Explanation:
It is given that,
The speed of light in vacuum is, c = 299,792,458 m/s
The permeability constant of vacuum is, [tex]\mu_o=4\pi\times 10^{-7}\ N.s^2/C^2[/tex]
Let [tex]\epsilon_o[/tex] is the permittivity of free space. The relation between [tex]\mu_o,\epsilon_o\ and\ c[/tex] is given by :
[tex]c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}[/tex]
[tex]\epsilon_o=\dfrac{1}{c^2u_o}[/tex]
[tex]\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}[/tex]
[tex]\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2[/tex]
Hence, this is the required solution.
The permittivity of free space is about :
[tex]\texttt{ }[/tex]
Let's recall the speed of light formula in vacuum as follows:
[tex]\boxed {c = \frac{1}{\sqrt{\mu_o \varepsilon_o}}}[/tex]
where:
c = speed of light in vacuum ( m/s )
μo = the permeability constant of vacuum
εo = the permittivity of free space
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
speed of light in vacuum = c = 299 792 458 m/s
the permeability constant of vacuum = μo = 4π × 10⁻⁷ N s²/C²
Asked:
the permittivity of free space = εo = ?
Solution:
We could calculate the value of εo by using following formula :
[tex]c = \frac{1}{\sqrt{\mu_o \varepsilon_o}}[/tex]
[tex]\sqrt{\mu_o \varepsilon_o} = \frac{1}{c}[/tex]
[tex]\mu_o \varepsilon_o = = \frac{1}{c^2}[/tex]
[tex]\varepsilon_o = \frac{1}{\,u_o c^2}[/tex]
[tex]\varepsilon_o = \frac{1}{4\pi \times 10^{-7} \times (299 \ 792 \ 458)^2}[/tex]
[tex]\boxed {\varepsilon_o = 8.8541878 \times 10^{-12} \texttt{ C}^2/\texttt{Nm}^2}[/tex] → (rounded to eight significant figures)
[tex]\texttt{ }[/tex]
The permittivity of free space is about :
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Light
