Electromagnetic Waves • Topic 1 of 3

Displacement Current & Maxwell's Equations

An electromagnetic wave is a self-sustaining ripple of electric and magnetic fields that travels through empty space at the speed of light. Before James Clerk Maxwell, the laws of electricity and magnetism were a collection of separate experimental rules. Maxwell's genius was to spot a missing piece, patch it, and in doing so reveal that light itself is an electromagnetic wave. The story begins with a subtle flaw in Ampère's law.

The problem with Ampère's circuital law. The original Ampère's law says that the line integral of the magnetic field around a closed loop equals $\mu_0$ times the current passing through any surface bounded by that loop: $\oint \vec{B}\cdot d\vec{l}=\mu_0 I$. Maxwell asked a sharp question using a charging capacitor. Imagine a wire carrying current $I$ into a parallel-plate capacitor. Choose a loop around the wire. If you stretch a flat surface across the loop, the conduction current $I$ passes through it. But if you balloon the surface so it dips between the plates, no charge actually crosses the gap — so the enclosed current is zero. The same loop gives two different answers. Something is missing.

Maxwell's fix: displacement current. Between the plates there is no conduction current, but there is a changing electric field as the capacitor charges. Maxwell proposed that a changing electric flux acts like a current and produces a magnetic field just as a real current does. He called it the displacement current:

  • $I_d=\epsilon_0\frac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux through the surface.
  • It carries no moving charge — it is the changing electric field itself acting as a source of magnetic field.
  • In the gap, $I_d$ exactly equals the conduction current $I$ in the wire, so the law gives one consistent answer for any surface.

The Ampère–Maxwell law. Combining the two, the corrected law is $\oint \vec{B}\cdot d\vec{l}=\mu_0(I+I_d)=\mu_0 I+\mu_0\epsilon_0\frac{d\Phi_E}{dt}$. The crucial new term tells us that a changing electric field produces a magnetic field — the partner of Faraday's discovery that a changing magnetic field produces an electric field.

Maxwell's four equations (conceptual form). The whole of classical electromagnetism is captured in four statements:

  • Gauss's law for electricity: $\oint \vec{E}\cdot d\vec{A}=\frac{q}{\epsilon_0}$ — electric field lines begin and end on charges.
  • Gauss's law for magnetism: $\oint \vec{B}\cdot d\vec{A}=0$ — magnetic field lines form closed loops; there are no isolated magnetic poles (monopoles).
  • Faraday's law: $\oint \vec{E}\cdot d\vec{l}=-\frac{d\Phi_B}{dt}$ — a changing magnetic field creates an electric field.
  • Ampère–Maxwell law: $\oint \vec{B}\cdot d\vec{l}=\mu_0 I+\mu_0\epsilon_0\frac{d\Phi_E}{dt}$ — currents and changing electric fields create a magnetic field.

How a wave sustains itself. Faraday's law and the Ampère–Maxwell law together create a feedback loop in empty space. A changing $\vec{E}$ generates a changing $\vec{B}$, which in turn generates a changing $\vec{E}$, and so on — the two fields keep regenerating each other and the disturbance marches forward as a wave. The source of an electromagnetic wave is an accelerating (or oscillating) charge: a charge moving at constant velocity gives steady fields, but an accelerating charge radiates energy as EM waves. The frequency of the wave equals the frequency of oscillation of the charge.

Displacement current between capacitor platesI (conduction)E (changing)II_d through the gapsame magnetic effect as a real current
1
Worked Example
A parallel-plate capacitor is being charged so that the electric flux between its plates changes at a rate of $5\times10^{8}\ \text{V·m/s}$. Find the displacement current. ($\epsilon_0=8.85\times10^{-12}\ \text{F/m}$.)
Solution
  1. Step 1: Use $I_d=\epsilon_0\frac{d\Phi_E}{dt}$.
  2. Step 2: Substitute: $I_d=8.85\times10^{-12}\times5\times10^{8}$.
  3. Step 3: Compute: $I_d=4.43\times10^{-3}\ \text{A}$.

Answer: $I_d\approx4.43\ \text{mA}$.

2
Worked Example
The conduction current charging a capacitor is 2.0 A. What is the displacement current between its plates?
Solution
  1. Step 1: For a charging capacitor, the displacement current in the gap equals the conduction current in the wire.
  2. Step 2: This is required so that the Ampère–Maxwell law gives the same magnetic field for any surface bounded by the loop.
  3. Step 3: Therefore $I_d=I=2.0\ \text{A}$.

