---
title: "EM wave in dielectric from Maxwell's equations"
subject: "Engineering Physics"
chapter: 9
key: "semester_1/engineering_physics:maxwell-em-wave-dielectric-velocity"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/maxwell-em-wave-dielectric-velocity
site: BCT Engineering
---

# EM wave in dielectric from Maxwell's equations

Engineering Physics · Chapter 9: Electromagnetic Waves

- Asked in: 2074 Asoj, 5 marks; 2071 Shrawan, 5 marks (asked 2 times)
- Video: [Deriving speed of light using Maxwell's equations | EM waves | Physics | Khan Academy](https://www.youtube.com/watch?v=O9YioslUtyI) — Khan Academy India - English
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/maxwell-em-wave-dielectric-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-9

## Question

What are Maxwell's equations? Using Maxwell equations derive the electromagnetic (em) wave equation (plane wave equation for E-field and B-field) in dielectric (non-conducting) medium. Prove that the em wave travels with velocity less than the velocity of light in such medium.

## Short answer

Starting from Maxwell's equations in a dielectric (ρ=0, J=0), taking curls yields ∇²E = με ∂²E/∂t² and ∇²B = με ∂²B/∂t². Wave speed v = 1/√(με) = c/√(μ_r ε_r) < c because ε_r > 1.

## Answer

#### Maxwell's equations
Four basic laws of electromagnetism: Gauss's law for E, Gauss's law for B, Faraday's law and the Ampere-Maxwell law.

For a **non-conducting dielectric** (ρ = 0, J = 0, permittivity ε = ε₀ε_r, permeability μ = μ₀μ_r):

1. ∇·E = 0
2. ∇·B = 0
3. ∇×E = −∂B/∂t
4. ∇×B = με ∂E/∂t

#### Wave equation for E
1. Take the curl of (3): ∇×(∇×E) = −∂/∂t (∇×B).
2. Use ∇×(∇×E) = ∇(∇·E) − ∇²E = −∇²E, and put (4) on the right side:

   −∇²E = −με ∂²E/∂t²
3. So **∇²E = με ∂²E/∂t²**.

#### Wave equation for B
Take the curl of (4) and use (3) and ∇·B = 0 in the same way:

**∇²B = με ∂²B/∂t²**

#### Plane-wave form
For a wave travelling along x, with E along y and B along z:

∂²E_y/∂x² = με ∂²E_y/∂t², ∂²B_z/∂x² = με ∂²B_z/∂t²

Comparing with the standard wave equation ∂²y/∂x² = (1/v²) ∂²y/∂t²:

**v = 1/√(με)**

Solutions: E_y = E₀ sin(kx − ωt), B_z = B₀ sin(kx − ωt), with E₀/B₀ = v.

#### Speed less than c
```text
c = 1/√(μ₀ε₀)
v = 1/√(μ₀μ_r ε₀ε_r) = c/√(μ_r ε_r) = c/n,   n = √(μ_r ε_r)
```
For a dielectric ε_r > 1 and μ_r ≈ 1, so n > 1. Therefore **v < c**: the wave is slower in the medium (e.g. glass n ≈ 1.5, so v ≈ 2×10⁸ m/s).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which set applies in a non-conducting dielectric with ρ = 0 and J = 0?
   A. ∇×B = με ∂E/∂t and ∇·E = 0
   B. ∇×B = μ₀J and ∇·E = ρ/ε
   C. ∇×B = 0 and ∇·E = ∂B/∂t
   D. ∇×E = με ∂B/∂t and ∇·B = ρ
   Answer: A. With no free charge or current the Ampere-Maxwell law keeps only the displacement term, με ∂E/∂t. (Key point: Equations for dielectric)

2. Which vector identity is used in the derivation of ∇²E = με ∂²E/∂t²?
   A. ∇×(∇×E) = ∇²E + ∇·E
   B. ∇×(∇×E) = ∇(∇·E) − ∇²E, with ∇·E = 0
   C. ∇·(∇×E) = ∇²E
   D. ∇×(∇×E) = (∇·E)²
   Answer: B. Taking the curl of Faraday's law and using this identity with ∇·E = 0 gives −∇²E on the left. (Key point: Deriving E wave equation)

3. Comparing ∂²E/∂x² = με ∂²E/∂t² with the standard wave equation gives the speed:
   A. v = √(με)
   B. v = 1/(με)
   C. v = 1/√(με)
   D. v = με
   Answer: C. The standard form has 1/v² as the coefficient, so 1/v² = με. (Key point: Wave speed in medium)

4. Why is the speed of the EM wave in a dielectric less than c?
   A. Because ε_r < 1 in a dielectric
   B. Because μ_r is much less than 1
   C. Because the frequency decreases to zero
   D. v = c/n with n = √(μ_r ε_r) > 1 because ε_r > 1 and μ_r ≈ 1
   Answer: D. v = c/√(μ_r ε_r), and ε_r > 1 for a dielectric makes n > 1. (Key point: Why v < c)

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