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The flaw in Ampère's law, Maxwell's displacement current $\varepsilon_0\,d\Phi_E/dt$, the magnetic field of a charging capacitor, and the four equations together.
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By the end of this lesson you will be able to compute displacement currents and the fields they make, and state what each of Maxwell's equations says.
You now have Gauss's law, Gauss's law for magnetism, Faraday's law and Ampère's law, each studied on its own. This lesson finds the one flaw in the set, fixes it as Maxwell did in 1865, and assembles the four into the complete theory of electricity and magnetism, ready for the prediction of light in the next lesson.
| Term | What it means |
|---|---|
| Displacement current | $i_d = \varepsilon_0\,d\Phi_E/dt$, the term Maxwell added to Ampère's law. |
| Electric flux | $\Phi_E = \int \vec{E}\cdot d\vec{A}$, through a surface. |
| Ampère–Maxwell law | $\oint \vec{B}\cdot d\vec{l} = \mu_0(I + \varepsilon_0\,d\Phi_E/dt)$. |
| Maxwell's equations | The four laws that together describe all electromagnetic fields. |
| Conduction current | Current carried by moving charges in a conductor. |
| Continuity of current | Conduction plus displacement current is the same through every cross section. |
| Field mill | An instrument that measures the atmospheric electric field. |
Ampère's law says the circulation of $\vec{B}$ around a loop equals $\mu_0$ times the current through any surface bounded by the loop. For a charging capacitor that gives two answers: a flat surface pierced by the wire encloses the current $I$, while a bulging surface passing between the plates encloses none. Maxwell resolved this by adding a term:
$$\oint \vec{B}\cdot d\vec{l} = \mu_0I + \mu_0\varepsilon_0\frac{d\Phi_E}{dt}.$$
The displacement current $\varepsilon_0\,d\Phi_E/dt$ is not moving charge, but a changing electric field. Between the plates it equals the current in the wires exactly, so both surfaces now give the same answer. With it, Maxwell's equations are complete: charges make $\vec{E}$, there are no magnetic charges, changing $\vec{B}$ makes $\vec{E}$, and currents and changing $\vec{E}$ make $\vec{B}$.
Another way: picture
Picture current flowing down a wire into a capacitor plate. At the plate the charges stop, but the field between the plates grows, and it grows at exactly the rate that keeps the story going. The magnetic field circling the wire does not stop at the gap; it keeps circling the gap, sustained by the changing electric field inside. Current and changing field hand off to each other seamlessly.
Another way: steps
Ampère's law was derived for steady currents, which flow in closed loops. A charging capacitor is not steady: charge piles up on the plates. Take a loop around the wire leading to one plate. A flat disk bounded by the loop is pierced by the wire and encloses $I$. A bag-shaped surface bounded by the same loop, passing between the plates, is pierced by nothing.
Ampère's law cannot give two different answers for the same loop, so something is missing. The missing piece must be present between the plates, and the only thing there is a growing electric field. Maxwell's insight was to count it.
Between the plates, $E = Q/\varepsilon_0A$ and the flux through the gap is $\Phi_E = Q/\varepsilon_0$. So $\varepsilon_0\,d\Phi_E/dt = dQ/dt = I$. The displacement current in the gap equals the conduction current in the wires, exactly, whatever the plates' size or spacing. Adding it to Ampère's law makes the flat and bag-shaped surfaces agree.
The name is historical: Maxwell pictured a medium whose parts were displaced. No charge moves across the gap. But the magnetic field it produces is entirely real: between the plates of a charging capacitor, a sensitive magnetometer finds a field circling the axis, growing from zero at the center to $\mu_0I/2\pi R$ at the edge.
Gauss's law, $\oint \vec{E}\cdot d\vec{A} = Q_{\text{enc}}/\varepsilon_0$, says charges are sources of electric field. Gauss's law for magnetism, $\oint \vec{B}\cdot d\vec{A} = 0$, says there are no magnetic sources. Faraday's law, $\oint \vec{E}\cdot d\vec{l} = -d\Phi_B/dt$, says a changing magnetic field circulates an electric field. The Ampère–Maxwell law says currents and changing electric fields circulate a magnetic field.
