For a spherical system, the electric field has the same value at all points a distance r from the center of the distribution and points radially outward or inward. The

electric flux out of a Gaussian surface of radius r is just the surface area of the sphere multiplied by the field (phi). = 4(pi)r²E(r). Apply Gauss' law to this flux \phi_{e}=\frac{Q_{\text {enclosed }}}{\varepsilon_{0}}=4 \pi r^{2} E(r) Solve for the electric field and put the unit vector back in, \vec{E}(\vec{r})=\frac{Q_{\text {enclosed }}}{4 \pi \varepsilon_{0} r^{2}} \hat{r} This expression gives the electric field at a radius r for a spherically symmetric distribution if we can figure out how much charge is contained in the sphere of radius r.