Displacement current | PhysicsRank

A capacitor charges while its plates remain separated by an insulating gap. Current reaches one plate and leaves the other, yet ordinary conduction need not cross the gap. Following the changing electric field resolves the apparent break in the magnetic-field account.

We use the field equations to derive the vacuum wave speed, then connect direction, wavelength, intensity and radiation pressure with explicit model limits.

What you will learn to do

Explain why a charging capacitor requires a displacement-current term.
Calculate full and partial displacement current and the ideal gap magnetic field.
Obtain the vacuum wave equation and identify its speed.
Determine electric, magnetic and propagation directions consistently.
Carry frequency through a stationary interface while changing wavelength.
Relate peak and RMS fields to energy density and intensity.
Calculate radiation pressure for ideal normal absorption and reflection.

Useful before this chapterInduction, moving conductors and magnetic energy →Alternating circuits: phase, RMS and real power →Travelling waves, superposition and energy →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

As an ideal capacitor charges, what changes in its gap even when conduction through the dielectric is negligible?

Compare your reasoning

The electric field and electric flux change as charge builds on the plates. A changing field can therefore be relevant even where there is no ordinary conduction path.

Revisit the explanation →

01 / From changing electric fields to electromagnetic waves

A charging capacitor exposes a missing term in the magnetic law

A wire carrying current into a capacitor produces a magnetic field around the circuit. Consider a closed path encircling the wire. A surface spanning that path can cut the wire, so conduction current crosses it. Another spanning surface can bulge through the insulating gap between the plates, where no conduction charge crosses in the ideal capacitor.

If the magnetic circulation depended only on the conduction current through whichever surface we chose, the two surfaces would predict different answers for the same closed path. The physical magnetic field cannot depend on an arbitrary choice of mathematical surface.

The electric field in the capacitor gap is changing as charge accumulates. Maxwell’s correction adds the time derivative of electric flux, multiplied by ε₀, to the conduction current in the vacuum form of the law. This makes the accounting consistent with charge conservation.

The added quantity is called displacement current. In vacuum it is not a stream of material charge crossing the gap. It is a field-change contribution with the same units as current and the same role in the magnetic circulation equation.

\oint_C\mathbf B\cdot d\boldsymbol\ell=\mu_0\left(I_{\mathrm{cond},S}+\epsilon_0\frac{d\Phi_{E,S}}{dt}\right)

Vacuum microscopic form, with the current and electric flux evaluated through a surface S bounded by the same oriented path C. The current includes all actual charge motion in this formulation.

STOP & THINK

In the ideal vacuum gap of a charging capacitor, displacement current means:

Electrons cross the vacuum gap at the wire-current rate
A changing electric-flux contribution enters the magnetic circulation law
Electric charge ceases to be conserved

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: A charging capacitor exposes a missing term in the magnetic law
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
\oint_C\mathbf B\cdot d\boldsymbol\ell=\mu_0\left(I_{\mathrm{cond},S}+\epsilon_0\frac{d\Phi_{E,S}}{dt}\right)
Meaning: Vacuum microscopic form, with the current and electric flux evaluated through a surface S bounded by the same oriented path C. The current includes all actual charge motion in this formulation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

02 / From changing electric fields to electromagnetic waves

The displacement-current value follows from electric flux

For a uniform electric field perpendicular to a fixed area A, electric flux is EA. The displacement current through that area is Id = ε₀A dE/dt. Its sign depends on the chosen surface normal and on whether the field component is increasing or decreasing.

In the ideal parallel-plate vacuum capacitor, E = Q/(ε₀A) when fringing is neglected. Therefore ε₀A dE/dt = dQ/dt. The displacement current through the full gap area equals the wire current charging the plate.

This equality concerns the whole plate area. A smaller coaxial area inside a uniformly charging circular capacitor intercepts only the corresponding area fraction of the displacement current. Using the full wire current for every smaller loop gives the wrong magnetic field.

For material dielectrics, a common macroscopic form uses the displacement field D and a free-current term. Do not mix that convention with the vacuum ε₀E expression while forgetting polarization currents. The examples here use vacuum or air treated as vacuum.

I_d=\epsilon_0\frac{d\Phi_E}{dt},\qquad I_d=\epsilon_0A\frac{dE}{dt}=\frac{dQ}{dt}\quad\text{for the full ideal plate area}

The area is fixed and the field is uniform in the capacitor approximation. Changing area or nonuniform field requires the full flux derivative.