With the Lorentz force $q(\vec{E} + \vec{v} \times \vec{B})$ saying how fields push charges, these equations describe every electric and magnetic phenomenon outside the quantum world: circuits, motors, radio, light, and the chemistry of materials at large scale.
Before Maxwell, a changing magnetic field made an electric field, but not the reverse. The displacement current gives the reverse: a changing electric field makes a magnetic field. The equations become nearly symmetric between $\vec{E}$ and $\vec{B}$, differing only because there are electric charges but no magnetic ones.
That symmetry has a dramatic consequence. In empty space, with no charges or currents, a changing $\vec{E}$ makes $\vec{B}$, and a changing $\vec{B}$ makes $\vec{E}$. The two can sustain each other and travel together, a self-propagating wave. The next lesson finds its speed, $1/\sqrt{\mu_0\varepsilon_0}$, which is the speed of light.
Checking an answer. Conduction plus displacement current through any closed surface must be zero: what flows in as current must appear as growing field. The magnetic field just outside a capacitor's gap must match that around the wire at the same distance. Units: $\varepsilon_0$ times V·m per second is A.
The displacement current is required by charge conservation. Gauss's law says the flux of $\vec{E}$ out of a closed surface is $Q/\varepsilon_0$; its rate of change is $(dQ/dt)/\varepsilon_0$, and $dQ/dt$ is the net current flowing in. So $I_{\text{in}} = \varepsilon_0\,d\Phi_E/dt$ through any closed surface. Ampère's law without the new term would contradict this whenever charge accumulates.
Maxwell's term was a theoretical necessity before it was an experimental fact. Heinrich Hertz's detection of radio waves in 1887, which the term predicted, confirmed it, and every radio since has depended on it.
In most circuits the displacement current is invisible except inside capacitors. Its size is $\varepsilon_0$ times the rate of change of electric flux, and $\varepsilon_0$ is tiny. A field changing by $1$ MV/m in a second through a square meter makes only $9$ μA of displacement current.
It becomes important where fields change fast: at radio frequencies, where a field reverses millions or billions of times a second, and in light, where it reverses hundreds of trillions of times. That is why electromagnetic waves were not noticed until Maxwell's theory predicted them. In a circuit at the $60$ Hz of household power, the term is too small to measure outside a capacitor.
James Clerk Maxwell published his equations in 1865, when the telegraph was the high technology of the day. Within a century they had given the world radio, television, radar and microwave ovens. Einstein kept a picture of Maxwell in his study, and special relativity grew directly from the question of how Maxwell's equations look to moving observers.
Maxwell's equations are also the model for later theories. The weak and strong nuclear forces are described by field equations built on the same pattern, and the search for a theory that includes gravity starts from the same ideas. In the quantum version, quantum electrodynamics, developed in part by Richard Feynman at Cornell and Caltech, Maxwell's fields become photons, and its predictions agree with experiment to better than a part in a billion.
Each of the four equations has been a working tool in an earlier lesson, and it helps to see them as a set of questions to ask about any situation. Is there charge? Gauss's law, with a symmetric surface, gives the electric field it makes. Is there current, or an electric field that changes in time? The Ampère–Maxwell law, with a symmetric loop, gives the magnetic field. Is a magnetic flux changing? Faraday's law gives the circulating electric field and the emf in any circuit. And the magnetic field must always close on itself, which fixes its shape wherever a symmetry argument alone leaves it uncertain.
The equations are linked in pairs. Gauss's law and the Ampère–Maxwell law together contain charge conservation, as this lesson showed. Faraday's law and Gauss's law for magnetism are consistent for the same reason in reverse: the flux of $\vec{B}$ through a closed surface is always zero, so its rate of change is zero too, and no closed surface can have a net circulating electric field around it.
Most problems use one equation at a time, as the lessons of this unit did. The next lesson uses two together, Faraday's law and the Ampère–Maxwell law in empty space, and finds that each feeds the other. That coupling, invisible in any single equation, is what turns electricity and magnetism into light. The final lesson of the course then asks you to choose among all four, the way real problems require.
