Maxwell's Equations — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: Gauss’s law for electricity, Gauss’s law for magnetism, Faraday’s law of induction, the Ampère-Maxwell law, displacement current, integral forms of all four Maxwell equations, and properties of electromagnetic waves derived from the equations.

You should already know: Gauss’s law for static electric fields, Faraday’s law of electromagnetic induction, Ampère’s law for static magnetic fields.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: E&M style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Maxwell's Equations?

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. The key innovation from Maxwell was adding the displacement current term to Ampère’s law, resolving the contradiction between steady-state Ampère’s law and time-varying systems like charging capacitors. According to the AP Physics C: E&M Course and Exam Description (CED), this topic accounts for 10-15% of the total exam score, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections. For AP CEM, you only need to master the integral form of the equations; the differential form is not required by the CED. We use standard notation: $\vec{E}$ for electric field, $\vec{B}$ for magnetic field, ${ϵ}_0$ for permittivity of free space, ${\mu }_0$ for permeability of free space, $Q_{enc}$ for enclosed charge, and $I_{enc}$ for enclosed conduction current. Maxwell’s equations describe all observable electromagnetic phenomena, from static fields to radio waves to light, making them the core result of E&M.

2. Gauss's Laws for Electricity and Magnetism

Gauss’s laws are the simplest of the four Maxwell equations, and they extend the rules you already learned for static fields 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: $$\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{{ϵ}_0}$$ This confirms that electric field lines originate and terminate on electric charges — net flux is non-zero only when there is net charge enclosed.

Gauss’s law for magnetism is structurally similar, but with a key difference: there are no isolated magnetic monopoles (point magnetic charges) 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 $\vec{B}=B_0\hat{x}$. 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 ${\Phi }_E=Q_{enc}/{ϵ}_0=Q/{ϵ}_0$.
2. The shape of the enclosed charge (cube vs sphere) does not affect Gauss’s law, so we do not need to adjust for the cube shape.
3. 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.
4. Final result: ${\Phi }_E=\frac{Q}{{ϵ}_0}$, ${\Phi }_B=0$.

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

The two remaining Maxwell equations describe how changing fields generate other fields. Faraday’s law, which you first learned 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 ($\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{enc}$) only works when nothing changes 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: $$\oint \vec{B}\cdot d\vec{l}={\mu }_0\left(I_{enc}+{ϵ}_0\frac{d{\Phi }_E}{dt}\right)$$ The term $I_d={ϵ}_0\frac{d{\Phi }_E}{dt}$ is called displacement current. It is not a flow of charged particles like conduction current $I_{enc}$, but it produces a magnetic field the same way conduction current does. This is critical for describing time-varying systems like charging capacitors, where there is no conduction current between the plates but there is a changing electric flux.

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 $I_{enc}=0$, and only displacement current contributes to the magnetic field.
2. The total electric flux between the plates is ${\Phi }_E=\frac{Q}{{ϵ}_0}$, so the total displacement current is $I_d={ϵ}_0\frac{d{\Phi }_E}{dt}=\frac{dQ}{dt}=I$, which equals the total conduction current charging the capacitor.
3. For an Amperian loop of radius $r>R$, all of the displacement current is enclosed, so the right-hand side of Ampère-Maxwell becomes ${\mu }_0{I}_d={\mu }_{0}I$.
4. The magnetic field is tangential to the loop and uniform in magnitude, so $\oint \vec{B}\cdot d\vec{l}=B(2\pi r)={\mu }_{0}I$. Solving for $B$ gives $B=\frac{{\mu }_{0}I}{2\pi r}$.

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

In a region of free space with no charges and no conduction currents ($Q_{enc}=0$, $I_{enc}=0$), 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 $v=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}$, which equals the measured speed of light $c$, proving that light is an electromagnetic wave.

Key properties of plane EM waves in vacuum:

1. They are transverse: $\vec{E}$ and $\vec{B}$ are both perpendicular to the direction of propagation.
2. $\vec{E}$, $\vec{B}$, and the direction of propagation are mutually perpendicular, with propagation in the direction of $\vec{E}\times \vec{B}$.
3. The amplitudes of the fields are related by $E_0=c{B}_0$, a relation that holds for all plane EM waves, regardless of frequency or wavelength.

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 $1.5\times 10^{-6}$ T. Find the amplitude and direction of the electric field at that point.

1. Use the amplitude relation $E_0=c{B}_0$. Substitute $c=3\times 10^8$ m/s and $B_0=1.5\times 10^{-6}$ T: $E_0=(3\times 10^8)(1.5\times 10^{-6})=450$ N/C.
2. The direction of propagation ($+y$) equals the direction of $\vec{E}\times \vec{B}$. We know propagation is $\hat{y}$, and $\vec{B}$ is $-\hat{z}$, so we solve for $\vec{E}$ direction: $\vec{E}\times (-\hat{z})=\hat{y}$.
3. Using cross product rules: $\hat{x}\times (-\hat{z})=-(\hat{x}\times \hat{z})=-(-\hat{y})=\hat{y}$, which matches the required propagation direction.
4. Final result: amplitude is 450 N/C, direction is $+x$.

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

5. Common Pitfalls (and how to avoid them)

• 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 $\vec{B}$. 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.

6. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

Which of the following statements correctly matches a Maxwell equation to its physical implication? A) Gauss’s law for magnetism implies that changing magnetic flux produces an electric field B) The Ampère-Maxwell law implies that changing electric flux produces a magnetic field C) Gauss’s law for electricity implies that there are no isolated magnetic monopoles D) Faraday’s law implies that net electric flux through a closed surface is proportional to enclosed charge

Worked Solution: Eliminate incorrect options one by one. Option A is wrong: changing magnetic flux producing an electric field is Faraday’s law, not Gauss’s law for magnetism. Option C is wrong: the claim that there are no magnetic monopoles is Gauss’s law for magnetism, not Gauss’s law for electricity. Option D is wrong: net electric flux proportional to enclosed charge is Gauss’s law for electricity, not Faraday’s law. Only Option B is correct: the displacement current term added to Ampère’s law by Maxwell describes magnetic fields produced by changing electric flux. The correct answer is B.

Question 2 (Free Response)

A parallel-plate capacitor with circular plates of radius $R=0.2$ m is charged by a time-varying conduction current $I(t)=3t$ A for $t\ge 0$. There is vacuum between the plates. (a) Show that the total displacement current between the plates equals $I(t)$ as a function of time. (b) Find the magnitude of the magnetic field at $r=0.1$ m from the central axis between the plates, as a function of time. (c) If the magnetic field between the plates is uniform and points along the central axis, find the magnitude of the induced electric field at $r=0.1$ m at $t=2$ s.

Worked Solution: (a) The total charge on the positive plate is $Q(t)={\int }_0^{t}I(t^{\prime })d{t}^{\prime }={\int }_0^{t}3t^{\prime }d{t}^{\prime }=\frac{3}{2}{t}^2$. The electric field between the plates is $E(t)=\sigma /{ϵ}_0=Q(t)/({ϵ}_0\pi R^2)$, so total electric flux is ${\Phi }_E=E(t)\pi R^2=Q(t)/{ϵ}_0$. Displacement current is $I_d={ϵ}_{0}d{\Phi }_E/dt=dQ/dt=3t=I(t)$, as required. (b) For $r=0.1$ m $<R=0.2$ m, the enclosed displacement current is $I_{d,enc}=I_d\cdot (\pi r^2/\pi R^2)=3t(r^2/R^2)=3t(0.01/0.04)=0.75t$. Apply Ampère-Maxwell: $B(2\pi r)={\mu }_0{I}_{d,enc}⟹B(2\pi (0.1))={\mu }_0(0.75t)⟹B=\frac{0.75{\mu }_{0}t}{0.2\pi }=\frac{3.75{\mu }_{0}t}{\pi }\approx 1.2\times 10^{-6}t$ T. (c) Magnetic flux through a loop of radius $r$ is ${\Phi }_B=B(t)\pi r^2=\left(\frac{3.75{\mu }_{0}t}{\pi }\right)\pi (0.1{)}^2=0.0375{\mu }_{0}t$. By Faraday’s law, $E(2\pi r)=|d{\Phi }_B/dt|=0.0375{\mu }_0⟹E=\frac{0.0375{\mu }_0}{2\pi (0.1)}=\frac{0.1875{\mu }_0}{\pi }$. Substitute ${\mu }_0=4\pi \times 10^{-7}$: $E=0.1875(4\pi \times 10^{-7})/\pi =7.5\times 10^{-8}$ N/C at $t=2$ s.

Question 3 (Application / Real-World Style)

A 5G cellular communications tower transmits radio waves with a frequency of 3.5 GHz. The maximum electric field amplitude measured 1 km from the tower is $1.2\times 10^{-2}$ N/C. Assuming the waves travel in vacuum, find the amplitude of the magnetic field of the transmitted wave at this position, and the wavelength of the 5G signal.

Worked Solution:

1. For all EM waves in vacuum, $E_0=c{B}_0$, so $B_0=E_0/c$. Substitute values: $E_0=1.2\times 10^{-2}$ N/C, $c=3\times 10^8$ m/s, so $B_0=(1.2\times 10^{-2})/(3\times 10^8)=4\times 10^{-11}$ T.
2. Wavelength is given by $c=f\lambda$, so $\lambda =c/f$. Convert frequency to Hz: $3.5$ GHz = $3.5\times 10^9$ Hz. So $\lambda =(3\times 10^8)/(3.5\times 10^9)\approx 0.086$ m = 8.6 cm.
3. In context: this 8.6 cm wavelength is shorter than common 4G wavelengths, allowing 5G antennas to be smaller and carry more data per unit time, consistent with real-world 5G design.

7. Quick Reference Cheatsheet

8. What's Next

Maxwell's equations are the culmination of all AP Physics C: E&M, and they are the prerequisite for understanding electromagnetic wave propagation, energy carried by EM waves, and the Poynting vector, which are 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 time-varying systems that include capacitors. This topic also reinforces all earlier E&M concepts: Gauss's law from electrostatics, Ampère's law from magnetostatics, and Faraday's law from induction, so reviewing this chapter will strengthen your preparation for the entire exam.

Next topics to study: Electromagnetic Plane Waves Poynting Vector and EM Wave Energy Time-Varying Capacitor Fields