Displacement current — Maxwell's correction to Ampere's law

Question

What is displacement current? Explain the inconsistency in Ampere’s circuital law that led Maxwell to introduce the concept. Derive the expression for displacement current through a parallel plate capacitor being charged.

(JEE Main 2023, similar pattern)

Solution — Step by Step

Ampere’s law states: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$ .

Consider a parallel plate capacitor being charged. Draw an Amperian loop around the wire. If we choose a flat surface through the wire, $I_{enc} = I$ (the conduction current). But if we choose a bulging surface that passes between the plates (where no charge flows), $I_{enc} = 0$ .

Same loop, two surfaces, two different answers — Ampere’s law gives a contradiction.

Maxwell resolved this by adding a new term. Between the plates, although no charge flows, the electric field is changing as the capacitor charges. This changing electric field acts like a current — the displacement current:

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

where $\Phi_E$ is the electric flux between the plates.

For a parallel plate capacitor with plate area $A$ :

Electric field between plates: $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$

Electric flux: $\Phi_E = EA = \frac{Q}{\varepsilon_0}$

Displacement current:

I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \cdot \frac{1}{\varepsilon_0}\frac{dQ}{dt} = \frac{dQ}{dt} = I

The displacement current between the plates equals the conduction current in the wire.

The corrected law:

\boxed{\oint \vec{B} \cdot d\vec{l} = \mu_0(I_c + I_d) = \mu_0 I_c + \mu_0\varepsilon_0\frac{d\Phi_E}{dt}}

Now both surfaces give the same answer: the flat surface sees $I_c$ , the bulging surface sees $I_d$ , and both equal $I$ . The contradiction is resolved.

Why This Works

Displacement current is not a “real” current — no charges flow between the plates. But the changing electric field produces magnetic effects identical to a real current. Maxwell’s insight was that electric and magnetic fields are symmetric: a changing magnetic field produces an electric field (Faraday’s law), and a changing electric field produces a magnetic field (this correction to Ampere’s law).

This symmetry is what led Maxwell to predict electromagnetic waves — one of the greatest achievements in physics.

Alternative Method

You can also see displacement current through the continuity equation. Conservation of charge requires $\nabla \cdot \vec{J} + \frac{\partial\rho}{\partial t} = 0$ . Ampere’s original law ( $\nabla \times \vec{B} = \mu_0 \vec{J}$ ) implies $\nabla \cdot \vec{J} = 0$ (divergence of curl is zero), which contradicts charge conservation when $\rho$ changes. Adding $\mu_0\varepsilon_0 \frac{\partial\vec{E}}{\partial t}$ fixes this.

For JEE, the key takeaway: displacement current $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ , and inside a charging capacitor, $I_d$ equals the conduction current $I_c$ . You do not need the full vector calculus derivation — the parallel plate example is sufficient for JEE Main.

Common Mistake

Students often say “displacement current flows between the plates” as if charges are moving. No charges flow — the changing electric field itself produces the magnetic field. The term “current” is misleading; it is better understood as $\varepsilon_0 \frac{d\Phi_E}{dt}$ , which has the units of current (Amperes) but involves no charge transport.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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