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Volume charge density w/o surface charge density

  1. Mar 27, 2014 #1
    Im confused by a concept i have run across in Griffiths electrodynamics.

    [itex] E_{out} - E_{in} = \frac{\sigma_{free}}{\epsilon_0}[/itex]

    However, in the case of a uniform, circular charge density,
    [itex]\vec{E_{in}} = \frac{\rho r}{3\epsilon_0}\hat{r}[/itex]
    [itex]\vec{E_{out}} = \frac{\rho R^3}{3\epsilon_0 r^2}\hat{r}[/itex]

    But this electric field is continuous @ r=R. If a volume has a charge density, doesnt it have to have some sort of a surface charge? How can there be no surface charge?
     
  2. jcsd
  3. Mar 27, 2014 #2
    Silly question. Realized my mistake. Sorry. I forgot that the normal to the electric field is in opposite directions when drawing a 'pillbox.' So sorry.

    Answered my own question.
     
    Last edited: Mar 27, 2014
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