Question

Show that units of $\dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} = m{s^{ - 1}}$?

Answer 1

Hint: In this question, we will first know the S.I units of each term i.e., permittivity and permeability. Now, by using these S.I units and substituting in the given expression, we will get the required R.H.S. further, we will know the basic definition of each term and we will know the difference between speed and velocity. Also, at last we will see the short trick to prove the given expression easily.

Formula used:
$\dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} = m{s^{ - 1}}$

Complete step by step solution:
As we know that ${\varepsilon _0}$ is the permittivity of free space and the S.I unit is given as farad per meter: $F{m^{ - 1}}$

Now, we also know that the unit of farad in S.I units‎ is given a‎s ${s^4}{A^2}{m^{ - 2}}k{g^{ - 1}}$

So, now when we substitute this value in the S.I units of permittivity, we get:
${s^4}{A^2}{m^{ - 3}}k{g^{ - 1}}........(1)$

As we know that${\mu _0}$ is the permeability of free space and the S.I unit is given as farad per meter:$H{m^{ - 1}}$

Now, we also know that the unit of farad in S.I units‎ is given a‎s $kg.{m^2}.{s^{ - 2}}.{A^{ - 2}}$.

So, now when we substitute this value in the S.I units of permittivity, we get:
$kg.m.{s^{ - 2}}.{A^{ - 2}}.......(2)$

Now, we will substitute these units in the L.H.S of the given equation:
$\dfrac{1}{{\sqrt {\left( {{s^4}{A^2}{m^{ - 3}}k{g^{ - 1}}} \right)\left( {kg.m.{s^{ - 2}}.{A^{ - 2}}} \right)} }}$
$ \Rightarrow \dfrac{1}{{\sqrt {{m^2}.{s^{ - 2}}} }}$
$\therefore m{s^{ - 1}}$

Therefore, we have proved L.H.S=R.H.S i.e.,$\dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} = m{s^{ - 1}}$ , and hence obtained the expression.

Additional information:
As we know that the permittivity can be defined as the measure of the opposition that is offered by any material against the formation of an electric field.
Also, we know that the permeability can be defined as the measure of the extent up to which the magnetic field lines can enter any substance or material.
We already know the basic difference between speed and velocity i.e., speed is given as the measure of how fast an object can travel, whereas velocity gives us the direction of this speed. Also, speed is a scalar quantity which means that it has only magnitude, whereas velocity is a vector quantity which means that it has both magnitude and direction. The S.I unit of both speed and velocity is meter per second (m/sec).

Note:
We can also, solve this question as we know that $\dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} = c$, and c is speed of light. Here, we know that the unit of speed is m/s, therefore we can directly answer the question. Also, we should remember that no other speed can be equal or faster than the speed of light.