Maxwell's Equations

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Electromagnetism· 14 min read

1. Introduction to Maxwell's Equations★★☆☆☆⏱ 2 min

Maxwell’s equations are a set of four fundamental relations that unify the description of electricity, magnetism, and light, first compiled by James Clerk Maxwell in the 1860s. Maxwell’s key innovation was adding the displacement current term to Ampère’s law, resolving a contradiction between steady-state Ampère’s law and time-varying systems like charging capacitors.

For AP Physics C: E&M, you only need to master the integral form of the equations (differential form is not required by the CED). This topic accounts for 10-15% of your total exam score, appearing in both multiple-choice and free-response sections. Standard notation used throughout this guide: $E$ for electric field, $B$ for magnetic field, $ϵ_{0}$ for permittivity of free space, $μ_{0}$ for permeability of free space, $Q_{e n c}$ for enclosed charge, and $I_{e n c}$ for enclosed conduction current.

2. Gauss's Laws for Electricity and Magnetism★★☆☆☆⏱ 3 min

Gauss’s laws extend the static field rules you already learned to all time-varying fields. Gauss’s law for electricity states that the net electric flux through any closed surface is proportional to the total charge enclosed by that surface:

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}

This confirms that electric field lines originate and terminate on electric charges; net flux is non-zero only when there is net enclosed charge. Next, Gauss’s law for magnetism reflects that no isolated magnetic monopoles (point magnetic charges) have been observed in nature. All magnetic sources are dipoles, so magnetic field lines are always closed loops. This gives Gauss’s law for magnetism:

\oint \vec{B} \cdot d\vec{A} = 0

The net magnetic flux through any closed surface is always zero, because any field line that enters the surface must also exit it.

📐 Worked Example

A uniformly charged cube of side length $L$ with total charge $Q$ sits inside a uniform external magnetic field $B = B_{0} \overset{x}{^}$ . What is the net electric flux through a closed spherical surface that fully encloses the cube, and what is the net magnetic flux through the same sphere?

1
By Gauss’s law for electricity, net electric flux depends only on the total enclosed charge. The enclosed charge is $Q$ , so:
2
$\Phi_E = \frac{Q_{enc}}{\epsilon_0} = \frac{Q}{\epsilon_0}$
3
The shape of the enclosed charge (cube vs any other shape) does not affect Gauss’s law, so no adjustment for the cube shape is needed.
4
By Gauss’s law for magnetism, the net magnetic flux through any closed surface is always zero, regardless of the external magnetic field or enclosed magnetic dipoles.
5
Final result:
6
$\Phi_E = \frac{Q}{\epsilon_0}, \quad \Phi_B = 0$

Exam tip:

If a multiple-choice question asks for the net magnetic flux through any closed surface, the answer is always zero — Gauss’s law for magnetism applies to all configurations, time-varying or not.

3. Faraday's Law and the Ampère-Maxwell Law★★★☆☆⏱ 4 min

The two remaining Maxwell equations describe how changing fields generate other fields. Faraday’s law, first introduced for electromagnetic induction, is formalized as:

\oint \vec{E} \cdot d\vec{l} = - \frac{d \Phi_B}{dt}

This states that a changing magnetic flux through an open loop creates a curly, non-conservative electric field around the loop. The negative sign encodes Lenz’s law: the direction of the induced electric field always opposes the change in magnetic flux that created it.

Original Ampère’s law for steady currents ( $\oint B \cdot d l = μ_{0} I_{e n c}$ ) only works when fields do not change with time. Maxwell fixed this by adding the displacement current term, which accounts for magnetic fields generated by changing electric flux. The full Ampère-Maxwell law is:

\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{enc} + \epsilon_0 \frac{d \Phi_E}{dt} \right)

📘 Definition

Displacement Current

I_d

The term accounting for magnetic fields produced by changing electric flux, given by $I_{d} = ϵ_{0} \frac{d Φ _{E}}{d t}$ . Displacement current is not a flow of charged particles, but produces a magnetic field identically to conduction current.

