Question

If $\mu_0$ is the permeability of free space and $\varepsilon_0$ is the permittivity of free space, then the dimension for $(\mu_0 \varepsilon_0)$ is:

• A. $[L^{-1}T]$
• B. $[MLT^{-1}]$
• C. $[L^{-2}T^{2}]$
• D. $[ML^{-1}T]$

Hint:

Instead of deriving the complicated individual dimensions of $\mu_0$ and $\varepsilon_0$ separately and multiplying them, look for an equation that links them together. The speed of light relation $c^2 = \frac{1}{\mu_0\varepsilon_0}$ makes this a $5$-second problem!

Answer 1

The Correct Option is C

Concept: From Maxwell's electromagnetic wave equations, the velocity of light ($c$) propagating through a vacuum medium is related to the fundamental constants of space by: \[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \] Squaring both sides of this equation allows us to express the product of permeability and permittivity in terms of velocity.

Step 1: Isolate the target variable product expression.
\[ c^2 = \frac{1}{\mu_0 \varepsilon_0} \implies \mu_0 \varepsilon_0 = \frac{1}{c^2} = c^{-2} \]

Step 2: Apply dimensional notation tracking.
The dimensions of velocity ($c$) are $[LT^{-1}]$. Let's find the dimensional formula for $c^{-2}$: \[ [\mu_0 \varepsilon_0] = [LT^{-1}]^{-2} = [L^{-2}T^{2}] \]