WORKED EXAMPLE

Infer a changing field from a charging current

A vacuum parallel-plate capacitor has plate area 0.020 m² and is being charged at 3.0 μA. Neglect fringing. Find the rate of increase of field magnitude between the plates. Use ε₀ = 8.85 × 10⁻¹² F/m.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: The displacement-current value follows from electric flux
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
I_d=\epsilon_0\frac{d\Phi_E}{dt},\qquad I_d=\epsilon_0A\frac{dE}{dt}=\frac{dQ}{dt}\quad\text{for the full ideal plate area}
Meaning: The area is fixed and the field is uniform in the capacitor approximation. Changing area or nonuniform field requires the full flux derivative.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

03 / From changing electric fields to electromagnetic waves

An Amperian loop samples the enclosed changing flux

Take circular capacitor plates of radius a, with a nearly uniform changing axial field between them. For an interior coaxial circular path of radius r < a, the enclosed displacement current is I r²/a² when the full-area charging current is I.

Rotational symmetry makes B tangential and equal in magnitude around that path. Thus B(2πr) = μ₀I r²/a², or B = μ₀Ir/(2πa²). It grows linearly from zero at the axis.

In the idealized extension outside the plate radius, a loop encloses the full displacement current, giving B = μ₀I/(2πr). The two expressions agree at r = a. Actual plate edges, feed wires and fringe fields limit the accuracy of this simple exterior model.

The direction follows a right-hand rule around the oriented displacement current. If the charging rate reverses sign, the magnetic circulation reverses. A large but static capacitor field has no displacement-current contribution because its time derivative is zero.

B(r)=\begin{cases}\dfrac{\mu_0Ir}{2\pi a^2},&r<a,\\[4pt]\dfrac{\mu_0I}{2\pi r},&r>a\quad\text{(idealized exterior)}.\end{cases}

Uniform-field, rotationally symmetric, quasistatic capacitor model. Nearby return-current and edge effects must be considered in a real geometry.

WORKED EXAMPLE

Use the enclosed fraction, not the whole charging current

Circular capacitor plates have radius 0.10 m and charge at 0.20 A. Find the ideal magnetic-field magnitude at radius 0.050 m in the gap. Use μ₀/(4π) = 10⁻⁷ T·m/A.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: An Amperian loop samples the enclosed changing flux
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
B(r)=\begin{cases}\dfrac{\mu_0Ir}{2\pi a^2},&r<a,\\[4pt]\dfrac{\mu_0I}{2\pi r},&r>a\quad\text{(idealized exterior)}.\end{cases}
Meaning: Uniform-field, rotationally symmetric, quasistatic capacitor model. Nearby return-current and edge effects must be considered in a real geometry.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

04 / From changing electric fields to electromagnetic waves

The field equations admit waves without a material medium

Faraday’s law connects a changing magnetic field with a circulating electric field. The Ampère–Maxwell law connects a changing electric field with a circulating magnetic field, even where conduction current is zero. Together with the electric and magnetic divergence laws, these relations admit travelling electromagnetic disturbances.

For a source-free vacuum region, consider a plane field with Ey depending on x and t and Bz depending on x and t. Faraday’s law gives ∂Ey/∂x = −∂Bz/∂t. The Ampère–Maxwell law gives ∂Bz/∂x = −μ₀ε₀ ∂Ey/∂t.

Differentiate the first relation with respect to x and substitute the time derivative of the second. The result is ∂²Ey/∂x² = μ₀ε₀ ∂²Ey/∂t². Comparing this with the standard wave equation identifies the speed as 1/√(μ₀ε₀).

This argument does not require a mechanical substance to oscillate. The electric and magnetic fields are the dynamical quantities. Sources establish the disturbance and boundary conditions, but propagation through vacuum requires no material medium.

\frac{\partial^2 E_y}{\partial x^2}=\mu_0\epsilon_0\frac{\partial^2 E_y}{\partial t^2},\qquad c=\frac1{\sqrt{\mu_0\epsilon_0}}

Source-free vacuum plane-wave derivation. In numerical examples, use c ≈ 3.00 × 10⁸ m/s unless greater precision is specified.

WORKED EXAMPLE

Recover the vacuum wave speed from the constants

Use μ₀ ≈ 4π × 10⁻⁷ H/m and ε₀ ≈ 8.85 × 10⁻¹² F/m to estimate the electromagnetic wave speed.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: The field equations admit waves without a material medium
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
\frac{\partial^2 E_y}{\partial x^2}=\mu_0\epsilon_0\frac{\partial^2 E_y}{\partial t^2},\qquad c=\frac1{\sqrt{\mu_0\epsilon_0}}
Meaning: Source-free vacuum plane-wave derivation. In numerical examples, use c ≈ 3.00 × 10⁸ m/s unless greater precision is specified.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

05 / From changing electric fields to electromagnetic waves

The propagation direction fixes the relationship between the fields

For a plane wave travelling along +x in vacuum, choose E along y and B along z. The fields are transverse to the propagation direction and to each other. Their cross product E × B points along the direction of energy transport.

