Question

Show that \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\) gives the velocity of an electromagnetic wave in free space.

Hint:

You can also verify this using dimensional analysis. The dimensions of \(\varepsilon_0\) are \([\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2]\) and for \(\mu_0\) they are \([\text{M L T}^{-2}\text{A}^{-2}]\). Multiplying them gives \([\text{L}^{-2}\text{T}^2]\). Taking the reciprocal square root yields \([\text{L T}^{-1}]\), which matches the dimensions of velocity.

Answer 1

Concept: From Maxwell's electromagnetic equations, the speed of an electromagnetic wave through free space is fundamentally linked to the experimental constants of the medium: the permeability of free space (\(\mu_0\)) and the permittivity of free space (\(\varepsilon_0\)).

Step 1: Listing values of fundamental constants.
The standard universal values for these free-space constants in SI units are:
• Permittivity of free space: \(\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\cdot\text{N}^{-1}\cdot\text{m}^{-2}\)
• Permeability of free space: \(\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m}\cdot\text{A}^{-1}\) We know from Coulomb's constant that: \[ \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \quad \Rightarrow \quad \varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} \]

Step 2: Performing algebraic substitution.
Let us compute the product of \(\mu_0\) and \(\varepsilon_0\): \[ \mu_0 \varepsilon_0 = (4\pi \times 10^{-7}) \times \left( \frac{1}{4\pi \times 9 \times 10^9} \right) \] Canceling the common factor of \(4\pi\) from both the numerator and the denominator: \[ \mu_0 \varepsilon_0 = \frac{10^{-7}}{9 \times 10^9} = \frac{1}{9 \times 10^9 \times 10^7} = \frac{1}{9 \times 10^{16}} \]

Step 3: Taking the square root and reciprocal.
Now, let's take the square root of this product: \[ \sqrt{\mu_0 \varepsilon_0} = \sqrt{\frac{1}{9 \times 10^{16}}} = \frac{1}{3 \times 10^8} \] Taking the reciprocal of this value gives: \[ \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3 \times 10^8 \] The numerical value \(3 \times 10^8 \text{ m/s}\) corresponds exactly to the experimentally measured speed of light \(c\) in a vacuum. Hence, this proves that \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}} = c\).