1.1 Gauss's law for electric fields

Gauss's law statement

Gauss's law connects the electric flux through a closed surface to the total charge enclosed by that surface. It's one of Maxwell's four equations, and at the Electromagnetism II level, you need fluency in both its integral and differential forms and a clear sense of when each is the right tool.

The integral form gives you a global relationship between flux and enclosed charge. The differential form gives you a local, point-by-point relationship between the electric field's divergence and the charge density. Together, they're equivalent statements of the same physics.

Integral form

The integral form says: the total electric flux through any closed surface equals the enclosed charge divided by $\epsilon_0$.

$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$

• $\vec{E}$ is the electric field at each point on the surface.
• $d\vec{A}$ is an outward-pointing infinitesimal area element.
• $Q_{\text{enc}}$ is the total charge enclosed by the surface.

The convention matters: $d\vec{A}$ always points outward from the enclosed volume. This fixes the sign of the flux. Positive enclosed charge produces positive (outward) net flux; negative enclosed charge produces negative (inward) net flux.

This form is most useful when symmetry lets you pull $\vec{E}$ out of the integral, turning a surface integral into a simple algebraic equation.

Differential form

The differential form relates the divergence of $\vec{E}$ at a single point to the local charge density $\rho$ at that point:

$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

You can derive this from the integral form using the divergence theorem ($\oint \vec{E} \cdot d\vec{A} = \int_V (\nabla \cdot \vec{E})\, dV$). Since the integral form holds for any closed surface, the integrands must be equal, giving you the differential form.

This version is essential for deriving Poisson's equation ($\nabla^2 V = -\rho/\epsilon_0$) and Laplace's equation ($\nabla^2 V = 0$ where $\rho = 0$). You'll rely on it heavily when solving boundary-value problems later in the course.

Gaussian surfaces

A Gaussian surface is an imaginary closed surface you construct to exploit symmetry when applying the integral form of Gauss's law. The surface itself has no physical reality; it's a mathematical tool you choose to make the integral tractable.

Closed surfaces

The surface must be closed, meaning it fully encloses a volume with no gaps. The outward normal convention for $d\vec{A}$ is standard. Any closed surface works in principle, but a poorly chosen one won't simplify the integral at all.

Symmetry of charge distributions

The whole power of Gauss's law in integral form rests on choosing a Gaussian surface that matches the symmetry of the charge distribution. When you do this correctly, $\vec{E} \cdot d\vec{A}$ becomes constant over the surface (or zero on parts of it), and the integral collapses.

Three canonical symmetries come up repeatedly:

• Spherical symmetry (point charges, uniformly charged spheres): Use a concentric spherical Gaussian surface. $\vec{E}$ is radial and constant in magnitude over the sphere, so $\oint \vec{E} \cdot d\vec{A} = E(4\pi r^2)$.
• Cylindrical symmetry (infinite line charges, infinite cylindrical shells): Use a coaxial cylindrical Gaussian surface. $\vec{E}$ is radial and constant on the curved surface; flux through the end caps is zero by symmetry, so $\oint \vec{E} \cdot d\vec{A} = E(2\pi r L)$.
• Planar symmetry (infinite sheets of charge): Use a pillbox (short rectangular box or cylinder) straddling the plane. $\vec{E}$ is perpendicular to the plane and passes through the two flat faces; no flux passes through the sides.

If the charge distribution doesn't have one of these symmetries, the integral form of Gauss's law is still true, but it won't give you a shortcut to finding $\vec{E}$.

Electric flux

Electric flux $\Phi_E$ quantifies how much electric field passes through a surface. Geometrically, you can think of it as counting the net number of field lines piercing the surface (outward minus inward). Its SI unit is V·m (equivalently, N·m²/C).

Flux through closed surfaces

For a closed surface, Gauss's law directly ties the net flux to the enclosed charge:

$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$

• If $Q_{\text{enc}} = 0$, the net flux is zero. Field lines entering the surface must also exit it.
• If $Q_{\text{enc}} > 0$, there's net outward flux.
• If $Q_{\text{enc}} < 0$, there's net inward flux.

Charges outside the closed surface contribute zero net flux. Their field lines enter and exit, canceling exactly.

Flux and electric field relationship

For a flat surface in a uniform field, the flux simplifies to:

$\Phi_E = \vec{E} \cdot \vec{A} = EA\cos\theta$

Here $\theta$ is the angle between $\vec{E}$ and the outward normal $\hat{n}$ to the surface. When the field is parallel to the surface ($\theta = 90°$), the flux is zero. When the field is perpendicular ($\theta = 0°$), the flux is maximized at $EA$.

