UY1: Standing Electromagnetic Waves

This page gives the UY1 working model/result for Standing Electromagnetic Waves. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Module path: Electromagnetism (UY1)
Practice: UY1 Electromagnetism Quiz
Full routing: UY1 Assessment Map
Math toolkit: Mathematics for Undergraduate Physics

1) At a glance

A standing EM wave forms when an incident wave and reflected wave superpose, creating fixed nodes and antinodes.
For a perfect conductor boundary, the tangential electric field must vanish at the surface: Eₜ = 0 at the conductor.
In the 1D plane-wave model used here:
Adjacent electric-field nodes are separated by λ/2.
Between two conducting planes separated by L, allowed wavelengths and frequencies are:

λₙ = 2L/n, fₙ = nc/2L (n = 1,2,3,dots)

Prerequisites: Electromagnetic Spectrum & Sinusoidal EM Plane Waves, Energy & Momentum In Electromagnetic Waves

2) Setup

Assume:

Perfect conductor at x = 0.
Incident plane wave travels in + x with electric field along haty.
Tangential electric field at the conductor surface must be zero.

Write

vecEᵢ = E₀cos(kx-ω t) haty, vecBᵢ = E₀/ccos(kx-ω t) hatz

Reflected field from a perfect conductor:

vecEᵣ = -E₀cos(kx + ω t) haty, vecBᵣ = E₀/ccos(kx + ω t) hatz

For a perfect conductor, the tangential electric field at the surface must be zero, so the conductor sits at an E node.
Don’t put a B node at the conductor in this simple model; E nodes coincide with B antinodes and vice versa.
If the wave is in a medium (not vacuum), replace c by the wave speed in that medium.

3) Core derivation/explanation

Superposition gives:

vecE = vecEᵢ + vecEᵣ = -2E₀sin(kx)sin(ω t) haty

vecB = vecBᵢ + vecBᵣ = 2E₀/ccos(kx)cos(ω t) hatz

So:

Electric-field nodes: sin(kx) = 0 Rightarrow x = nλ/2.
Magnetic-field nodes: cos(kx) = 0 Rightarrow x = (2n + 1)λ/4.

Thus E nodes coincide with B antinodes, and vice versa.

If a second conducting plane is placed at x = L, it must also be an E node:

L = nλ/2 (n = 1,2,3,dots)

Hence allowed modes are

λₙ = 2L/n, fₙ = c/λₙ = nc/2L

Checks (sanity)

n = 1 gives the fundamental mode: L = λ/2.
Higher modes scale linearly: fₙ = n f₁.

4) Worked example(s)

Two conducting plates are separated by L = 0.30 m.

Fundamental frequency: f₁ = dfracc2L = dfrac3.0 × 10⁸0.60 = 5.0 × 10⁸ Hz.
Third mode: f₃ = 3f₁ = 1.5 × 10⁹ Hz.

5) Practice set (with hints + answers)

For L = 0.50 m, find f₁ and f₂. Hint: use fₙ = n c/(2L). Answer: f₁ = 3.0 × 10⁸ Hz, f₂ = 6.0 × 10⁸ Hz.
In a standing wave, an E-field node is observed at x = 0.75 m and another at x = 1.25 m. Find λ. Hint: adjacent E nodes are λ/2 apart. Answer: λ = 1.0 m.
If E has a node at some plane, what does B have there? Hint: compare sine vs cosine factors. Answer: a B antinode.

6) Summary + next steps

Standing EM waves are boundary-condition problems: enforce Eₜ = 0 at conductor surfaces, then only discrete wavelengths and frequencies survive.

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