A sinusoidal travelling solution can be written Ey = E0 cos(kx − ωt + α) and Bz = B0 cos(kx − ωt + α), with ω/k = c and E0 = cB0. The electric and magnetic components are in phase in this travelling plane wave.

When the cosine changes sign, both fields reverse, so their cross product still points along +x. Changing the propagation direction requires changing the relative orientation, not simply reversing both fields together.

The relation E/B = c is a travelling plane-wave relation in vacuum. It does not apply to arbitrary static fields, all near-source fields or every standing-wave point. State the wave model before using it.

\mathbf B=\frac1c\hat{\mathbf n}\times\mathbf E,\qquad E_0=cB_0

A travelling plane wave in vacuum has propagation unit vector n̂. Peak values must be compared with peak values, and RMS with RMS.

WORKED EXAMPLE

Find a field magnitude and check the propagation direction

A plane wave travels along +x. At an instant its electric field is +90 ŷ V/m. Find its magnetic field at that instant using c = 3.00 × 10⁸ m/s.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: The propagation direction fixes the relationship between the fields
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
\mathbf B=\frac1c\hat{\mathbf n}\times\mathbf E,\qquad E_0=cB_0
Meaning: A travelling plane wave in vacuum has propagation unit vector n̂. Peak values must be compared with peak values, and RMS with RMS.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

06 / From changing electric fields to electromagnetic waves

Frequency and wavelength play different roles at a boundary

For a monochromatic wave, speed, frequency and wavelength obey v = fλ. In vacuum v = c. Different frequencies therefore have different wavelengths while sharing the same vacuum speed.

At a stationary interface between time-independent media, the oscillations must match in time on both sides, so frequency remains the same. The wavelength changes when the phase speed changes.

A simple transparent medium is described by refractive index n = c/vphase at the specified frequency. Thus λmedium = λvacuum/n. In a dispersive medium, n depends on frequency; a pulse can also have a group speed different from its phase speed.

The expression v = 1/√(με) applies to the appropriate ideal homogeneous, linear, nondispersive, lossless model. Real materials can absorb and disperse waves. We will use their refractive index at the relevant frequency for elementary refraction problems.

WORKED EXAMPLE

Carry frequency unchanged through a stationary interface

Light has vacuum wavelength 600 nm and enters a transparent medium whose refractive index at that frequency is 1.50. Find frequency, phase speed and wavelength inside. Use c = 3.00 × 10⁸ m/s.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: Frequency and wavelength play different roles at a boundary
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

07 / From changing electric fields to electromagnetic waves

Electric and magnetic fields share the wave energy

In vacuum, electric energy density is ε₀E²/2 and magnetic energy density is B²/(2μ₀). For a travelling plane wave with E = cB and c² = 1/(μ₀ε₀), the two contributions are equal at every point and instant.

The total instantaneous energy density is therefore ε₀E². The energy-flow vector is S = (E × B)/μ₀; for the travelling vacuum plane wave its magnitude is cu. This says that energy density times propagation speed gives energy crossing unit area per unit time.

For a sinusoid with peak electric amplitude E0, average intensity is I = ε₀cE0²/2. In terms of RMS electric field it is ε₀cErms². A missing factor one half often signals that peak and RMS amplitudes have been mixed.

Intensity scales with field amplitude squared. Doubling the amplitude quadruples intensity. The inverse-square distance law additionally requires a source spreading power uniformly over growing spherical area without loss; it does not follow for every beam or plane-wave geometry.

\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B,\qquad I=\frac12\epsilon_0cE_0^2=\epsilon_0cE_{\mathrm{rms}}^2

I is cycle-averaged energy flux for a sinusoidal travelling plane wave in vacuum, measured in W/m².

Fields reverse together while energy density remains nonnegative

Inspect a plotted point: x / λ = 0

E / E₀: 1
cB / E₀: 1
u / (ε₀E₀²): 1

The normalized electric and magnetic components coincide for this travelling plane-wave snapshot. Total energy density follows their square. Position is measured in wavelengths. Lines connect the plotted samples.

