Volume charge density w/o surface charge density (1 Viewer)

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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?
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:

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