- #1
vikasagartha
- 15
- 0
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, doesn't it have to have some sort of a surface charge? How can there be no surface charge?
[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, doesn't it have to have some sort of a surface charge? How can there be no surface charge?