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E3.4 · Describe electromagnetic radiation from an oscillating electric dipole

Learn to describe electromagnetic radiation from an oscillating electric dipole through clear examples and targeted practice.

Ontario Grade 12 Physics

The Wave Nature of Light

How changing charge motion produces waves that carry energy through space

An electric dipole has two equal and opposite charges separated by a distance. If the charges move back and forth, the dipole oscillates. This changing motion produces changing electric and magnetic fields that can travel outward as electromagnetic radiation. This lesson describes that process using Grade 12 wave and electricity ideas.

What you will learn

1. Prerequisite bridge: charges, fields, and waves

A charge is a property of matter that causes electric interactions. Opposite charges attract, and like charges repel. An electric field is a region where an electric charge experiences an electric force. The electric field has a direction: it points in the direction of the force on a small positive test charge.
A dipole consists of two equal and opposite charges separated by a distance. The line joining the charges is the dipole axis. When the charges stay still, their electric field is steady. When they move back and forth, the electric field changes over time.
A wave transfers energy from place to place. In a travelling wave, the disturbance moves through space. Electromagnetic radiation is a travelling pattern of changing electric and magnetic fields. It does not need air or another material medium to travel.
For this lesson, the physical system is an oscillating dipole and the electromagnetic radiation it produces. Use a reference frame in which the dipole is at rest on average. Choose the dipole axis as the vertical direction; the positive direction points from the negative charge toward the positive charge at the instant shown. The electric field is a vector, so it has magnitude and direction. Frequency and wavelength are scalars; each has magnitude but no direction.
c=fλc=f\lambda

2. How an oscillating dipole radiates

Imagine the charges moving in opposite directions along the dipole axis. Their separation and the dipole's electric field change as they move. Because the charges repeatedly change speed and direction, their motion is changing. In the Grade 12 model, changing charge motion produces electromagnetic radiation.
The changing electric field is linked to a changing magnetic field. Together, these changing fields travel outward from the dipole. In the travelling radiation, the electric field and magnetic field are at right angles to each other. Both are at right angles to the direction the wave travels. This is a transverse wave.
A useful field sketch has the dipole axis vertical, the wave travelling horizontally away from the dipole, and the electric and magnetic fields perpendicular to the travel direction and to each other. The electric-field direction is related to the dipole's oscillation direction. A complete drawing must show these three directions clearly; the fields are not arrows pointing along the direction of travel.
Radiation is strongest in directions perpendicular to the dipole axis. It is weakest along the axis, where the ideal dipole model has no radiation directed outward. This pattern is why the direction an antenna points can affect how well it sends or receives a signal. This description concerns the radiation pattern, not a claim that the electric field itself has the same strength everywhere.
The radiation frequency matches the oscillation frequency of the dipole. Frequency is the number of complete oscillations per second, measured in hertz. Wavelength is the distance between matching points on successive wave cycles, measured in metres. In vacuum, electromagnetic radiation travels at approximately 3.00×108 m/s3.00\times10^8\ \mathrm{m/s}, so frequency and wavelength are related by wave speed.
v=fλv=f\lambda

3. Describing the radiation with wave quantities

The relationship v=fλv=f\lambda applies to waves. Here, vv is wave speed in metres per second, ff is frequency in hertz, and λ\lambda is wavelength in metres. For electromagnetic radiation in vacuum, use v=cv=c, where cc is the speed of light in vacuum.
An oscillation period is the time for one complete cycle. It is measured in seconds. Frequency and period are related: a higher frequency means more cycles each second and therefore a shorter period. This relationship helps connect the dipole's motion to its emitted wave.
When solving a problem, identify whether it asks about the dipole's motion, the radiation's direction, or a wave quantity. Use the dipole axis to decide the radiation direction. Use the wave equation to calculate speed, frequency, or wavelength. Keep units in the substitution, and check that the answer makes sense. For example, at fixed wave speed, increasing frequency must reduce wavelength.
T=1fT=\frac{1}{f}

4. Reading a dipole radiation pattern

A radiation pattern tells how the emitted radiation depends on direction. For an ideal oscillating dipole, picture a ring around the dipole axis: directions out through the sides correspond to strong radiation, while directions along the axis correspond to the minimum. The pattern is symmetric around the axis.
The pattern does not mean the dipole sends a beam only in one direction. It radiates into many directions, with different strengths depending on direction. The axis direction is a special case with no outward radiation in the ideal model.
When describing a field diagram, label the dipole axis, the direction of wave travel, the electric field, and the magnetic field. Separate direction statements from magnitude statements. For instance, 'the field points this way' describes direction, while 'radiation is strongest here' compares strength.

Worked example

Finding the wavelength

An ideal dipole oscillates at 1.20×108 Hz1.20\times10^8\ \mathrm{Hz}. Find the wavelength of its radiation in vacuum.
  1. Set the system and known values
    The system is the dipole and its emitted radiation, viewed from the frame in which the dipole is at rest on average. The unknown is wavelength. Use the positive dipole direction only to describe orientation; this calculation uses scalar wave quantities.
    f=1.20×108 Hz,c=3.00×108 m/sf=1.20\times10^8\ \mathrm{Hz},\quad c=3.00\times10^8\ \mathrm{m/s}
  2. Choose the wave relationship
    In vacuum, the radiation speed is the speed of light. Rearrange the wave relationship to solve for wavelength.
    λ=cf\lambda=\frac{c}{f}
  3. Substitute and calculate
    Substitute the values with their units. The units reduce to metres.
    λ=3.00×108 m/s1.20×108 s−1=2.50 m\lambda=\frac{3.00\times10^8\ \mathrm{m/s}}{1.20\times10^8\ \mathrm{s^{-1}}}=2.50\ \mathrm{m}
Answer: The wavelength is 2.50 m2.50\ \mathrm{m}.
Check: The units are metres. A frequency below the speed-of-light value expressed in metres per second gives a wavelength of a few metres here, which is consistent with the calculation.

