Any surface, the same current
Ampere's circuital law, from the chapter Sources of the Magnetic Field, links the field round a closed loop to the current through it: \(\oint\vec B\cdot d\vec l = \mu_0I_\text{enc}\). The current is counted through a surface whose edge is the loop. Think of a soap film on a wire ring. For a steady current it does not matter which film you choose. A flat film and a bulging one on the same ring are crossed by the same current. Charge cannot pile up anywhere between them.
A capacitor breaks the rule
Now charge a capacitor from a cell through a resistor. While it charges, a current \(I\) flows in its leads. Draw a loop round the left lead. Take first a flat disc bounded by the loop, which the wire pierces: the current through it is \(I\). Take next a surface bounded by the same loop that bulges round the left plate and passes between the plates. No wire crosses the bulge, and no charge crosses the gap. The current through it is zero.
So the same loop, with the same magnetic field round it, would give \(\mu_0I\) through one surface and zero through the other. The law cannot be right as it stands. Something must happen between the plates that the law has missed. There is no current there, but there is an electric field, and while the capacitor charges, that field grows.
Ampere's law, \(\oint\vec B\cdot d\vec l = \mu_0I_\text{enc}\), is applied to a loop round the lead of a charging capacitor. Why does it give two different answers?
Maxwell's extra term
In the gap of a parallel-plate capacitor the field is \(E = q/\epsilon_0A\), as the chapter Capacitance and Dielectrics found. So the charge on a plate is \(q = \epsilon_0AE\). The product \(EA\) is the electric flux \(\Phi_E\) through the gap. So \(q = \epsilon_0\Phi_E\), and the current in the lead, \(dq/dt\), equals \(\epsilon_0\,d\Phi_E/dt\). A quantity in the gap matches the current in the wire exactly.
The displacement current through a surface is \(\epsilon_0\) times the rate of change of the electric flux through it, \(I_d = \epsilon_0\,d\Phi_E/dt\). It makes a magnetic field exactly as a conduction current of the same size does.
\[ E = \dfrac{\sigma}{\epsilon_0} = \dfrac{q}{\epsilon_0A} \]
Why. the field between the plates of a parallel-plate capacitor with charge \(\pm q\) on area \(A\).
\[ \Phi_E = EA = \dfrac{q}{\epsilon_0} \]
Why. the field is uniform across the gap and at right angles to the plates.
\[ \epsilon_0\dfrac{d\Phi_E}{dt} = \dfrac{dq}{dt} \]
Why. differentiate \(q = \epsilon_0\Phi_E\) with respect to time.
\[ I_d = I \]
Read it. the left side is the displacement current in the gap; the right side is the current in the lead.
James Clerk Maxwell added this displacement current to the current in Ampere's law. The new law is the Ampere-Maxwell law. Now the flat disc counts the current \(I\) in the wire, and the bulge counts the displacement current \(I_d\) in the gap. The two are equal at every instant, so both surfaces give the same answer.
The displacement current has a direction as well as a size. It points along the change of \(\vec E\). While the capacitor charges, \(\vec E\) grows from the positive plate to the negative plate. So \(I_d\) runs across the gap in the same direction as the current arriving at the positive plate. While it discharges, \(\vec E\) shrinks, and \(I_d\) runs from the negative plate back to the positive one. Either way, the current never stops at the gap.
The displacement current is not a flow of charge. No electron crosses the gap of a capacitor. The name is old, and only the magnetic effect is real: a changing electric field makes a magnetic field, exactly as a current does.
(a) The plates of a capacitor have area \(0.1\,\text{m}^2\). The field between them grows at \(10^{12}\,\text{V/m}\) per second. Find the displacement current, and the current in the leads. (b) The voltage across a \(2\,\mu\text{F}\) capacitor rises at \(10^6\,\text{V/s}\). Find the displacement current between its plates.
(a) \(\Phi_E = EA\), so \(I_d = \epsilon_0A\,dE/dt\) \(= 8.85\times10^{-12}\times0.1\times10^{12} = 0.885\,\text{A}\). The current in the leads is also \(0.885\,\text{A}\): the lead current \(dq/dt\) equals \(I_d\) exactly.
