- Using mass(M) , length(L) ,time(T) and current (A) as fundamental quantities, the dimension of permeability is
Let us consider Coulomb’s law : F = q2 /(4πε0×r2) ;
q is charge in Coulomb, r is distance in metre, F is force in Newton and ε0 is permeability.
dimenison of ε0 = dimension of q2/(F×r2) = [Coulomb]2 /[Newton× square metre]…(1)
dimension of Coulomb = dimension of (Current×Time) = [ A T ]
dimension of Newton = [ MLT-2]
substitute the above in (1), we have
dimension of ε0 = [ A2T2 ]/ [MLT-2 × L2] = [ A2 M-1 L-3 T4 ]