📐 Worked Example

A parallel-plate capacitor with circular plates of radius $R$ is being charged by a constant conduction current $I$ . What is the magnitude of the magnetic field at a distance $r > R$ from the central axis, outside the plates?

1
There is no conduction current outside the plates, so $I_{e n c} = 0$ , and only displacement current contributes to the magnetic field.
2
The total electric flux between the plates is $Φ_{E} = \frac{Q}{ϵ _{0}}$ , so total displacement current is:
3
$I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \frac{dQ}{dt} = I$
4
For an Amperian loop of radius $r > R$ , all of the displacement current is enclosed, so the right-hand side of Ampère-Maxwell becomes $μ_{0} I_{d} = μ_{0} I$ .
5
The magnetic field is tangential to the loop and uniform in magnitude, so:
6
$\oint \vec{B} \cdot d\vec{l} = B(2\pi r) = \mu_0 I$
7
Solving for $B$ gives the final result:
8
$B = \frac{\mu_0 I}{2\pi r}$

Exam tip:

Always check whether you have conduction current enclosed by your Amperian loop between capacitor plates: the conduction current is zero there, so only the displacement current term contributes.

4. Electromagnetic Waves from Maxwell's Equations★★★★☆⏱ 3 min

In a region of free space with no charges and no conduction currents ( $Q_{e n c} = 0$ , $I_{e n c} = 0$ ), Maxwell’s equations simplify to a set of coupled equations that predict wave solutions called electromagnetic (EM) waves. Substituting the equations into each other gives a wave equation that travels at speed $v = \frac{1}{μ _{0} ϵ _{0}}$ , which equals the measured speed of light $c$ , proving that light is an electromagnetic wave.

EM waves are transverse: $E$ and $B$ are both perpendicular to the direction of propagation.
$E$ , $B$ , and the direction of propagation are mutually perpendicular, with propagation in the direction of $E \times B$ .
The amplitudes of the fields are related by $E_{0} = c B_{0}$ , a relation that holds for all plane EM waves in vacuum, regardless of frequency or waveform.

📐 Worked Example

A plane EM wave travels in vacuum in the $+ y$ direction. At a given point in space and time, the magnetic field of the wave points in the $- z$ direction with amplitude $1.5 \times 1 0^{- 6}$ T. Find the amplitude and direction of the electric field at that point.

1
Use the amplitude relation $E_{0} = c B_{0}$ . Substitute $c = 3 \times 1 0^{8}$ m/s and $B_{0} = 1.5 \times 1 0^{- 6}$ T:
2
$E_0 = (3 \times 10^8)(1.5 \times 10^{-6}) = 450 \text{ N/C}$
3
The direction of propagation ( $+ y$ ) equals the direction of $E \times B$ . We know propagation direction is $\overset{y}{^}$ , and $B$ is $- \overset{z}{^}$ , so we solve for the direction of $E$ : $E \times (- \overset{z}{^}) = \overset{y}{^}$
4
Using cross product rules: $\overset{x}{^} \times (- \overset{z}{^}) = - (\overset{x}{^} \times \overset{z}{^}) = - (- \overset{y}{^}) = \overset{y}{^}$ , which matches the required propagation direction.
5
Final result: amplitude is 450 N/C, direction is $+ x$ .

Exam tip:

When asked for the direction of $B$ or $E$ given the propagation direction, always use the cross product rule for $E \times B$ to confirm orientation — don’t rely on memory of common orientations.

5. AP-Style Practice Problems★★★★☆⏱ 4 min

✓ Quick check

Test your understanding of core concepts

Which of the following statements correctly matches a Maxwell equation to its physical implication?
- Gauss’s law for magnetism implies that changing magnetic flux produces an electric field
- The Ampère-Maxwell law implies that changing electric flux produces a magnetic field
- Gauss’s law for electricity implies that there are no isolated magnetic monopoles
- Faraday’s law implies that net electric flux through a closed surface is proportional to enclosed charge
Reveal answer
1 —
Incorrect options: A describes Faraday's law, C describes Gauss's law for magnetism, D describes Gauss's law for electricity. Only B is correct.