Thunderstorms around the world pump negative charge to the ground, and fair-weather regions let it leak back up through the slightly conducting air, forming a global circuit. Field mills at Kennedy Space Center in Florida, the lightning capital of the United States, watch the field above the launch pads, and launches wait when it is too strong.
When lightning strikes, the field changes by kilovolts per meter in a fraction of a second, and the displacement current in the air, $\varepsilon_0\,dE/dt$, carries the circuit's current through regions where no charge moves. Atmospheric scientists need Maxwell's term to balance the current budget of a storm.
Every radio transmitter is an antenna in which charge sloshes back and forth millions of times a second, so the electric field around it changes very fast. The displacement current of that changing field creates a changing magnetic field, which by Faraday's law creates a changing electric field farther out, and so on.
Without Maxwell's term, this chain would stop at the antenna, and no signal could travel through empty space. Guglielmo Marconi's transatlantic radio in 1901, the broadcasts from KDKA in Pittsburgh in 1920, and every phone call and Wi-Fi packet since are consequences of the displacement current.
Since no charge crosses the gap, it seems there can be no magnetic field between the plates. But the changing electric field there acts as a current in the Ampère–Maxwell law, and the magnetic field circles the gap just as it circles the wire. Outside the plates, at the same distance from the axis, the field is exactly the same as around the wire.
A related error is to think displacement current is a flow of something. It is a changing field, not moving charge, but its magnetic effect is identical to a real current's.
A capacitor with circular plates of radius $5.0$ cm is charged by a steady $2.0$ A. Find the displacement current in the gap.
$i_d = 2.0\ \text{A}$
It equals the conduction current.
Find the magnetic field at the plates' edge.
$B = \dfrac{\mu_0i_d}{2\pi R} = \dfrac{2 \times 10^{-7} \times 2.0}{0.050} = 8.0\ \mu\text{T}$
Ampère–Maxwell with a circle at the edge.
Find the field at $2.5$ cm from the axis.
$B = 8.0 \times \dfrac{2.5}{5.0} = 4.0\ \mu\text{T}$
Linear inside.
Find the field at $10$ cm, outside the plates.
$B = \dfrac{2 \times 10^{-7} \times 2.0}{0.10} = 4.0\ \mu\text{T}$
All the displacement current is enclosed.
Compare with the wire.
$\text{the same } 4.0\ \mu\text{T at } 10\ \text{cm}$
The field does not notice the gap.
Plates of area $0.010$ m² hold a field rising at $4.0 \times 10^{12}$ V/m per second. Find the rate of change of flux.
$\dfrac{d\Phi_E}{dt} = 0.010 \times 4.0 \times 10^{12} = 4.0 \times 10^{10}\ \text{V·m/s}$
$A\,dE/dt$.
Find the displacement current.
$i_d = 8.85 \times 10^{-12} \times 4.0 \times 10^{10} = 0.354\ \text{A}$
$\varepsilon_0\,d\Phi_E/dt$.
Find the charging current in the wires.
$I = 0.354\ \text{A}$
Equal.
Find the plate spacing if $C = 44$ pF.
$d = \dfrac{\varepsilon_0A}{C} = \dfrac{8.85 \times 10^{-14}}{44 \times 10^{-12}} = 2.0\ \text{mm}$
$C = \varepsilon_0A/d$.
Find how fast the voltage rises.
$\dfrac{dV}{dt} = d\dfrac{dE}{dt} = 0.0020 \times 4.0 \times 10^{12} = 8.0 \times 10^9\ \text{V/s}$
$V = Ed$.
Check with $I/C$.
$\dfrac{0.354}{44 \times 10^{-12}} = 8.0 \times 10^9\ \text{V/s}$
Consistent.
A current of $3.0$ A flows into a small metal sphere through a thin wire. Find how fast its charge grows.
$\dfrac{dQ}{dt} = 3.0\ \text{C/s}$
No current leaves.
Find the rate of change of electric flux through a sphere around it.
