Gauss's law is a cornerstone of electromagnetism and one of Maxwell's equations. It relates the distribution of electric charge to the resulting electric field. The law states that the net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that surface.
This closed surface is often called a Gaussian surface. Gauss's law is incredibly powerful for calculating electric fields of highly symmetric charge distributions (spherical, cylindrical, or planar) where Coulomb's law would require complex integration.
Consider a point charge $q$ located at the origin. The electric field at a distance $r$ is:
$$ \mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{r} $$Let's calculate the electric flux $\Phi_E$ through a spherical surface of radius $r$ centered on the charge. The area element $d\mathbf{a}$ points radially outward, so $d\mathbf{a} = r^2 \sin\theta d\theta d\phi \hat{r}$.
$$ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{a} = \oint \left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{r}\right) \cdot (r^2 \sin\theta d\theta d\phi \hat{r}) $$Since $\hat{r} \cdot \hat{r} = 1$, the $r^2$ terms cancel out:
$$ \Phi_E = \frac{q}{4\pi\varepsilon_0} \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta = \frac{q}{4\pi\varepsilon_0} (2\pi)(2) = \frac{q}{\varepsilon_0} $$By the principle of superposition, this result holds for any arrangement of enclosed charges, leading to the integral form of Gauss's Law.
The total enclosed charge can be expressed as a volume integral of charge density $\rho$: $Q_{enc} = \int_V \rho d\tau$.
Applying the Divergence theorem to the left side: $\oint_S \mathbf{E} \cdot d\mathbf{a} = \int_V (\nabla \cdot \mathbf{E}) d\tau$.
Equating the two volume integrals:
$$ \int_V (\nabla \cdot \mathbf{E}) d\tau = \int_V \frac{\rho}{\varepsilon_0} d\tau $$Since this must hold for any arbitrary volume $V$, the integrands must be equal. This gives the differential form of Gauss's Law.
Outside ($r > R$): Construct a spherical Gaussian surface of radius $r$. By symmetry, $\mathbf{E}$ is radial and constant over the surface.
$$ \oint \mathbf{E} \cdot d\mathbf{a} = E(4\pi r^2) $$The total enclosed charge is $Q_{enc} = \rho \left(\frac{4}{3}\pi R^3\right)$. Using Gauss's law:
$$ E(4\pi r^2) = \frac{\rho \frac{4}{3}\pi R^3}{\varepsilon_0} \implies E = \frac{\rho R^3}{3\varepsilon_0 r^2} $$Inside ($r < R$): Construct a spherical Gaussian surface of radius $r$. The enclosed charge is only the charge within radius $r$.
$$ Q_{enc} = \rho \left(\frac{4}{3}\pi r^3\right) $$ $$ E(4\pi r^2) = \frac{\rho \frac{4}{3}\pi r^3}{\varepsilon_0} \implies E = \frac{\rho r}{3\varepsilon_0} $$Inside, the field increases linearly with $r$. Outside, it falls off as $1/r^2$.
Use the differential form of Gauss's Law: $\nabla \cdot \mathbf{E} = \rho/\varepsilon_0$.
The divergence in spherical coordinates for a purely radial field is:
$$ \nabla \cdot \mathbf{E} = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 E_r) $$Given $E_r = k r^3$:
$$ \nabla \cdot \mathbf{E} = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \cdot k r^3) = \frac{1}{r^2}\frac{\partial}{\partial r}(k r^5) = \frac{1}{r^2} (5k r^4) = 5k r^2 $$Therefore, $\rho(r) = \varepsilon_0 \nabla \cdot \mathbf{E} = 5\varepsilon_0 k r^2$.