Ampère's law says $\oint\vec{B}\cdot d\vec{\ell} = \mu_0I_{\text{enc}}$, where $I_{\text{enc}}$ is the current through any surface bounded by the path. Try it on a wire feeding a charging capacitor, and it breaks. Fixing it was Maxwell's great step, and it led straight to light.
Why it works
Draw a circle around the wire near the capacitor. A flat disk bounded by that circle is pierced by the wire: $I_{\text{enc}} = I$. Now stretch the surface into a bag that passes between the plates, still bounded by the same circle. No charge crosses the gap, so $I_{\text{enc}} = 0$. Same path, two different answers: the law as written cannot be right. What does pass through the bag is a growing electric field. Between the plates $E = \dfrac{q}{\varepsilon_0A}$, so $\varepsilon_0\dfrac{d\Phi_E}{dt} = \varepsilon_0A\dfrac{dE}{dt} = \dfrac{dq}{dt} = I$: exactly the missing current.
Key idea
The displacement current $I_d = \varepsilon_0\dfrac{d\Phi_E}{dt}$ makes magnetic field just as a real current does:
$$\oint\vec{B}\cdot d\vec{\ell} = \mu_0\left(I_{\text{enc}} + \varepsilon_0\frac{d\Phi_E}{dt}\right).$$
In a charging capacitor, $I_d$ in the gap equals the current in the wires.
Worked examples
Worked example: How fast the field grows
A capacitor with circular plates of radius $10\ \text{cm}$ is charged by a current of $2.0\ \text{A}$. How fast does the field between the plates grow?
- $\varepsilon_0A\dfrac{dE}{dt} = I$, so $\dfrac{dE}{dt} = \dfrac{2.0}{(8.85 \times 10^{-12})\pi(0.10)^2} = 7.19 \times 10^{12}\ \text{V/(m·s)}$.
Answer: $\dfrac{dE}{dt} \approx 7.19 \times 10^{12}\ \text{V/(m·s)}$
Worked example: Magnetic field in the gap
For the same capacitor, find $B$ between the plates at $5.0\ \text{cm}$ from the axis, and at the plates' edge.
- The displacement current is spread evenly over the plates, so a circle of radius $r$ encloses $I\dfrac{r^2}{R^2}$.
- $B(2\pi r) = \mu_0I\dfrac{r^2}{R^2}$, so $B = \dfrac{\mu_0Ir}{2\pi R^2} = \dfrac{(4\pi \times 10^{-7})(2.0)(0.050)}{2\pi(0.10)^2} = 2.00\ \text{μT}$.
- At the edge, $B = \dfrac{\mu_0I}{2\pi R} = 4.00\ \text{μT}$, the same as next to the wire at that distance.
Answer: $2.00\ \text{μT}$ and $4.00\ \text{μT}$
Worked example: In terms of voltage
Show that the displacement current in a parallel-plate capacitor is $C\,dV/dt$.
- $E = V/d$, so $I_d = \varepsilon_0A\dfrac{d}{dt}\left(\dfrac{V}{d}\right) = \dfrac{\varepsilon_0A}{d}\dfrac{dV}{dt} = C\dfrac{dV}{dt}$.
- That is also $dq/dt$, the current in the wires.
Answer: $I_d = C\,dV/dt$
Common mistake
"Charge flows across the gap of a charging capacitor." No charge crosses an ideal capacitor's gap: charge piles up on one plate and drains from the other. What fills the gap is a changing electric field, and that change acts like a current in producing a magnetic field. Check: if charge crossed, the capacitor would not charge at all.
Maxwell added this term in the 1860s, partly on grounds of consistency. It meant that a changing electric field makes a magnetic field and a changing magnetic field makes an electric field, so each can keep the other going through empty space: an electromagnetic wave. Heinrich Hertz made and detected such waves in the late 1880s.