Question

The speed of light in vacuum is defined to be c = 299,792,458 m/s = 1 μ0ε0 . The permeability constant of vacuum is defined to be μ0 = 4π ✕ 10−7 N · s2/C2. Use these definitions to calculate the value of ε0, the permittivity of free space, to at least eight significant figures. (Enter your answer in C2/(N · m2).) HINT

Answer 1

The permittivity of free space is about :

8.8541878 × 10⁻¹² C²/Nm²

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Further explanation

Let's recall the speed of light formula in vacuum as follows:

$\boxed {c = \frac{1}{\sqrt{\mu_o \varepsilon_o}}}$

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!

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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 :

$c = \frac{1}{\sqrt{\mu_o \varepsilon_o}}$

$\sqrt{\mu_o \varepsilon_o} = \frac{1}{c}$

$\mu_o \varepsilon_o = = \frac{1}{c^2}$

$\varepsilon_o = \frac{1}{\,u_o c^2}$

$\varepsilon_o = \frac{1}{4\pi \times 10^{-7} \times (299 \ 792 \ 458)^2}$

$\boxed {\varepsilon_o = 8.8541878 \times 10^{-12} \texttt{ C}^2/\texttt{Nm}^2}$ → (rounded to eight significant figures)

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Conclusion :

The permittivity of free space is about :

8.8541878 × 10⁻¹² C²/Nm²

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Answer details

Grade: High School

Subject: Physics

Chapter: Light

Answer 2

Explanation:

It is given that,

The speed of light in vacuum is, c = 299,792,458 m/s

The permeability constant of vacuum is, $\mu_o=4\pi\times 10^{-7}\ N.s^2/C^2$

Let $\epsilon_o$ is the permittivity of free space. The relation between $\mu_o,\epsilon_o\ and\ c$ is given by :

$c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}$

$\epsilon_o=\dfrac{1}{c^2u_o}$

$\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}$

$\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2$

Hence, this is the required solution.