Question

What is the dimensional formula of the permittivity of a vacuum?

Answer 1

Step 1: Definition

The substance's permittivity is a characteristic of all materials that quantifies the resistance to the production of an electric field.

The Greek alphabet $\mathrm{\epsilon }$ is used to represent it. It indicates how many charges are necessary to create one unit of electric flux throughout the specified medium.

$\mathrm{Farad}/\mathrm{meter}$ is the SI unit of permittivity.

Step 2: Dimensional formula

Since,

$\mathrm{Force}={\left[4\mathrm{\pi \epsilon }\right]}^{-1}\times \left[{\mathrm{Charge}}^2\right]\times {\left[\mathrm{Distance}\right]}^{-2}$

∴ $\mathrm{Permittivity}\left(\mathrm{\epsilon }\right)=\left[{\mathrm{Charge}}^2\right]\times {\left[\mathrm{Force}\right]}^{-1}\times {\left[\mathrm{Distance}\right]}^{-2}$…(equation $1$)

$\Rightarrow \mathrm{Charge}=\mathrm{current}\times \mathrm{time}$

∴ The dimensional formula of charge is:

$\left[{\mathrm{I}}^1{\mathrm{T}}^1\right]$…(equation $2$)

The dimensional formula of distance is:

$\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]$ …(equation $3$)

$\Rightarrow \mathrm{Force}=\mathrm{Mass}\times \mathrm{Acceleration}$

∴ The dimensional formula of force is:

$\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]$ …(equation $4$)

By substituting the values of equation $2,3,$ and $4$ in equation $1$ we get,

$\mathrm{Permittivity}={\left[\mathrm{Charge}\right]}^2\times {\left[\mathrm{Force}\right]}^{-1}\times {\left[\mathrm{Distance}\right]}^{-2}$

Or,

$$\begin{gathered} \mathrm{\epsilon }={\left[{\mathrm{I}}^1{\mathrm{T}}^1\right]}^2\times \left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]–1\times {\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]}^{-2} \\ =\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{I}}^2\right] \end{gathered}$$

Therefore, the dimensional formula of permittivity of vacuum will be $\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right]$.