Question

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

Hint:

Using the fractional form of $\varepsilon_0$ (via $\frac{1}{4\pi\varepsilon_0}$) makes this calculation incredibly fast and avoids dealing with the messy decimal $8.854 \times 10^{-12}$.

Answer 1

Concept:
• According to Maxwell's electromagnetic theory, the speed of light (an electromagnetic wave) in a vacuum is dictated by the fundamental electric and magnetic properties of free space.
• These properties are the permittivity of free space ($\varepsilon_0$) and the permeability of free space ($\mu_0$).

Step 1: Identify the fundamental constants
The permeability of free space, representing the magnetic capability of vacuum, has a standard value: \[ \mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1} \]
The permittivity of free space, representing the electrostatic capability of vacuum, is derived from Coulomb's constant ($\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{C}^{-2}$): \[ \varepsilon_0 = \frac{1}{4\pi \times (9 \times 10^9)} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2} = \frac{1}{36\pi \times 10^9} \text{ F m}^{-1} \]

Step 2: Substitute values into the expression
We need to evaluate the expression $\frac{1}{\sqrt{\varepsilon_0 \mu_0}}$. Let's substitute the known values: \[ \varepsilon_0 \mu_0 = \left( \frac{1}{36\pi \times 10^9} \right) \times (4\pi \times 10^{-7}) \]

Step 3: Simplify the product
Cancel out the $\pi$ terms and simplify the fraction: \[ \varepsilon_0 \mu_0 = \frac{4\pi \times 10^{-7}}{36\pi \times 10^9} \]
\[ \varepsilon_0 \mu_0 = \frac{4}{36} \times \frac{10^{-7}}{10^9} \]
\[ \varepsilon_0 \mu_0 = \frac{1}{9} \times 10^{-16} \]

Step 4: Calculate the final velocity
Now, take the square root of this product and invert it: \[ \sqrt{\varepsilon_0 \mu_0} = \sqrt{\frac{1}{9} \times 10^{-16}} \]
\[ \sqrt{\varepsilon_0 \mu_0} = \frac{1}{3} \times 10^{-8} \]
Finally, find the reciprocal: \[ v = \frac{1}{\sqrt{\varepsilon_0 \mu_0}} = \frac{1}{\frac{1}{3} \times 10^{-8}} \]
\[ v = 3 \times 10^8 \text{ m/s} \]

Step 5: Conclusion
The resulting value, $3 \times 10^8 \text{ m/s}$, perfectly matches the known velocity of an electromagnetic wave (light) in free space, conventionally denoted as $c$. Thus, it is proved that $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$.