5.2.1 Why the term must be there: one loop, two surfaces

To see that the static Ampère law cannot be right for unsteady currents, consider the wire feeding a capacitor and draw a closed curve $C$ around the wire (Figure 5.2.2). The integral form of Ampère's law, obtained from $\nabla \times H = J$ by Stokes's theorem (Section 2.4),

\oint_{C} H \cdot d l = \int_{S} J \cdot d S,

holds for any open surface $S$ whose rim is $C$ : Stokes's theorem does not care which surface we stretch over the curve. Take two of them.

The flat disc $S_{1}$ is cut by the wire. The current $i (t)$ passes through it, so $\int_{S_{1}} J \cdot d S = i (t)$ .
The bag $S_{2}$ has the same rim $C$ but bulges to the right and passes between the plates of the capacitor. No wire cuts it, and no charge crosses the gap, so $\int_{S_{2}} J \cdot d S = 0$ .

One left-hand side, $\oint_{C} H \cdot d l$ , and two different right-hand sides, $i (t)$ and $0$ . Ampère's law is inconsistent as soon as the current is unsteady. For a steady current the contradiction never arises, because a steady current cannot stop at a plate: the charge on the plate would grow without limit. The capacitor can only be charging, $q (t)$ growing, and this is precisely the situation in which $D = ε E$ between the plates is changing.

One loop, two surfaces. The curve C encircles the wire that feeds a capacitor. The flat disc S_1 is cut by the wire and carries the conduction current i(t). The bag S_2 has the same rim but passes between the plates, where \vec J=0. The static Ampère law gives two different values for the same circulation \oint_C\vec H\cdot d\vec l. The displacement current \int(\partial\vec D/\partial t)\cdot d\vec S through S_2 is exactly i(t) and removes the contradiction. — **Figure 5.2.2.** One loop, two surfaces. The curve $C$ encircles the wire that feeds a capacitor. The flat disc $S_{1}$ is cut by the wire and carries the conduction current $i (t)$ . The bag $S_{2}$ has the same rim but passes between the plates, where $J = 0$ . The static Ampère law gives two different values for the same circulation $\oint_{C} H \cdot d l$ . The displacement current $\int (\partial D / \partial t) \cdot d S$ through $S_{2}$ is exactly $i (t)$ and removes the contradiction.

Now let us see what the changing flux amounts to. Apply Gauss's law to a closed surface drawn around the left-hand plate: it encloses the charge $q (t)$ , so the flux of $D$ out of it is $q$ . With fringing neglected, all of this flux leaves through the gap, where it crosses the bag $S_{2}$ . Hence $\int_{S_{2}} D \cdot d S = q (t)$ and

\int_{S_{2}} \frac{\partial D}{\partial t} \cdot d S = \frac{d q}{d t} = i (t) . (5.2.3)

The rate of change of the electric flux through $S_{2}$ is exactly the current that the wire delivers to the plate. If, therefore, we take $J + \partial D / \partial t$ instead of $J$ as the source of $\nabla \times H$ , both surfaces give $i (t)$ : through $S_{1}$ it is the conduction current, through $S_{2}$ it is the displacement current, and the law is consistent again.

Switch between the flat disc and the bag, and remove the Maxwell term: only with $\partial D / \partial t$ do both surfaces give $i (t)$ .

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There is a general reason behind this. The divergence of a curl is identically zero, so whatever stands on the right-hand side of a curl equation must be divergence-free. For the static law this requires $\nabla \cdot J = 0$ , which is true only for steady currents. For the corrected source, using Gauss's law and the continuity equation of Section 5.4,

\nabla \cdot (J + \frac{\partial D}{\partial t}) = \nabla \cdot J + \frac{\partial}{\partial t} (\nabla \cdot D) = \nabla \cdot J + \frac{\partial ρ _{v}}{\partial t} = 0 (5.2.4)

identically, whether the currents are steady or not. A divergence-free field has the same flux through every surface spanning a given curve (its field lines neither begin nor end, so the lines crossing $S_{1}$ are the lines crossing $S_{2}$ ), and the circulation of $H$ around $C$ is then well defined.

The generalised Ampère law

\oint_{C} H \cdot d l \nabla \times H = \int_{S} (J + \frac{\partial D}{\partial t}) \cdot d S, = J + \frac{\partial D}{\partial t} . (5.2.5)

The right-hand side is the same for every surface $S$ spanning $C$ , as a curl law requires. The displacement current is the one term of the four equations that came from consistency rather than from an experiment.

In plain words

Where a conduction current stops, a changing electric flux takes over, and the magnetic field cannot tell the difference. The total current, conduction plus displacement, never stops anywhere: it flows in closed loops, through the capacitor gap as well as through the wire.

Common mistake · Using the static Ampère law with an unsteady current

Writing $\oint_{C} H \cdot d l = i_{enc}$ for a loop near a charging capacitor and then choosing the surface through the gap, which gives $\oint_{C} H \cdot d l = 0$ and no magnetic field around a wire that plainly carries a current. The static law is only consistent when $\nabla \cdot J = 0$ . As soon as charge piles up anywhere, the displacement current must be included on every surface that passes through a region where $D$ changes.