Q.Write the unit and dimensional formula of permittivity of free space.

From Coulomb's law, ε₀ = q₁q₂/(4πF r²) → units C²/(N·m²) = C² N⁻¹ m⁻².

Coulomb's law is $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$, so

$\varepsilon_0 = \dfrac{q_1 q_2}{4\pi F r^2}$

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The permittivity of free space (epsilon_0) is the fundamental constant appearing in Coulomb's law; its correct SI value is 8.854 x 10^-12 C^2 N^-1 m^-2.

Coulomb's law is written as $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2} = k\dfrac{q_1q_2}{r^2}$, where $k = \dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}$.

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$\varepsilon_0 = 8.85\times10^{-12}$ F/m.

The permittivity of free space (electric constant) appears in Coulomb's law and in the relation $c = 1/\sqrt{\mu_0\varepsilon_0}$. Its accepted value is

$\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2} = 8.85\times10^{-12}\ \text{F/m}.$

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Relative permittivity ε_r = ε/ε₀, a dimensionless ratio.

The absolute permittivity ε of a medium measures how easily it permits electric field lines through it. To compare a medium with vacuum, we define the relative permittivity (also called the dielectric constant K):

$\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$

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The relative permittivity (dielectric constant) of air ≈ vacuum ≈ 1.

Relative permittivity $\varepsilon_r = \varepsilon/\varepsilon_0$ is the ratio of a medium's permittivity to that of free space. It measures how much the medium reduces the electric force (or field) compared with vacuum:

$F = \frac{1}{4\pi\varepsilon_0 \varepsilon_r}\frac{q_1 q_2}{r^2}.$

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