Maxwell's Equations - Complete Interactive Lesson

Part 1: Maxwell's Equations Overview

🌐 Maxwell's Equations

Part 1 of 7 — The Four Laws of Electromagnetism

The Four Equations

#	Name	Integral Form
1	Gauss's Law (E)	$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$
2	Gauss's Law (B)	$\oint \vec{B} \cdot d\vec{A} = 0$
3	Faraday's Law	$\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$
4	Ampere-Maxwell	$\oint \vec{B} \cdot d\vec{l} = \mu_0\left(I + \varepsilon_0 \frac{d\Phi_E}{dt}\right)$

Physical Meaning

Equation	Says...
Gauss (E)	Electric charges create electric fields
Gauss (B)	No magnetic monopoles
Faraday	Changing B creates E
Ampere-Maxwell	Currents AND changing E create B

🔑 Maxwell's equations unify electricity and magnetism and predict electromagnetic waves.

📝 Worked Example — Gauss's Law with Calculus

A point charge $Q = +4.0 \text{ nC}$ sits at the center of a sphere of radius $r = 0.20 \text{ m}$ . Find the electric flux through the sphere, then the field magnitude on its surface.

Step 1 — Apply Gauss's Law. The closed-surface integral gives the enclosed charge over $\varepsilon_0$ :

Concept Check 🎯

Part 2: Displacement Current

🔄 Displacement Current

Part 2 of 7 — Maxwell's Key Insight

The Problem with Ampere's Law

Consider a charging capacitor — current flows in the wire but not between the plates. Ampere's law gives different answers depending on which surface you choose!

Maxwell's Fix: Displacement Current

$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$

Part 3: Electromagnetic Waves

🌊 Electromagnetic Waves

Part 3 of 7 — Light as an EM Wave

EM Wave Properties

$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3.0 \times 10^8 \text{ m/s}$

Part 4: EM Spectrum

🌈 The Electromagnetic Spectrum

Part 4 of 7 — Types of EM Radiation

The Spectrum

Type	Wavelength	Frequency
Radio	> 1 m	< 300 MHz
Microwave	1 mm – 1 m	300 MHz – 300 GHz
Infrared	700 nm – 1 mm
Visible	400 – 700 nm
Ultraviolet	10 – 400 nm
X-ray	0.01 – 10 nm
Gamma	< 0.01 nm	> $10^{19}$ Hz

Part 5: Energy & Momentum of EM Waves

💡 Energy and Momentum of EM Waves

Part 5 of 7 — Poynting Vector and Radiation Pressure

Poynting Vector

$\vec{S} = \frac{1}{\mu_0}\vec{E} \times \vec{B}$

Part 6: Problem-Solving Workshop

🛠️ Maxwell Workshop

Part 6 of 7 — Practice

AP Physics C E&M: Maxwell Topics

Concept	What to Know
Identify which equation applies	Match to symmetry and context
Displacement current	$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$

Part 7: Review & Applications

📋 Maxwell's Equations Review

Part 7 of 7 — Final Summary

The Big Picture

Maxwell's four equations describe ALL of classical electromagnetism:

Gauss (E): Charges → E fields
Gauss (B): No magnetic monopoles
Faraday: Changing B → E
Ampere-Maxwell: Currents + changing E → B

Together they predict EM waves traveling at $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$ .

📝 Worked Example — Displacement Current in a Capacitor

A parallel-plate capacitor with circular plates of radius $R = 5.0 \text{ cm}$ is charged so the electric field between the plates increases at $\frac{dE}{dt} = 2.0 \times 10^{12} \text{ V/m per s}$ . Find the displacement current.

Step 1 — Write the electric flux. For a uniform field perpendicular to plates of area $A = \pi R^2$ :

$\Phi_E = E A = E\,\pi R^2$

Step 2 — Differentiate with respect to time. Since $A$ is constant, the derivative passes through:

$\frac{d\Phi_E}{dt} = \pi R^2 \frac{dE}{dt}$

Step 3 — Multiply by $\varepsilon_0$ . The displacement current is

$I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \pi R^2 \frac{dE}{dt}$

$I_d = (8.85 \times 10^{-12})\,\pi (0.050)^2 (2.0 \times 10^{12}) \approx 0.14 \text{ A}$

🔑 The displacement current between the plates exactly equals the conduction current in the wire — that is what makes Ampere-Maxwell consistent for any surface.

📝 Worked Example — Speed from $\mu_0$ and $\varepsilon_0$

Maxwell's equations predict that a plane wave $E(x,t) = E_0 \sin(kx - \omega t)$ must satisfy the wave equation $\frac{\partial^2 E}{\partial x^2} = \mu_0\varepsilon_0 \frac{\partial^2 E}{\partial t^2}$ . Show the propagation speed and evaluate it.

Step 1 — Take the spatial derivatives. Differentiating twice in $x$ pulls down $k^2$ :

$\frac{\partial^2 E}{\partial x^2} = -k^2 E_0 \sin(kx - \omega t)$

Step 2 — Take the time derivatives. Differentiating twice in $t$ pulls down $\omega^2$ :

$\frac{\partial^2 E}{\partial t^2} = -\omega^2 E_0 \sin(kx - \omega t)$

Step 3 — Match coefficients. Substituting into the wave equation gives $k^2 = \mu_0\varepsilon_0\,\omega^2$ , so the speed is

$v = \frac{\omega}{k} = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = \frac{1}{\sqrt{(4\pi \times 10^{-7})(8.85 \times 10^{-12})}} \approx 3.0 \times 10^8 \text{ m/s}$

🔑 The wave speed contains only the universal constants $\mu_0$ and $\varepsilon_0$ — so every EM wave in vacuum travels at $c$ , independent of frequency.