WORKED EXAMPLE

Use the peak amplitude once, with its required averaging factor

A sinusoidal plane wave in vacuum has peak electric field 100 V/m. Estimate its average intensity using ε₀ = 8.85 × 10⁻¹² F/m and c = 3.00 × 10⁸ m/s.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: Electric and magnetic fields share the wave energy
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B,\qquad I=\frac12\epsilon_0cE_0^2=\epsilon_0cE_{\mathrm{rms}}^2
Meaning: I is cycle-averaged energy flux for a sinusoidal travelling plane wave in vacuum, measured in W/m².

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

08 / From changing electric fields to electromagnetic waves

Radiation also transfers momentum to a surface

Electromagnetic radiation carries momentum. For a normally incident beam absorbed by a surface, the momentum delivered per unit time equals the absorbed power divided by c, giving radiation pressure I/c for complete absorption.

For ideal specular reflection straight back along the incident direction, the momentum change doubles and pressure is 2I/c. The factor two comes from reversing momentum, not from absorbing twice as much energy.

These expressions assume normal incidence, a stationary surface and the stated absorption or reflection model. Oblique incidence, partial reflection or transmission requires a momentum balance with vector components.

Radiation pressure is often small at ordinary intensities, but it provides a useful consistency check that electromagnetic waves transport both energy and momentum. The units W/m² divided by m/s reduce to N/m².

WORKED EXAMPLE

Distinguish absorption from reversal of momentum

A normally incident beam has average intensity 600 W/m². Find the radiation pressure for a completely absorbing surface and for an ideal mirror that reflects it straight back. Use c = 3.00 × 10⁸ m/s.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: From changing electric fields to electromagnetic waves
Focus: Radiation also transfers momentum to a surface
Learning goals:
- Explain why a charging capacitor requires a displacement-current term.
- Calculate full and partial displacement current and the ideal gap magnetic field.
- Obtain the vacuum wave equation and identify its speed.
- Determine electric, magnetic and propagation directions consistently.
- Carry frequency through a stationary interface while changing wavelength.
- Relate peak and RMS fields to energy density and intensity.
- Calculate radiation pressure for ideal normal absorption and reflection.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

Displacement current is a changing-flux term, not necessarily charge crossing a gap.
A smaller loop encloses only part of a uniformly distributed displacement current.
The source-free vacuum field equations give c = 1/√(μ₀ε₀).
In a travelling vacuum plane wave, E and B are in phase and E × B gives energy-flow direction.
Frequency stays fixed at a stationary time-independent interface while speed and wavelength can change.
Intensity is quadratic in field amplitude and requires a peak-to-RMS distinction.
Radiation pressure follows the change of momentum flux.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

A uniformly charging circular capacitor carries 0.80 A in its leads. What displacement current is enclosed by a coaxial loop of half the plate radius?

Answer in A.

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SELF CHECK 2

A wave has vacuum wavelength 540 nm and enters a stationary medium with refractive index 1.80 at that frequency. What is its wavelength inside?

Answer in nm.

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SELF CHECK 3

A travelling vacuum plane wave has peak E = 60 V/m. Using c = 3 × 10⁸ m/s, what is peak B in microteslas?

Answer in μT.

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SELF CHECK 4

A travelling sinusoidal wave’s electric-field amplitude triples. What is the ratio of its new intensity to its old intensity?

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SELF CHECK 5

A capacitor has a large but time-independent electric field. Its displacement-current contribution is:

Large because E is large
Zero because the electric flux is constant
Equal to its stored charge

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SELF CHECK 6

A vacuum plane wave has E along +y and B along +z at an instant. Which way does its energy flow?

+x
−x
+y

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SELF CHECK 7

In a travelling plane wave in vacuum, electric and magnetic field magnitudes reach their maxima:

A quarter period apart
At the same phase
With no fixed relation

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

Frequency from a radio wavelengthIntermediate · Free sample · Hints and guided reasoning Follow changing flux across the gapIntermediate · Hints and guided reasoning The magnetic part of a light waveIntermediate · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

Why is displacement current not a claim that electrons cross an ideal vacuum capacitor gap?

Compare your reasoning

It is the field-change term in the magnetic circulation law. It keeps the field account consistent for different spanning surfaces, without introducing a fictitious conduction flow through the gap.

Revisit the explanation →

A few days later: change the situation

An electromagnetic wave enters a stationary linear medium. Which stays fixed across the boundary: frequency or wavelength?

Compare your reasoning

Frequency stays fixed by the source and continuity in time. Speed and wavelength can change together in the medium. This transmission assumption differs from frequency-changing scattering processes.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

Back to chapter contents ↑