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I Units of polarizability

  1. Oct 2, 2016 #1
    Can someone explain how it is possible for polarizability to have units of volume?

    uinduced = αEapplied

    so when i divide u_induced by E_applied I get units of (C*m)/(V/m) = (C*m2)/(J/C) = (C2m2)/J
     
  2. jcsd
  3. Oct 3, 2016 #2
    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
     
  4. Oct 3, 2016 #3
    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?
     
  5. Oct 3, 2016 #4
    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−1


    In 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
     
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