Calculating Fields with Gauss's Law

Let $S$ be a closed surface, and let $Q_{ins}$ be the total charge in the interior of $S$. Gauss's Law states that: $$\int_{S}\vec{E}\cdot\vec{dA}=\frac{Q_{ins}}{\varepsilon_0}$$ where $\vec{dA}$ is the outward normal vector of $S$ and $\vec{E}$ is the electric field vector.

Gauss's Law can be very useful in charge configurations with a high degree of symmetry. If the surface integral on the left can be simplified to something nice, and if $Q_{ins}$ on the right can be expressed in a simple form, then computing electric fields can be made significantly easier by Gauss's Law.

Note that the above equation requires a surface. Thus, our first step is to find a convenient choice of surface, called a Gaussian surface. Often, the type of symmetry present in the problem will provide a clue as to which Gaussian surface to use. Then, find the charge $Q_{ins}$ inside the Gaussian surface and use the symmetry to simplify the surface integral. Finally, plug all of these into the above equation to find $\vec{E}$ everywhere.

Let us examine a situation in which we have a high degree of symmetry. Suppose that we have a solid ball of radius $R$ and charge $Q$, which is spread uniformly throughout the ball. We wish to find the electric field $\vec{E}$ for all distances $r\ge 0$ from the center of the ball.

In this case, we are dealing with spherical symmetry, so it makes sense to make our Gaussian surfaces spheres centered on the center of the ball. Notice that, due to the spherical symmetry, $\vec{E}$ must point directly outward everywhere on the sphere, and $|\vec{E}|$ must be a constant everywhere on the sphere. Because of this, our surface integral just becomes: $$\int_{S}\vec{E}\cdot\vec{dA}=E\times 4\pi r^2=4\pi Er^2$$ where $E=|\vec{E}|$.

Now, we need to find the charge contained inside the sphere. If $r\ge R$, then the entire sphere is contained inside our Gaussian surface. Thus, $Q_{ins}=Q$. If $r\le R$, then: $$Q_{ins}=\frac{Q}{\frac{4}{3}\pi R^3}\times\frac{4}{3}\pi r^3$$ $$Q_{ins}=\frac{r^3}{R^3}Q$$ Now, we use Gauss's Law: $$\int_{S}\vec{E}\cdot\vec{dA}=\frac{Q_{ins}}{\varepsilon_0}$$ $$r\le R\implies 4\pi Er^2=\frac{Qr^3}{R^3\varepsilon_0}$$ $$r\le R\implies E=\frac{Q}{4\pi\varepsilon_0 R^3}r$$ $$r\ge R\implies 4\pi Er^2=\frac{Q}{\varepsilon_0}$$ $$r\ge R\implies E=\frac{Q}{4\pi\varepsilon_0 r^2}$$ Finally, we re-introduce the vector to finish the problem. Let $\vec{r}$ denote the vector from the center of the ball to a desired point. Then, since $\vec{E}\parallel\vec{r}$, we get the following: $$r\le R\implies E=\frac{Q}{4\pi\varepsilon_0 R^3}\vec{r}$$ $$r\ge R\implies E=\frac{Q}{4\pi\varepsilon_0 r^3}\vec{r}$$ This is the electric field produced by a uniformly charged ball.