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

Options

A. `[varepsilon_0]=[M^(-1)L^(-3)T^(4)A^(2)]`
B. `[varepsilon_0]=[M^(-1)L^(2)T^(-1)A^(-2)]`
C. `[varepsilon_0]=[M^(-1)L^(2)T^(-1)A]`
D. `[varepsilon_0]=[M^(-1)L^(-3)T^(2)A]`

Answer

A

Explanation

From the coulomb.s law
`varepsilon_(0)=(1)/(4pi)((q^2)/(Fr^2))`
`=(1)/(4pi)((I^(2)*t^(2))/(F*r^(2)))`
`=(A^(2)T^(2))/(MLT^(-2)L^(2))`
`varepsilon_(0)= [M^(-1)L^(-3)T^(4)A^(2)]`