By expressing \(\epsilon_{0}\) and \(\mu_{0}\) in base SI units $(\mathrm{kg}, \mathrm{m}, \mathrm{s}, \mathrm{A})$ prove that the only combination of the two with dimensions of speed is $\left(\boldsymbol{\epsilon}_{0} \mu_{0}\right)^{-1 / 2}$.

Recall the definitions of the constants: - Permittivity of free space, \(\epsilon_0 = 8.854 \times 10^{-12} \frac{\mathrm{C}^2}{\mathrm{N} \cdot \mathrm{m}^2}\) - Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \frac{\mathrm{N}}{\mathrm{A}^2}\) Convert the ratio of Coulombs squared per Newton meter squared (C²/N·m²) for permittivity and Newtons per Ampere squared (N/A²) for permeability to base SI units: - \(\epsilon_0 = 8.854 \times 10^{-12} \frac{(\mathrm{A} \cdot \mathrm{s})^2}{\mathrm{kg} \cdot \mathrm{m}/\mathrm{s}^2 \cdot \mathrm{m}^2}\) - \(\mu_0 = 4\pi \times 10^{-7} \frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2\cdot \mathrm{A}^2}\) Now, the dimensions of these constants in base SI units are: - \(\epsilon_0 = \frac{\mathrm{A}^2 \cdot \mathrm{s}^2}{\mathrm{kg} \cdot \mathrm{m}}\) - \(\mu_0 = \frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2\cdot \mathrm{A}^2}\)

According to the given expression to find, we have: Speed = \(\left(\epsilon_{0} \mu_{0}\right)^{-1 / 2}\) Let's analyze the dimensions of the product \(\epsilon_{0} \mu_{0}\): \(\epsilon_{0} \mu_{0} = \left(\frac{\mathrm{A}^2 \cdot \mathrm{s}^2}{\mathrm{kg} \cdot \mathrm{m}}\right) \left(\frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2\cdot \mathrm{A}^2}\right)\) Simplify the expression: \(\epsilon_{0} \mu_{0} = \frac{\mathrm{s}^2}{\mathrm{s}^2} = 1\) Now, let's analyze the dimensions of the expression \(\left(\epsilon_{0} \mu_{0}\right)^{-1 / 2}\): \(=\left(1\right)^{-1 / 2} = 1\) However, since dimensional analysis does not provide an expression for the speed dimensions, this does not prove the given expression. We should try to find a combination of \(\epsilon_0\) and \(\mu_0\) that results in the dimensions of speed. Let speed = \(k \cdot (\epsilon_{0} \mu_{0})^n\) where k is a constant and n is the power to be determined. We want to find a value for n such that the dimensions of the expression are equal to the dimensions of speed (m/s). Let's analyze the dimensions of the expression \((\epsilon_{0} \mu_{0})^n\): \((\epsilon_{0} \mu_{0})^n = \left(\frac{\mathrm{A}^2 \cdot \mathrm{s}^2}{\mathrm{kg} \cdot \mathrm{m}} \cdot \frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2\cdot \mathrm{A}^2}\right)^n\) \(= \left(\frac{\mathrm{s}^2}{\mathrm{s}^2}\right)^n = 1^n = 1\) We see that the dimensions of the expression \((\epsilon_{0} \mu_{0})^n\) do not depend on n. Therefore, the given expression, \(\left(\epsilon_{0} \mu_{0}\right)^{-1 / 2}\) does not have dimensions of speed. The given exercise does not seem to be correct.