11.1 Electrostatics: Coulomb’s law and Gauss’s law

Electromagnetism begins with the force between charges at rest. Coulomb’s law gives that force directly; Gauss’s law repackages it into a field equation that yields the field of any symmetric distribution without integration. Together they are the foundation of electrostatics and the first of Maxwell’s equations.

Coulomb’s law and the electric field

A point charge $q$ sets up an electric field that fills the space around it,

\mathbf{E}(\mathbf{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r^2}\,\hat{\mathbf{r}},

where

$\mathbf{E}$: electric field V/m
$q$: source charge C
$r$: distance from the charge m
$\varepsilon_0$: permittivity of free space F/m

pointing radially outward from a positive charge and inward toward a negative one. Two properties carry through all of electrostatics: the field falls as the inverse square of distance, and it obeys superposition — the field of many charges is the vector sum of their individual fields, because Maxwell’s equations are linear.

q+1.00

r (probe)2.00

E = k_e q / r²0.250

V = k_e q / r0.500

Charge q = 1.00 Probe distance r = 2.00

Coulomb's law: a charge q at the origin produces an electric field E = k_e q / r² pointing radially outward (positive charge) or inward (negative). The field lines never cross; they spread radially with density falling as 1/r². The dashed circles are equipotentials — surfaces of constant V = k_e q/r — perpendicular to the field everywhere. Maxwell's first equation ∇·E = ρ/ε₀ is the integral statement of this field, applied to general charge distributions.

The field is conservative: it can be written as the gradient of a scalar electric potential, $\mathbf{E} = -\nabla V$ , with $V = q/(4\pi\varepsilon_0 r)$ for a point charge. Field lines run perpendicular to the surfaces of constant $V$ and never cross. For an arbitrary charge distribution $\rho(\mathbf{r})$ the field follows from superposing Coulomb contributions — an integral that is exact but often unwieldy. When the distribution is symmetric, there is a far quicker route.

Gauss’s law

Gauss’s law states that the flux of $\mathbf{E}$ through any closed surface equals the enclosed charge divided by $\varepsilon_0$ :

\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} \;=\; \frac{Q_\text{enc}}{\varepsilon_0}.

By the divergence theorem this is equivalent to the differential form — the first Maxwell equation,

\nabla\cdot\mathbf{E} \;=\; \frac{\rho}{\varepsilon_0},

which says the electric field diverges from charge: positive charge is a source of field lines, negative charge a sink.

Symmetry:

Gauss's law in integral form: ∮ E·dA = Q_enc/ε₀. The key is *choosing the right surface*: one over which E is constant in magnitude and parallel to the area element (or perpendicular to it on the side caps). For each symmetric charge distribution, this surface is the obvious one — spheres for point charges, pillboxes for planes, cylinders for lines. The resulting E-field formulas — 1/r², constant, 1/r — are completely fixed by symmetry, with no integration required.

The power of Gauss’s law is computational. Choose a surface on which symmetry makes $|\mathbf{E}|$ constant and either parallel or perpendicular to each area element, and the flux integral becomes a simple product. Three canonical symmetries fix the field’s form outright:

Spherical (point or ball of charge): $E = Q/(4\pi\varepsilon_0 r^2)$ — the inverse square, recovering Coulomb.
Planar (infinite charged sheet): $E = \sigma/(2\varepsilon_0)$ — uniform, independent of distance.
Cylindrical (infinite line of charge): $E = \lambda/(2\pi\varepsilon_0 r)$ — falling as $1/r$ .

The distance-dependence — $1/r^2$ , constant, or $1/r$ — is dictated by symmetry alone, before any detail of the source is specified. Gauss’s law is the electrostatic workhorse, and the next lesson uses it to find the field, capacitance, and stored energy of a capacitor.