Show that, when SI units for \({\mu _0}\) and \({\varepsilon _0}\) are entered, the units given by the right-hand side of the equation in the problem above are \(m/s\).

Short Answer

Expert verified

As a result, the unit of \(\frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}\) is confirmed to be the same as the unit of speed.

Step by step solution

01

Definition of electromagnetic wave

Examples of waves transmitted by concurrent periodic changes in the strength of the electric and magnetic fields include radio waves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

02

Find the unit of speed

The speed of light permittivity and the permeability of open space are known to be connected by a formula.

\(c = \frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}\)

Where\({\varepsilon _0} = \)permeability of free space,\({\mu _0} = \)the permittivity of free space, and\(c = \)speed of light

Here SI unit of\({\varepsilon _0} = \frac{{{C^2}}}{{\;N \times {m^2}}}\)and SI unit\({\mu _0} = \frac{{T \times m}}{A}\).

In the above equation, replace\({\varepsilon _0}\) and\({\mu _0}\).

\(\begin{aligned} \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}{\rm{ }} &= \frac{1}{{\sqrt {\left( {\frac{{{C^2}}}{{\;N \times {m^2}}}} \right)\left( {\frac{{T \times m}}{A}} \right)} }}\\ &= \frac{1}{{\sqrt {\left( {\frac{{{C^2}}}{{\;N \times m}}} \right)\left( {\frac{T}{A}} \right)} }}\\ &= \sqrt {\frac{A}{{T \times F}}} \end{aligned}\)

Now substitute,\(F = \frac{{{C^2}}}{J},T = \frac{{N \times s}}{{C \times m}}\)and\(A = \frac{C}{s}\)in the above equation –

\(\begin{aligned} \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }} &= \sqrt {\frac{{\frac{C}{{\;s}}}}{{\left( {\frac{{N \times s}}{{C \times m}}} \right)\left( {\frac{{{C^2}}}{{\;J}}} \right)}}} \\ &= \sqrt {\frac{{J \times m}}{{N \times {s^2}}}} \\ &= \sqrt {\frac{{(N \times m)m}}{{N \times {s^2}}}} \\ &= \sqrt {\frac{{{m^2}}}{{{s^2}}}} \\ &= \frac{m}{s}\end{aligned}\)

Therefore, the unit of\(\frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}\)is same as the unit of speed.

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