Surface Charge of Uniformly charged sphere In the section on the discontinuity of the electric field (Griffiths Introduction to electrodynamics), there is the following statement:

Where there is no surface charge, the perpendicular component of the electric field (to the surface) is continuous, as for instance at the surface of a uniformly charged solid sphere.

Why does the surface of a uniformly charged solid sphere have no charge on it?
 A: Surface charge refers to a thin layer of charge at a surface.  For example, an ideal charged conducting sphere would have a layer of charge on the surface whose thickness would be infinitesimal.  It would be represented by a delta function.
In the case of an idealized non-conducting uniformly charged sphere whose boundary is discreet, the charge density simply goes from a finite value to zero.  No surface charge.
If we relax the restriction to ideal spheres, the conducting sphere has a charge density that spikes at the surface, while for a "uniformly charged" sphere, the charge density would drop smoothly to zero without a spike.  (But it might have some structure depending on what model you use for the material and the electrons.)
A: Take a uniformly charged solid sphere of radius $R$ and total charge $Q$. Electric field $\mathbf E$ at a radial distance $r$ is radially directed as there is no other unique direction. So,
$$\mathbf E(\mathbf r)~=\begin{cases}\dfrac1{4\pi\varepsilon_0}\dfrac{Q}{r^2}~\hat{\mathbf r}~~~~r\geq R\\ \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qr}{R^3}~\hat{\mathbf r}~~~r\leq R\end{cases}$$
What happens when $r=R\,?$ 
$$\mathbf E_\textrm{above} =\dfrac1{4\pi\varepsilon_0}\dfrac{Q}{R^2}~\hat{\mathbf R}$$
and $$\mathbf E_\textrm{below} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{QR}{R^3}~\hat{\mathbf R}= \dfrac1{4\pi\varepsilon_0}\dfrac{Q}{R^2}~\hat{\mathbf R}.$$
Thus, $$\mathbf E_\textrm{above}- \mathbf E_\textrm{below} = \mathbf 0.$$
What does that mean?
A: The citation of Griffith obviously refers to the boundary condition for the electric field at the interface between uniformly charged solid sphere (with permittivity $\epsilon_0$) and vacuum (also permittivity $\epsilon_0$). If you consider a uniformly charged insulator sphere with absolute permittivity $\epsilon_1$ embedded into an outer insulating dielectric with $\epsilon_2$ (this can also be vacuum) you could, in principle, also have an immobile surface charge on the insulator with surface charge density $\rho_s$. Then the boundary condition for the electric field is $$\epsilon_1 E_1+\rho_s=\epsilon_2 E_2$$ This boundary condition would also hold if the sphere was a conducting sphere with mobile surface charge. In case of no surface charge, the boundary condition reduces to the continuity of the dielectric displacement $$\epsilon_1 E_1=\epsilon_2 E_2$$ A uniformly charged insulating dielectric does, in general, have no surface charge, unless you purposely put a charge on the surface. This in contrast to a conductor where any charges reside always on the surface as surface charges.