(b) \(q = CV\), so \(I_d = dq/dt = C\,dV/dt\) \(= 2\times10^{-6}\times10^6 = 2\,\text{A}\).
Values used: \(\epsilon_0 = 8.85\times10^{-12}\,\text{C}^2/(\text{N m}^2)\).
A magnetic field in the gap
Because the displacement current makes a magnetic field, a field circles the gap of a charging capacitor too. Take circular plates of radius \(R\), charged by a current \(I\). A loop of radius \(r < R\) in the gap encloses only the flux through its own area, a fraction \(r^2/R^2\) of the whole. Ampere-Maxwell then gives \(B\cdot2\pi r = \mu_0Ir^2/R^2\).
\[ \begin{gathered} B = \dfrac{\mu_0Ir}{2\pi R^2}\quad(r \le R) \\ B = \dfrac{\mu_0I}{2\pi r}\quad(r \ge R) \end{gathered} \]
Inside the gap the field grows in proportion to \(r\), like the field inside a solid wire. Beyond the edge it falls as \(1/r\), exactly like the field of the lead. For plates of radius \(10\,\text{cm}\) charged by \(2\,\text{A}\), the field at the edge is \(\mu_0I/2\pi R = 4\,\mu\text{T}\).
The voltage across a \(5\,\text{nF}\) capacitor rises at \(2\times10^8\,\text{V/s}\). What is the displacement current between its plates, in amperes?
A charged capacitor discharges through a resistor. Its upper plate is positive. Which way does the displacement current run across the gap?
Circular plates of radius \(10\,\text{cm}\) are charged by a current of \(2\,\text{A}\). What is the magnetic field between the plates, \(5\,\text{cm}\) from their axis, in microtesla?
Four laws for every field
With the displacement current in place, four laws describe every electric and magnetic field. Together they are Maxwell's equations. Each says in one sentence what makes a field, or what a field cannot do.
Gauss's law: the electric flux out of a closed surface is the charge inside divided by \(\epsilon_0\). Gauss's law for magnetism: the magnetic flux out of any closed surface is zero, since no magnetic pole exists alone. Faraday's law: a changing magnetic flux makes an electric field round it. The Ampere-Maxwell law: currents and a changing electric flux make a magnetic field round them.
In integral form the four laws read as follows. They go beyond the syllabus, and this book uses them only in words. \[ \begin{gathered} \oint\vec E\cdot d\vec A = \dfrac{q_\text{enc}}{\epsilon_0} \\ \oint\vec B\cdot d\vec A = 0 \\ \oint\vec E\cdot d\vec l = -\dfrac{d\Phi_B}{dt} \\ \oint\vec B\cdot d\vec l = \mu_0I + \mu_0\epsilon_0\dfrac{d\Phi_E}{dt} \end{gathered} \] The first two are Gauss's laws for \(\vec E\) and \(\vec B\), the third is Faraday's law, and the fourth is the Ampere-Maxwell law.
Look at the last two laws side by side. A changing magnetic field makes an electric field. A changing electric field makes a magnetic field. Each change can keep the other going, with no charge or current nearby. Maxwell saw that the two together could travel through empty space as a wave. The next lesson follows that wave.
Gauss's law for magnetism says that the magnetic flux out of every closed surface is zero. What does this tell us?
James Clerk Maxwell was a Scottish physicist who worked at King's College London in the early 1860s. He gathered the known laws of electricity and magnetism and found that Ampere's law failed wherever charge built up. He added the displacement current to mend it. In 1865 he published the full theory. His equations allowed waves of electric and magnetic field, and their speed came out as \(1/\sqrt{\mu_0\epsilon_0}\). From electrical measurements alone, that speed was close to the measured speed of light. Maxwell concluded that light itself is an electromagnetic wave.
- Thinking the displacement current is charge crossing the gap.No charge crosses the gap; a changing electric field makes a magnetic field as a current would.
- Using the field instead of its rate of change.\(I_d = \epsilon_0\,d\Phi_E/dt\); a large but steady field gives no displacement current.
- Taking the whole displacement current for a small loop inside the gap.A loop of radius \(r < R\) encloses only the fraction \(r^2/R^2\) of it.