Value of Vacuum Permittivity/Proof 2

Theorem

The value of the vacuum permittivity is calculated as:

$\varepsilon_0 = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \, \mathrm F \, \mathrm m^{-1}$ (farads per metre)

with a relative uncertainty of $1 \cdotp 5 \times 10^{-10}$.

Proof

The vacuum permittivity is the physical constant denoted $\varepsilon_0$ defined as:

$\varepsilon_0 := \dfrac 1 {\mu_0 c^2}$

where:

$\mu_0$ is the vacuum permeability defined in $\mathrm H \, \mathrm m^{-1}$ (henries per metre)

$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$

$\mu_0$ has the value determined experimentally as:

$\mu_0 \approx 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6} \, \mathrm H \, \mathrm m^{-1}$

$c$ is defined precisely as:

$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$

Hence $\varepsilon_0$ can be calculated as:

\(\ds \varepsilon_0\)

\(=\)

\(\ds \dfrac 1 {\mu_0 c^2}\)

\(\ds \dfrac 1 {\mathrm H \, \mathrm m^{-1} \times \paren {\mathrm m \, \mathrm s^{-1} }^2}\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 {1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6} \times \paren {299 \, 792 \, 458}^2}\)

\(\ds \dfrac 1 {\frac {\mathrm {kg} \times \mathrm m^2} {\mathrm s^2 \times \mathrm A^2} \, \mathrm m^{-1} \times \paren {\mathrm m \, \mathrm s^{-1} }^2}\)

Base Units of Henry

\(\ds \)

\(=\)

\(\ds 8 \cdotp 85418 \, 78128 (13) \times 10^{-12}\)

\(\ds \dfrac {\mathrm A^2 \times \mathrm s^4} {\mathrm {kg} \times \mathrm m^3}\)

\(\ds \)

\(=\)

\(\ds 8 \cdotp 85418 \, 78128 (13) \times 10^{-12}\)

\(\ds \dfrac {\mathrm F} {\mathrm m}\)

Base Units of Farad

$\blacksquare$