How do the units of polarizability differ between SI and Gaussian systems?

[Steven Hanna]

Can someone explain how it is possible for polarizability to have units of volume?

u_induced = αE_applied

so when i divide u_induced by E_applied I get units of (C*m)/(V/m) = (C*m²)/(J/C) = (C²m²)/J

 

[drvrm]

Can someone explain how it is possible for polarizability to have units of volume?

polarizability can not be visualized as a volume element ...in cgs system of units it may come out as cube of a length but in SI system it has physical more rational units as its a ratio of dipole moment induced by the external field ; one can represent it by the ratio as far as its units are concerned.
one can use C m^2. V^-1 as SI units

 

[Steven Hanna]

polarizability can not be visualized as a volume element ...in cgs system of units it may come out as cube of a length but in SI system it has physical more rational units as its a ratio of dipole moment induced by the external field ; one can represent it by the ratio as far as its units are concerned.
one can use C m^2. V^-1 as SI units

can you explain how it is possible for the units to work out to cm^3 in CGS but not to m^3 in SI?

 

[drvrm]

can you explain how it is possible for the units to work out to cm^3 in CGS but not to m^3 in SI?

The dipole moment has the dimension of charge times distance, which in SI units is C m (coulomb . meter).

In Gaussian units dipole moment is (stat coulomb .centimeter).

An electric field has dimension voltage divided by distance,

so that in SI units E has dimension V/m and in Gaussian units stat V/cm. Hence the dimension of α is In SI: C m^2 V−1In Gaussian: statC cm^2 .statV−1 = cm^3,

where one can use that in Gaussian units the dimension of V is equal to stat C/cm (as per Coulomb's law).
for discussion one can see
http://en.citizendium.org/wiki/Polarizability