According to Gauss's Law, for a uniformly charged spherical shell of radius $R$: 1) Inside the sphere ($r < R$), the electric field is zero because the Gaussian surface encloses no charge. 2) Outside the sphere ($r > R$), the shell behaves as if the entire charge is concentrated at the center, and the electric field behaves like that of a point charge, given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$. Thus, for $r > R$, the electric field decreases as $r$ increases ($E \propto 1/r^2$).