6. Common Pitfalls

Wrong move:

Forgetting the displacement current term when applying Ampère’s law between the plates of a charging capacitor, concluding the magnetic field is zero there.

Why:

Students associate current only with moving charge, and forget that changing electric flux acts as a source of $B$ .

Correct move:

Always check if the enclosed region has changing electric flux, even if there is no conduction current, add the displacement current term to your calculation.

Wrong move:

Interpreting Gauss’s law for magnetism as saying the magnetic field is zero everywhere inside a closed surface.

Why:

Students confuse net flux through a closed surface with the magnitude of the field at points inside the surface, mirroring a common Gauss’s law for E misconception.

Correct move:

Remember Gauss’s law for magnetism only constrains net flux through a closed surface, it does not say the magnetic field itself is zero inside the surface.

Wrong move:

Claiming displacement current between capacitor plates is a flow of charged particles.

Why:

The name "current" leads students to confuse it with conduction current.

Correct move:

Remember displacement current is a term accounting for the magnetic field produced by a changing electric flux, not moving charge.

Wrong move:

Dropping the negative sign in Faraday’s law and getting the wrong direction for the induced electric field.

Why:

Students often memorize only the magnitude of the induced field, and forget the negative sign corresponds directly to Lenz’s law.

Correct move:

Always keep the negative sign when writing Faraday’s law, and use it with Lenz’s law to confirm the direction of the induced field.

Wrong move:

Stating that $E_{0} = c B_{0}$ only applies to sinusoidal/harmonic EM waves.

Why:

Students learn the relation in the context of harmonic plane waves and incorrectly generalize it to only that case.

Correct move:

This relation holds for all plane EM waves in vacuum, regardless of their waveform.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Gauss's Law for E	$\oint E \cdot d A = \frac{Q _{e n c}}{ϵ _{0}}$	Applies to all electric fields, static or time-varying; net flux proportional to enclosed charge
Gauss's Law for B	$\oint B \cdot d A = 0$	Applies to all magnetic fields; no magnetic monopoles, so net flux through any closed surface is always zero
Faraday's Law	$\oint E \cdot d l = - \frac{d Φ _{B}}{d t}$	Negative sign encodes Lenz's law; changing magnetic flux produces non-conservative electric field
Ampère-Maxwell Law	$\oint B \cdot d l = μ_{0} (I_{e n c} + ϵ_{0} \frac{d Φ _{E}}{d t})$	$I_{e n c}$ is enclosed conduction current; displacement current term accounts for B from changing E flux
Displacement Current	$I_{d} = ϵ_{0} \frac{d Φ _{E}}{d t}$	Not a flow of charge; equals conduction current for charging capacitors
Speed of EM Waves in Vacuum	$c = \frac{1}{μ _{0} ϵ _{0}} = 3 \times 1 0^{8}$ m/s	Constant for all EM waves in vacuum, independent of frequency
EM Wave Amplitude Relation	$E_{0} = c B_{0}$	Holds for all plane EM waves in vacuum, any waveform
Propagation Direction	$v ∥ E \times B$	EM waves are transverse; E and B are perpendicular to propagation

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Displacement current in charging capacitor
2022 · FRQ
EM wave direction and amplitude calculation

What's Next

Maxwell's equations are the culmination of all AP Physics C: E&M content, and form the prerequisite for understanding electromagnetic wave propagation, energy carried by EM waves, and the Poynting vector — the final topics in Unit 5. Without mastering each of the four equations and the role of displacement current, you will not be able to solve problems involving energy flow in EM waves or complex time-varying systems that include capacitors. This topic also reinforces all earlier E&M concepts, so reviewing this sub-topic strengthens your preparation for the entire AP exam.