "Displacement current, electromagnetic waves and their characteristics". For the spectrum itself, see our post on electromagnetic waves and the spectrum.
Worked example 1: current through a capacitor
A 2.0 nF capacitor is charged so that its voltage rises at 5.0 × 10⁷ V/s. Find the displacement current between the plates.
Since Q = CV, i = dQ/dt = C dV/dt = 2.0 × 10⁻⁹ × 5.0 × 10⁷ = 0.10 A. The displacement current between the plates is also 0.10 A, equal to the conduction current in the wires, as it must be.
Worked example 2: magnetic field between the plates
Circular plates of radius R = 5.0 cm have an electric field between them rising at dE/dt = 1.0 × 10¹² V m⁻¹ s⁻¹.
Displacement current: i_d = ε₀ A dE/dt = 8.85 × 10⁻¹² × π(0.050)² × 1.0 × 10¹² = 0.070 A.
Field inside at r = 2.0 cm: only the part of i_d inside the loop counts, which is i_d(r²/R²). So B = μ₀i_d r/(2πR²) = 2 × 10⁻⁷ × 0.070 × 0.020/0.0025 = 1.1 × 10⁻⁷ T.
Field outside at r = 10 cm: the full i_d counts, so B = μ₀i_d/(2πr) = 2 × 10⁻⁷ × 0.070/0.10 = 1.4 × 10⁻⁷ T.
Inside, B grows linearly with r. That is not just theory: Bartlett and Corle (1985) measured the magnetic field inside a thin, circular capacitor as it charged, found it varied linearly with distance from the axis, and found the slope agreed with prediction to within 5%.
Common mistakes
Thinking charge flows across the gap. No charge crosses between the plates. Only the field changes.
Using the whole current inside the plates. For r < R, only the fraction r²/R² of the displacement current is enclosed.
Forgetting that i_d = 0 for a steady field. A fully charged capacitor has no displacement current.
Confusing ε₀ and μ₀. i_d uses ε₀; B uses μ₀.
A practice routine
Redraw the two-surface argument from memory and explain it aloud in under a minute.
Solve two problems using i_d = C dV/dt and two using ε₀ A dE/dt.
For one capacitor, plot B against r from the axis to twice the plate radius.
If the step from flux to current feels unclear, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise electric flux and Gauss's law before returning here.
Next, take worked example 2 and find the radius inside the plates where B equals the value at 10 cm outside.
