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

Options

A. `[epsilon _(0) ] = [M^(-1)L^(-3)T^(2)I]`
B. ` [epsilon_(0) ] = [M^(-1)L^(-3)T^(4)I^(2)]`
C. ` [mu_(0)] = [MLT^(-2)I^(-2)] `
D. `[mu_(0)] = [ML^(2)T^(-1)I]`

Answer

B, C

Explanation

` F = (1)/(4 pi epsilon _(0)) (q_(1) q_(2))/(r^(2))`
` [epsilon_(0)] = ([q_(1)][q_(2)])/([F][r^(2)]) = ([IT]^(2))/([MLT^(-2)][L^(2))]`
`= [M^(-1)L^(-2)T^(4)I^(2)]`
speed of light , `c= (1)/(sqrt epsilon _(0)mu_(0)) `
` :. [mu_(0)] = (1)/([epsilon _(0)] [c]^(2)) = (1)/([M^(-1)L^(-3)T^(4)I^(2)][LT^(-1)]^(2)) `
= [MLT^(-2)T^(-2)]`