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Let [epsilon(0)] denote the dimensional ...

Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of vacuum. If `M`=mass,`L`=length, `T`=time and `A`= electric current, then :

A

`[epsilon_(0)]=[M^(-1)L^(-3)T^(4)A^(2)]`

B

`[epsilon_(0)]=[M^(-1)L^(2)T^(-1)A^(-2)]`

C

`[epsilon_(0)]=[M^(-1)L^(2)T^(-1)A]`

D

`[epsilon_(0)]=[M^(-1)L^(-3)T^(2)A]`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(4piepsilon_(0))(q^(2))/(r^(2))=F`
`epsilon_(0)=([A^(2)T^(2)])/([MLT^(-2)L^(2)])=[M^(-1)L^(-3)A^(2)T^(4)]`
Hence correct choice is `(a)`.
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