Gauss's law

Gauss’s law is the statement that the total electric flux through any closed surface equals the total free charge enclosed:

${\oint }_S\mathbf{D}\cdot d\mathbf{s}=Q_{\text{enc}}\text{(integral form)}.$

Equivalently, by the Divergence theorem, the local form is

$\nabla \cdot \mathbf{D}={\rho }_v\text{(differential form)},$

where $\mathbf{D}=ϵ\mathbf{E}$ is the Electric flux density and ${\rho }_v$ is the free volume Charge density.

Differential form intuition: shrink the Gaussian surface to a point. The flux per unit volume — $\nabla \cdot \mathbf{D}$ — equals the local charge density ${\rho }_v$. The integral form is the same statement summed over the inside of a finite surface.

These two forms — integral and differential — are equivalent. Apply the divergence theorem to a small ball; the integral form turns into the differential form, and vice versa. Which to use depends on the problem: integral form is the practical tool when symmetry makes the flux integral easy; the differential form is the local, source-relating-to-field statement that fits into the differential Maxwell’s equations.

Why it’s true

Coulomb’s law for a point charge gives $\mathbf{E}=(q/4\pi ϵR^2)\hat{\mathbf{R}}$ — radial $1/R^2$ field. The flux of $\mathbf{E}$ through a sphere of radius $R$ centered on the charge is $|\mathbf{E}|\cdot 4\pi R^2=q/ϵ$. So ${\oint }_Sϵ\mathbf{E}\cdot d\mathbf{s}=q$.

For an arbitrary closed surface enclosing the charge, the same total flux comes through — the $1/R^2$ field is exactly the geometric “spreading out” of a fixed amount of flux. For surfaces not enclosing the charge, equal flux in and out cancels.

Superposition extends this to any distribution: total flux = total enclosed charge (divided by $ϵ$ if you used $\mathbf{E}$, exactly enclosed charge if you used $\mathbf{D}$).

When Gauss’s law beats direct integration

Direct integration of Coulomb’s law gives the field at any point — but the integrals are often hideous. Gauss’s law trades general applicability for tractability when symmetry is present: if you can guess the field’s direction and dependence by symmetry, you can pull the magnitude outside the surface integral.

The recipe:

1. Identify the symmetry (spherical, cylindrical, planar).
2. Pick a Gaussian surface that matches the symmetry — usually one where $|\mathbf{D}|$ is constant on it and $\mathbf{D}$ is parallel or perpendicular to $d\mathbf{s}$ everywhere.
3. The flux integral collapses to $|\mathbf{D}|\cdot A$, with $A$ the surface area.
4. Set equal to enclosed charge, solve for $|\mathbf{D}|$.

Worked example: point charge

A positive point charge $q$ in free space. Spherical symmetry → $\mathbf{D}=D_R\hat{\mathbf{R}}$ on a sphere of radius $R$, constant magnitude.

Gaussian surface: sphere of radius $R$.

${\oint }_S\mathbf{D}\cdot d\mathbf{s}={\int }_0^{\pi }{\int }_0^{2\pi }{D}_R{R}^2\sin \theta d\varphi d\theta =D_R\cdot 4\pi R^2.$

Enclosed charge is $q$. So $D_R\cdot 4\pi R^2=q$, giving

$D_R=\frac{q}{4\pi R^2},\mathbf{E}=\frac{\mathbf{D}}{{ϵ}_0}=\hat{\mathbf{R}}\frac{q}{4\pi {ϵ}_0{R}^2}.$

Coulomb’s law falls out of Gauss’s law plus spherical symmetry.

Worked example: infinite line charge

A line on the $z$-axis with line charge density ${\rho }_l$. Cylindrical symmetry → $\mathbf{D}=D_r\hat{\mathbf{r}}$, constant on a cylinder of radius $r$.

Gaussian surface: cylinder of radius $r$, height $h$. Flux through the curved side is $D_r\cdot 2\pi rh$. Flux through top and bottom caps is zero (field is perpendicular to $\hat{\mathbf{z}}$). Enclosed charge: ${\rho }_{l}h$.

$D_r\cdot 2\pi rh={\rho }_{l}h⟹D_r=\frac{{\rho }_l}{2\pi r},\mathbf{E}=\hat{\mathbf{r}}\frac{{\rho }_l}{2\pi {ϵ}_{0}r}.$

Notice the $1/r$ fall-off, not $1/r^2$ — line charges spread their flux over an area that grows linearly with $r$ (the cylindrical side), not quadratically.

Worked example: infinite plane of charge

By similar logic with a “pillbox” Gaussian surface straddling the plane:

$\mathbf{E}=\frac{{\rho }_s}{2{ϵ}_0}\hat{\mathbf{n}},$

independent of distance — flux doesn’t spread out at all (the area enclosed stays constant as you move away in 1D).

Why $\mathbf{D}$ and not $\mathbf{E}$

Gauss’s law in $\mathbf{D}$ form has free charge on the right side:

$\nabla \cdot \mathbf{D}={\rho }_v.$

The bound polarization charge inside dielectrics is absorbed into $\mathbf{D}=ϵ\mathbf{E}$. By contrast, $\nabla \cdot \mathbf{E}={\rho }_{\text{total}}/{ϵ}_0$ would put bound + free charges on the right, which you can’t easily track.

For free space or anywhere $ϵ={ϵ}_0$, the two forms differ only by a constant and either works.

In Maxwell’s equations

$\nabla \cdot \mathbf{D}={\rho }_v$ is one of the four Maxwell’s equations. The other three are $\nabla \cdot \mathbf{B}=0$ (no magnetic monopoles), $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ (Faraday’s law), and $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$ (Ampère’s law with displacement current). All four hold in static and dynamic problems alike.