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SI unit of permittivity of free space is...

SI unit of permittivity of free space is

A

`"Newton metre"^2// "Coulomb"^2`

B

`"Coulomb"^2 //"Newton metre"^2`

C

`"Coulomb"^2// "(Newton metre)"^2`

D

Coulomb/Newton metre

Text Solution

Verified by Experts

The correct Answer is:
B
`F=(q_1q_2)/(4piepsilon_0r^2) rArr [epsilon_0]=("Coulomb"^2)/("Newton Meter"^2)`
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