Answer: $I_d=2.0\ \text{A}$.

3
Worked Example
A parallel-plate capacitor has plate area $0.02\ \text{m}^2$. The electric field between the plates changes at $1\times10^{12}\ \text{V/(m·s)}$. Find the displacement current. ($\epsilon_0=8.85\times10^{-12}\ \text{F/m}$.)
Solution
  1. Step 1: $\Phi_E=EA$, so $\frac{d\Phi_E}{dt}=A\frac{dE}{dt}$.
  2. Step 2: $I_d=\epsilon_0 A\frac{dE}{dt}=8.85\times10^{-12}\times0.02\times1\times10^{12}$.
  3. Step 3: Compute: $I_d=8.85\times10^{-12}\times2\times10^{10}=0.177\ \text{A}$.

Answer: $I_d\approx0.18\ \text{A}$.

4
Worked Example
State which Maxwell equation expresses the fact that isolated magnetic monopoles do not exist, and write it.
Solution
  1. Step 1: Magnetic field lines always form closed loops, so the net magnetic flux through any closed surface is zero.
  2. Step 2: This is Gauss's law for magnetism.
  3. Step 3: It is written $\oint \vec{B}\cdot d\vec{A}=0$.

Answer: Gauss's law for magnetism, $\oint \vec{B}\cdot d\vec{A}=0$.

5
Worked Example
A capacitor of capacitance $100\ \mu\text{F}$ is charged so that the voltage across it rises at $2000\ \text{V/s}$. Find the displacement current.
Solution
  1. Step 1: Charge $q=CV$, so the conduction current $I=\frac{dq}{dt}=C\frac{dV}{dt}$.
  2. Step 2: The displacement current equals this conduction current: $I_d=C\frac{dV}{dt}$.
  3. Step 3: $I_d=100\times10^{-6}\times2000=0.2\ \text{A}$.

Answer: $I_d=0.2\ \text{A}$.

6
Worked Example
Why does a steady (constant) current in a wire not produce an electromagnetic wave, while an oscillating current does?
Solution
  1. Step 1: EM waves are radiated only by accelerating charges; a steady current means charges drift at constant velocity, so there is no acceleration.
  2. Step 2: A steady current gives a static magnetic field and no changing electric field, so no self-sustaining $E$–$B$ feedback is set up.
  3. Step 3: An oscillating current means the charges accelerate back and forth, producing changing $\vec{E}$ and $\vec{B}$ fields that regenerate each other and propagate.

Answer: Only accelerating (oscillating) charges radiate; a steady current has no acceleration and no changing fields, so it produces no EM wave.

Key Points

  • Maxwell found Ampère's law incomplete for a charging capacitor and added the displacement current $I_d=\epsilon_0\frac{d\Phi_E}{dt}$.
  • The Ampère–Maxwell law $\oint \vec{B}\cdot d\vec{l}=\mu_0 I+\mu_0\epsilon_0\frac{d\Phi_E}{dt}$ shows a changing electric field produces a magnetic field.
  • The four Maxwell equations: Gauss for $E$ ($q/\epsilon_0$), Gauss for $B$ (=0, no monopoles), Faraday (changing $B\Rightarrow E$), Ampère–Maxwell (current & changing $E\Rightarrow B$).
  • A changing $E$ creates $B$ and a changing $B$ creates $E$; this mutual regeneration lets EM waves travel through vacuum.
  • The source of an electromagnetic wave is an accelerating or oscillating charge; the wave frequency equals the oscillation frequency.
Tap an option to check your answer0 / 4
Q1.The displacement current is given by:
Explanation: Maxwell defined the displacement current as $I_d=\epsilon_0\frac{d\Phi_E}{dt}$, arising from a changing electric flux.
Q2.Maxwell's correction to Ampère's law showed that a magnetic field can be produced by:
Explanation: The added term $\mu_0\epsilon_0\frac{d\Phi_E}{dt}$ means a changing electric field (like in a charging capacitor) produces a magnetic field.
Q3.Which Maxwell equation states that magnetic monopoles do not exist?
Explanation: Gauss's law for magnetism, $\oint \vec{B}\cdot d\vec{A}=0$, says net magnetic flux is always zero — field lines are closed loops, so no isolated poles exist.
Q4.Electromagnetic waves are produced by:
Explanation: Only an accelerating (or oscillating) charge radiates electromagnetic waves; charges at rest or in uniform motion do not.