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. (a) `[epsilon_(0)] = M^(-1)l^(-3)T^(2)I`
B. (b) `[epsilon_(0)] = M^(-1)l^(-3)T^(4)I^(2)`
C. ( c ) `[mu_(0)] = MLT^(-2)I^(-2)`
D. (d) `[mu_(0)] = ML^(2)T^(-1)I`

Answer

B, C

Explanation

(b,c) By defination `F = (Q_(1)Q_(2))/((4piepsilon_(0))r^(2)) and (F)/(l) = (mu_(0)I_(1)I_(2))/(2piL)`
Hence,`[epsilon_(0)] = [Q^(2)]/([F][r^(2)]) = (I^(2)T^(2))/(MLT^(-2)L^(2)) = M^(-1)L^(-3)T^(4)I^(2)`
`[mu_(0)] = ([F])/([I]^(2)) = (MLT^(-2))/(I^(2)) = ML^(-2)T^(-2)`