Worked example

Finding oscillation frequency from period

A dipole completes one oscillation every 4.00×10−9 s4.00\times10^{-9}\ \mathrm{s}. Find its frequency and the wavelength of its radiation in vacuum.
  1. Identify the system and unknowns
    The system is the oscillating dipole and the radiation it emits, in the dipole's average-rest frame. The given period is a scalar time. Find frequency first, then wavelength.
    T=4.00×10−9 sT=4.00\times10^{-9}\ \mathrm{s}
  2. Calculate frequency
    Frequency is the reciprocal of the time for one cycle. This gives cycles per second, or hertz.
    f=1T=14.00×10−9 s=2.50×108 Hzf=\frac{1}{T}=\frac{1}{4.00\times10^{-9}\ \mathrm{s}}=2.50\times10^8\ \mathrm{Hz}
  3. Calculate wavelength
    Use the vacuum wave speed and the frequency. The result is a length.
    λ=cf=3.00×108 m/s2.50×108 s−1=1.20 m\lambda=\frac{c}{f}=\frac{3.00\times10^8\ \mathrm{m/s}}{2.50\times10^8\ \mathrm{s^{-1}}}=1.20\ \mathrm{m}
Answer: The frequency is 2.50×108 Hz2.50\times10^8\ \mathrm{Hz} and the wavelength is 1.20 m1.20\ \mathrm{m}.
Check: The reciprocal of seconds gives inverse seconds, equivalent to hertz. The wavelength has units of metres. The short period means a high frequency, so a wavelength of about one metre is reasonable at this wave speed.

Worked example

Comparing two observation directions

An ideal dipole oscillates vertically. Compare the radiation directed toward a point directly above it with radiation directed toward a point due east of it.
  1. Set the directions
    Take the dipole's average position as the origin. The vertical line is the dipole axis, and east is perpendicular to that axis. Direction is a vector idea, so the answer depends on the orientation relative to the axis. dipole axis: vertical\text{dipole axis: vertical}, east: perpendicular to axis
  2. Apply the dipole pattern
    An ideal oscillating dipole radiates least along its axis and most in directions perpendicular to it. Compare the two points using those relative directions.
    above: minimum,east: maximum\text{above: minimum},\quad \text{east: maximum}
Answer: The point directly above the dipole lies along its axis, so the ideal model predicts no radiation directed that way. The point due east is perpendicular to the axis, where radiation is strongest.
Check: The comparison uses direction relative to the dipole axis, not distance from the dipole. It describes radiation strength by direction and does not assign an unsupported numerical intensity.

Common mistakes and how to avoid them

Saying the electric and magnetic fields point along the direction the wave travels.
Correction: In the travelling electromagnetic wave, both fields are perpendicular to the travel direction and to each other.
Saying the dipole radiates most strongly along its axis.
Correction: The ideal dipole radiates most strongly perpendicular to its axis and least along the axis.
Using the dipole's direction of motion as the wave's direction.
Correction: The dipole charges move along the dipole axis, while the radiation travels outward in directions around the dipole.
Treating wavelength as a direction or frequency as a distance.
Correction: Wavelength is a scalar distance in metres. Frequency is a scalar number of cycles per second.

Lesson summary

Check your understanding

Question 1

An ideal dipole oscillates horizontally. In which direction is its radiation strongest?
  1. Horizontally, along the dipole axis
  2. Vertically, perpendicular to the dipole axis
  3. Only in the direction the positive charge moves at one instant
  4. correctIndex
Show answer and explanation
Vertically, perpendicular to the dipole axis
The ideal dipole radiates most strongly perpendicular to its axis. Its radiation is not restricted to the instantaneous motion direction of one charge.

Question 2

A dipole's oscillation frequency increases while the radiation remains in vacuum. What happens to the wavelength?
  1. It decreases
  2. It increases
  3. It stays fixed because the speed of light changes
  4. correctIndex
Show answer and explanation
It decreases
The speed in vacuum is fixed, and wavelength is speed divided by frequency. Increasing frequency therefore decreases wavelength.

Question 3

Which statement correctly describes the fields in the travelling radiation?
  1. The electric and magnetic fields are parallel to each other.
  2. The electric field is parallel to the wave's travel direction.
  3. The electric and magnetic fields are perpendicular to each other and to the travel direction.
  4. correctIndex
Show answer and explanation
The electric and magnetic fields are perpendicular to each other and to the travel direction.
Electromagnetic radiation is transverse: its electric field and magnetic field are mutually perpendicular and each is perpendicular to the direction of travel.

Key terms

Electric dipole
A pair of equal and opposite charges separated by a distance.
Dipole axis
The line joining the two charges of a dipole.
Oscillation
Repeated motion back and forth around an average position.
Electromagnetic radiation
Energy travelling through space as changing electric and magnetic fields.
Frequency
The number of complete cycles per second, measured in hertz.
Wavelength
The distance between corresponding points on successive wave cycles, measured in metres.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation E3.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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