Question

The unit of \(\dfrac{1}{\mu_0 \varepsilon_0}\) is:

• A. \(\text{m s}^{-1}\)
• B. \(\text{m}^{-1}\,\text{s}\)
• C. \(\text{m}^{2}\,\text{s}^{-2}\)
• D. \(\text{m}^{-2}\,\text{s}^{2}\)

Hint:

Recall \(c = 1/\sqrt{\mu_0\varepsilon_0}\), so \(1/(\mu_0\varepsilon_0)\) equals \(c^2\). Find the unit of speed squared.

Answer 1

The Correct Option is C

Step 1: Concept. The speed of light in vacuum is related to the permeability \(\mu_0\) and permittivity \(\varepsilon_0\) of free space by \[c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}.\]
Step 2: Square both sides. \[c^2 = \frac{1}{\mu_0 \varepsilon_0}.\] So \(\dfrac{1}{\mu_0 \varepsilon_0}\) has exactly the units of \(c^2\), i.e. the square of a speed.
Step 3: Substitute the unit of speed. Speed has unit \(\text{m s}^{-1}\), therefore \(c^2\) has unit \((\text{m s}^{-1})^2 = \text{m}^2\,\text{s}^{-2}\).
Step 4: Hence the unit of \(\dfrac{1}{\mu_0 \varepsilon_0}\) is \(\text{m}^2\,\text{s}^{-2}\), which is option (iii).
Why other options are wrong: Option (i) is the unit of speed \(c\) itself, not \(c^2\). Options (ii) and (iv) are reciprocal-type units that do not match \(c^2\).
\[\boxed{\text{m}^2\,\text{s}^{-2}}\]