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

By coulomb's law the electrostatic force
`F = (1)/(4 pi epsilon_(0)) xx (q_(1)q_(2))/(r^(2)) rArr epsilon_(0) = (1)/(4 pi) xx (q_(1)q_(2))/(r^(2)F)`
Substituting the unit for q, r,and F , we abtain unit of
`epsilon_(0) = ("coulomb" xx "coulomb")/("newton" - ("metre")^(2)) = (("coulumb")^(2))/("newton" - ("meter")^(2))`
`= C^(2) N -m^(2)`
Methed 2: `F = (1)/(4 pi epsilon_(0)) (Q_(1)Q_(2))/(r^(2)) rArr epsilon_(0) oo (Q^(1))/(F xx r^(2))`
So `epsilon_(0)` has unit of `"coulumb"^(2) "newton" - m^(2)`