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Units for flux density

  1. May 28, 2013 #1
    While I do understand the use for the displacement vector D = (e0)E + P, I don't quite understand why they label it as "flux density". The units for the displacement vector are C/m^2. The units for electric flux are Vm, so wouldn't electric flux density become Vm/m^2 = V/m which is just equal to the electric field?

    edit: C/m^2 just looks like surface charge density...
  2. jcsd
  3. May 28, 2013 #2


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    For this reason, I hate the SI. It's totally unintuitive. Anyway, let's work through the dimensional analysis. According to Gauß's law, which reads in differential form
    [tex]\vec{\nabla} \cdot \vec{D}=\rho,[/tex]
    where [itex]\rho[/itex] is the charge density (i.e., charge per unit volume), which has the SI unit [itex]\mathrm{C}/\mathrm{m}^3[/itex] (Coulomb per cubic metre). The Nabla operator is a differential operator wrt. to space coordinates and thus has the unit [itex]1/\mathrm{m}[/itex].

    This implies that [itex]\vec{D}[/itex] has the unit [itex]\mathrm{C}/\mathrm{m}^2[/itex].
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