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

Let |epsi_(0)| denote the dimensional formula of the permittivity of v

1.	Let \|epsi_(0)\| denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then
Answer» `[epsi_(0)]=[M^(-1)L^(-3)T^(2)A]` `[epsi_(0)]=[M^(-1)L^(-3)T^(4)A^(2)]` `[epsi_(0)]=[M^(-1)L^(2)T^(-1)A^(-2)]` `[epsi_(0)]=[M^(-1)L^(2)T^(-1)A]` SOLUTION :`F=(q_(1)q_(2))/(4PI epsi_(0)^(r)) rArr epsi_(0)=(q_(1)q_(2))/(4pi Fr^(2))` `:.[epsi_(0)]=([q_(1)][q_(2)])/([F][r^(2)]) "" [ :. 4pi` is dimensionless] `=((AT)(AT))/((M^(1)L^(1)T^(-2))(L^(2)))` `=M^(-1)L^(-3)T^(-4)A^(2)`