Question

calculate permittivity of free space

Answer 1

Suppose two point charges q₁ and q₂ are separated in vacuum by a distance r.
Then according to Coulomb's law :-
$F=\frac{k\left|q_1\right|\left|q_2\right|}{r^r}$ where k is electrostatic force constant = 9 x 10⁹ Nm²C^-2
Here, k = $\frac{1}{4\pi \epsilon 0}$ where ε₀ is called absolute electrical permittivity of the free space.
Therefore, ε_{0 = $\frac{1}{4\pi k}$= $\frac{1}{4X3.14X9X{10}^9}=8.85X{10}^{-12}{N}^{-1}{C}^2{m}^{-2}$}