The unit of permittivity of free space `epsilon_(0)` is:

Options

A. `"coulomb"/"newton-metre"`
B. `"newton-metre"^(2)/"coulomb"^(2)`
C. `"coulomb"^(2)/"newton-metre"^(2)`
D. `"coulomb"^(2)/("newton-metre")^(2)`

Answer

C

Explanation

According to Coulomb's law, the electrostatic force
`F = (1)/(4pi epsilon_(0)) xx (q_(1)q_(2))/(r^(2))`
`q_(1)` and `q_(2) =` charges, `r =` distance between charges and `epsilon_(0) =` permittivity of free space
`rArr epsilon_(0) = (1)/(4pi) xx (q_(1)q_(2))/(r^(2)F)`
Substituting the units for q,r and F, we obtain unit of `epsilon_(0)`
`=("coulomb" xx "coulomb")/("newton"-("metre")^(2)) = (("coulomb")^(2))/("newton"-("metre")^(2))`