Problem — Speed of light from the vacuum constants

Problem

Numerically verify Maxwell’s relation $c = 1/\sqrt{\mu_0\varepsilon_0}$, using $\mu_0 = 4\pi\cdot 10^{-7}\;\text{T·m/A}$ and $\varepsilon_0 = 8{,}85\cdot 10^{-12}\;\text{C}^2/(\text{N·m}^2)$. Also show that the result has the dimensions of a velocity.

Solution

Product of the constants.

$\mu_0\varepsilon_0 = (4\pi\cdot 10^{-7})\cdot(8{,}85\cdot 10^{-12}) \approx 1{,}11\cdot 10^{-17}$

Square root and reciprocal.

$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = \frac{1}{\sqrt{1{,}11\cdot 10^{-17}}} = \frac{1}{3{,}34\cdot 10^{-9}} \approx 3{,}00\cdot 10^8\;\text{m/s}$

$\ev{c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3{,}00\cdot 10^8\;\mathrm{m/s}}$

Dimensional check. From $\mu_0$ in T·m/A and $\varepsilon_0$ in C$^2$/(N·m$^2$), recalling that $1\;\text{T} = 1\;\text{N}/(\text{A·m})$ and $1\;\text{C} = 1\;\text{A·s}$:

$[\mu_0\varepsilon_0] = \frac{\text{N}}{\text{A}^2}\cdot\frac{\text{A}^2\text{s}^2}{\text{N·m}^2} = \frac{\text{s}^2}{\text{m}^2}$

Hence $[1/\sqrt{\mu_0\varepsilon_0}] = \text{m/s}$: it really is a velocity. The fact that this number coincides with the measured speed of light was, for Maxwell, proof that light is an electromagnetic wave.

Connections

Topics: Electromagnetic waves Concepts: Maxwell’s equations Skills: Dimensional analysis