$\dfrac{d\Phi_E}{dt} = \dfrac{1}{\varepsilon_0}\dfrac{dQ}{dt} = 3.4 \times 10^{11}\ \text{V·m/s}$
Gauss's law.
Find the outward displacement current.
$i_d = \varepsilon_0\dfrac{d\Phi_E}{dt} = 3.0\ \text{A}$
It carries the current onward.
Find the net current through the closed surface.
$-3.0 + 3.0 = 0$
Conduction in, displacement out.
State what this shows.
$\text{total current is conserved}$
The displacement current enforces charge conservation.
Find the field $10$ cm from the sphere after $1.0$ s.
$E = \dfrac{9.0 \times 10^9 \times 3.0}{0.010} = 2.7 \times 10^{12}\ \text{N/C}$
Far past breakdown: the charge would spark away long before.
Explain why real spheres cannot keep charging.
$\text{air breaks down near } 3 \times 10^6\ \text{N/C}$
A few microcoulombs at most on a small sphere.
State the displacement current.
$i_d = 1.0\ \text{A}$
Equal to the wire's current.
Apply the Ampère–Maxwell law.
$B = \dfrac{2 \times 10^{-7} \times 1.0}{0.040}$
A circle at the edge.
Evaluate the field.
A parallel-plate capacitor is being charged by a steady current of $4$ A in its wires. What is the displacement current in the gap between its plates?
Complete the worked solution: a $5$ μF parallel-plate capacitor with plates $2.0$ mm apart is charged by a steady $4$ mA current. Find how fast its voltage rises in kV/s, how fast the field between the plates rises in MV/m per second, and the charge on it after $4.0$ s in mC.
Divide the current by the capacitance.
$\dfrac{dV}{dt} = \dfrac{I}{C} =$ v
mA per μF is kV/s.
Divide by the plate spacing.
$\dfrac{dE}{dt} = \dfrac{1}{d}\dfrac{dV}{dt} =$ e
$E = V/d$ between the plates.
Multiply the current by the time.
$Q = It =$ q
A steady current.
Check the displacement current.
$\varepsilon_0A\dfrac{dE}{dt} = C\dfrac{dV}{dt} = I$
Since $C = \varepsilon_0A/d$.
Match each of Maxwell's equations to what it says.
| charges make electric fields | no magnetic monopoles | changing magnetic flux makes electric fields | currents and changing electric flux make magnetic fields | |
|---|---|---|---|---|
| Gauss's law | ||||
| Gauss's law for magnetism | ||||
| Faraday's law | ||||
| Ampère–Maxwell law |
A capacitor with circular plates of radius $2$ cm is charged by a steady $7$ A current. With $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, fill in the displacement current through the whole gap in A, the magnetic field at the plates' edge in μT, and the field halfway from the axis to the edge in μT.
| value | |
|---|---|
| displacement current (A) | |
| field at the edge (μT) | |
| field at half the radius (μT) |
A capacitor with circular plates of radius $2$ cm is charged by a steady $7$ A current. With $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, write the magnetic field between the plates, in μT, as a formula in the distance $r$ (cm) from the axis, for $r$ less than the radius.
Answer:
Between two plates of area $0.04$ m², the electric field is rising at $9 \times 10^{12}$ V/m per second. With $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m, what is the displacement current between them, in A?
Answer: A of displacement current
Field mills at Kennedy Space Center in Florida record the electric field at the ground under a thunderstorm. After a lightning flash, the field changes by $6$ kV/m over $0.5$ s. With $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m, what is the average displacement current density in the air, in nA/m²?
Answer: nA/m² in the air
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A capacitor with circular plates of radius $2$ cm is charged by a steady $2$ A current. With $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, write the magnetic field between the plates, in μT, as a formula in the distance $r$ (cm) from the axis, for $r$ less than the radius.
Answer:
You can use the Ampère–Maxwell law. Explain to someone why a charging capacitor has a magnetic field between its plates.
20. Your turn: a capacitor with plate radius $4.0$ cm is charged by $1.0$ A. Find the magnetic field at the plates' edge., step 3
$B = 5.0\ \mu\text{T}$
Circling the axis.