1.

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

Answer»

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

Solution :`(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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