What is the value of the permittivity of free space, \( \varepsilon_0 \), in SI units?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \)
The permittivity of free space \( \varepsilon_0 \) enters Coulomb’s law through \( k=1/4\pi\varepsilon_0 \). The number \( 9\times10^{9} \) (in \( \mathrm{N\,m^{2}\,C^{-2}} \)) is the value of \( k \).
NCERT (2026-27):
Coulomb’s Law – “… ε0 is called the permittivity of free space. The value of ε0 in SI units is”
\( \varepsilon_0=8.854\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \)