Consider a uniform nonconducting sphere with a charge \(\rho=3.57 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{3}\) and a radius \(R=1.72 \mathrm{~m}\). What is the magnitude of the electric field \(0.530 \mathrm{~m}\) from the center of the sphere?

We will define our Gaussian surface to be a concentric sphere with the same center as the charged sphere, but with a radius of equal to the distance from the center to the point of interest, \(r = 0.530\mathrm{~m}\).

Gauss's law states that \(\oint \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}\). Here, \(Q_{enc}\) is the charge enclosed by the Gaussian surface, and \(\vec{E}\) is the electric field at a point on the Gaussian surface. We'll determine the charge enclosed by the Gaussian surface and the area of the Gaussian surface.

Since \(\rho\) is constant, we can find the charge enclosed \(Q_{enc}\) by taking the integral of the charge density over the Gaussian surface: \begin{align*} Q_{enc} = \rho \cdot V = \rho \cdot \frac{4}{3}\pi r^3 \end{align*} Now plug in the values for \(\rho=3.57 \cdot 10^{-6} \mathrm{C/m^3}\) and \(r=0.530 \mathrm{~m}\): \begin{align*} Q_{enc} = (3.57 \cdot 10^{-6}\mathrm{C/m^3})\cdot \frac{4}{3}\pi (0.530\mathrm{~m})^3 = 2.554 \times 10^{-6} \mathrm{C} \end{align*}

Now return to Gauss's law and solve for the electric field: \begin{align*} \oint \vec{E}\cdot d\vec{A} &= \frac{Q_{enc}}{\varepsilon_0}\\ E \oint dA &= \frac{Q_{enc}}{\varepsilon_0} \end{align*} Since the electric field is parallel to the area vector at every point on the Gaussian surface, their dot product becomes a simple multiplication. Also, the electric field is uniform over the Gaussian surface, so it can be pulled out of the integral: \begin{align*} E \cdot 4\pi r^2 &= \frac{Q_{enc}}{\varepsilon_0}\\ \end{align*} Now, solve for \(E\) and plug in the values for \(Q_{enc}\), \(r\), and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\): \begin{align*} E &= \frac{Q_{enc}}{4\pi \varepsilon_0 r^2}\\ E &= \frac{2.554 \times 10^{-6}\mathrm{C}}{4\pi \cdot 8.85 \times 10^{-12}\mathrm{F/m} \cdot (0.530 \mathrm{m})^2} = 212.99 \mathrm{N/C} \end{align*}

The magnitude of the electric field at a distance of \(0.530 \mathrm{~m}\) from the center of the uniformly charged nonconducting sphere is \(212.99 \mathrm{N/C}\).