📝 Worked Example — Wavelength, Frequency, and Photon Energy

A green laser emits light of wavelength $\lambda = 532 \text{ nm}$ . Find its frequency and the energy of one photon.

Step 1 — Use $c = f\lambda$ . Solve for frequency:

$f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{532 \times 10^{-9}} \approx 5.6 \times 10^{14} \text{ Hz}$

Step 2 — Apply the Planck relation. Photon energy is $E = hf$ with $h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s}$ :

$E = hf = (6.63 \times 10^{-34})(5.6 \times 10^{14}) \approx 3.7 \times 10^{-19} \text{ J}$

Step 3 — Convert to electron-volts. Dividing by $1.6 \times 10^{-19} \text{ J/eV}$ :

$E \approx \frac{3.7 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.3 \text{ eV}$

🔑 Shorter wavelength means higher frequency and higher photon energy — that is why gamma rays are far more energetic than radio waves.

📝 Worked Example — Intensity and Radiation Force on a Solar Sail

Sunlight near Earth has intensity $I = 1.36 \times 10^3 \text{ W/m}^2$ . A reflective solar sail of area $A = 100 \text{ m}^2$ faces the Sun. Find the force on it and the peak electric field of the sunlight.

Step 1 — Radiation pressure on a reflector. A perfect reflector reverses the photon momentum, so

$P = \frac{2I}{c} = \frac{2(1.36 \times 10^3)}{3.0 \times 10^8} \approx 9.1 \times 10^{-6} \text{ Pa}$

Step 2 — Force from pressure. Multiply by the sail area:

$F = P A = (9.1 \times 10^{-6})(100) \approx 9.1 \times 10^{-4} \text{ N}$

Step 3 — Peak field from intensity. Invert $I = \frac{1}{2}c\varepsilon_0 E_0^2$ :

$E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} = \sqrt{\frac{2(1.36 \times 10^3)}{(3.0 \times 10^8)(8.85 \times 10^{-12})}} \approx 1.0 \times 10^3 \text{ N/C}$

🔑 Radiation force is tiny but continuous; over months it can meaningfully accelerate a low-mass spacecraft.

📝 Worked Example — Faraday's Law for an Induced Field

A uniform magnetic field fills a circular region of radius $R = 0.10 \text{ m}$ and increases at $\frac{dB}{dt} = 0.50 \text{ T/s}$ . Find the induced electric field magnitude at radius $r = 0.10 \text{ m}$ (the edge).

Step 1 — Write Faraday's Law. Around a circular loop of radius $r$ , symmetry makes $\vec{E}$ tangential and constant, so

$\oint \vec{E} \cdot d\vec{l} = E(2\pi r) = -\frac{d\Phi_B}{dt}$

Step 2 — Express the flux. For $r \le R$ the loop encloses the full field over area $\pi r^2$ :

$\Phi_B = B\,\pi r^2 \quad\Rightarrow\quad \frac{d\Phi_B}{dt} = \pi r^2 \frac{dB}{dt}$

Step 3 — Solve for $E$ . Taking magnitudes,

$E = \frac{r}{2}\frac{dB}{dt} = \frac{0.10}{2}(0.50) = 0.025 \text{ N/C}$

🔑 Inside the field region $E \propto r$ ; outside it ( $r > R$ ) the enclosed flux is fixed at $B\pi R^2$ , so .

📝 Worked Example — Synthesis: From Fields to Power

A laser beam in vacuum has peak electric field $E_0 = 2.0 \times 10^3 \text{ N/C}$ and a circular cross-section of radius $r = 1.0 \text{ mm}$ . Find its average intensity and total power.

Step 1 — Average intensity from $E_0$ . Using the Poynting time-average,

$I = \frac{1}{2}c\varepsilon_0 E_0^2 = \frac{1}{2}(3.0 \times 10^8)(8.85 \times 10^{-12})(2.0 \times 10^3)^2 \approx 5.3 \times 10^3 \text{ W/m}^2$

Step 2 — Beam cross-sectional area. A circle of radius $r$ :

$A = \pi r^2 = \pi (1.0 \times 10^{-3})^2 \approx 3.14 \times 10^{-6} \text{ m}^2$

Step 3 — Power = intensity × area.

$\mathcal{P} = I A = (5.3 \times 10^3)(3.14 \times 10^{-6}) \approx 1.7 \times 10^{-2} \text{ W}$

🔑 This single chain — $E_0 \to I \to \mathcal{P}$ — ties together the field, energy-density, and Poynting-vector ideas from the whole unit.

Property	Value
Speed	$c = 3 \times 10^8$ m/s (in vacuum)
$\vec{E} \perp \vec{B}$	E and B are perpendicular
$\vec{E} \perp \vec{v}$
$E/B = c$	Ratio of field amplitudes

Surface	Pressure
Perfect absorber	$P = \frac{I}{c}$
Perfect reflector	$P = \frac{2I}{c}$

Maxwell's Equations