Question 13 Marks
Write the S.I. unit and dimensional formula for the permittivity of free space ?
Answer
View full question & answer→The basic formula that relates permittivity of free space with the electrical force between two charges is :
$F=(1/4~\pi\in_{0}).q_{1}.q_{2}/r^{2}$
Where F is the electric force between two charges, $q_{1}$, $q_{2}$ are the charges and r is the distance between the charges.
$\in_{0}$ is the permittivity of free space.
$Unit=C^{2}/Nm^{2}$
The dimensional formula of force is [MLT-2].
The dimensional formula of charge (q) is [AT].
The dimensional formula of distance (r) is [L].
Assembling all the formulas, we get the dimensional formula of $\in_{0}$ is $[m^{-1}L^{-3}T^{4}A^{2}]$
$F=(1/4~\pi\in_{0}).q_{1}.q_{2}/r^{2}$
Where F is the electric force between two charges, $q_{1}$, $q_{2}$ are the charges and r is the distance between the charges.
$\in_{0}$ is the permittivity of free space.
$Unit=C^{2}/Nm^{2}$
The dimensional formula of force is [MLT-2].
The dimensional formula of charge (q) is [AT].
The dimensional formula of distance (r) is [L].
Assembling all the formulas, we get the dimensional formula of $\in_{0}$ is $[m^{-1}L^{-3}T^{4}A^{2}]$