How does the speed of light relate to Maxwell’s equations?

By Tutor GuyNo Comments


You may have seen in your physics class that the speed of light can be calculated from the permittivity of free space, \varepsilon_0 , and the permeability of free space, \mu_0 , as follows:

c = \dfrac{1}{\sqrt{\varepsilon_0 \mu_0}}

where  \varepsilon_0 = 8.85 \; C^2/N \cdot m^2 and  \mu_0 = 4 \pi \cdot 10^{-7} \; T \cdot m/A .

Sure enough, plug the numbers into your calculator and you do get the correct value for the speed of light. It’s a very famous result, and helped convince Maxwell that light is an electromagnetic wave. But the units just don’t seem to work out, do they? Well, actually they do. Let’s do the dimensional analysis. We want to show the following:

\dfrac{m}{s} = \dfrac{1}{\sqrt{\dfrac{C^2}{N \cdot m^2} \times \dfrac{T \cdot m}{A}}} \; \; (1)

From the equation for the magnetic force on a charged particle:

F = qvB \; \rightarrow \; B = \dfrac{F}{qv}

If we express this equation in terms of its units and simplify we have

T = \dfrac{N}{C \cdot m/s} = \dfrac{N \cdot s}{C \cdot m}

We also note that 1 A = 1 C/s, so we substitute for T and A in equation (1) above to get:

\dfrac{m}{s} = \dfrac{1}{\sqrt{\dfrac{C^2}{N \cdot m^2} \times \dfrac{N \cdot s}{C \cdot m} \dfrac{m \cdot s}{C}}} = \dfrac{1}{\sqrt{\dfrac{s^2}{m^2}}} = \dfrac{m}{s}

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