For non-uniform fields or curved surfaces, you must integrate: $\Phi_E = \int \vec{E} \cdot d\vec{A}$.

Applying Gauss's law

The general procedure for using Gauss's law to find $\vec{E}$:

1. Identify the symmetry of the charge distribution (spherical, cylindrical, or planar).
2. Determine the direction of $\vec{E}$ from that symmetry. For example, spherical symmetry forces $\vec{E}$ to be purely radial.
3. Choose a Gaussian surface that matches the symmetry so that $\vec{E} \cdot d\vec{A}$ is either constant or zero over each part of the surface.
4. Evaluate the flux integral. Because of your surface choice, this should reduce to $E$ times some geometric factor (area of a sphere, curved surface of a cylinder, etc.).
5. Find $Q_{\text{enc}}$ by integrating the charge density over the volume enclosed by your Gaussian surface.
6. Set flux equal to $Q_{\text{enc}}/\epsilon_0$ and solve for $E$.

Spherical charge distributions

Use a spherical Gaussian surface of radius $r$ centered on the charge distribution.

Point charge $q$: For $r > 0$, $E(4\pi r^2) = q/\epsilon_0$, giving $E = q/(4\pi\epsilon_0 r^2)$. This recovers Coulomb's law.

Uniformly charged solid sphere (total charge $Q$, radius $R$):

• Outside ($r > R$): $E = Q/(4\pi\epsilon_0 r^2)$. The sphere acts like a point charge.
• Inside ($r < R$): Only the charge within radius $r$ is enclosed. With uniform volume charge density $\rho = 3Q/(4\pi R^3)$, you get $Q_{\text{enc}} = Q(r/R)^3$, so $E = Qr/(4\pi\epsilon_0 R^3)$. The field grows linearly with $r$ inside.

Cylindrical charge distributions

Use a coaxial cylindrical Gaussian surface of radius $r$ and length $L$.

Infinite line charge (linear charge density $\lambda$): The flux passes only through the curved surface: $E(2\pi r L) = \lambda L/\epsilon_0$, giving $E = \lambda/(2\pi\epsilon_0 r)$.

Uniformly charged infinite cylinder (radius $R$, volume charge density $\rho$):

• Outside ($r > R$): $E = \rho R^2/(2\epsilon_0 r)$.
• Inside ($r < R$): $Q_{\text{enc}} = \rho \pi r^2 L$, so $E = \rho r/(2\epsilon_0)$. Again, the field grows linearly with $r$ inside.

Planar charge distributions

Infinite sheet (surface charge density $\sigma$): Use a pillbox straddling the sheet. Flux exits through both flat faces: $2EA = \sigma A/\epsilon_0$, giving $E = \sigma/(2\epsilon_0)$. The field is uniform and independent of distance from the sheet.

Parallel-plate capacitor (plates with $+\sigma$ and $-\sigma$): Superposing the fields from each sheet, the field between the plates is $E = \sigma/\epsilon_0$ and zero outside. This is the standard result used in capacitor analysis.

Gauss's law applications

Electric fields of conductors

In electrostatic equilibrium, three results follow directly from Gauss's law:

• The electric field inside a conductor is zero. If it weren't, free charges would move until it is.
• Any net charge resides on the surface. A Gaussian surface just inside the conductor encloses zero field, so it encloses zero charge.
• The field just outside the surface is $E = \sigma/\epsilon_0$, directed perpendicular to the surface. You can show this with a small pillbox Gaussian surface straddling the conductor's surface: one face is inside (zero field), the other is just outside.

Shielding and Faraday cages

A closed conducting shell screens its interior from external electric fields. Gauss's law shows why: in equilibrium, charges on the outer surface rearrange to cancel any external field inside the conductor. A Gaussian surface inside the conductor's bulk has zero flux, so the cavity inside is field-free regardless of what's happening outside.

This is the principle behind Faraday cages, used to shield sensitive electronics, protect occupants of vehicles struck by lightning, and isolate experiments from stray fields.

Capacitors and dielectrics

For a parallel-plate capacitor, Gauss's law gives the field between the plates as $E = \sigma/\epsilon_0$ (as derived above). The potential difference is then $V = Ed = \sigma d/\epsilon_0$, and the capacitance is $C = \epsilon_0 A/d$.

When a dielectric with permittivity $\epsilon = \kappa \epsilon_0$ fills the gap, bound charges on the dielectric surfaces partially cancel the free charge. Gauss's law in matter uses $\vec{D}$ (the displacement field) instead of $\vec{E}$:

$\oint \vec{D} \cdot d\vec{A} = Q_{\text{free, enc}}$

where $\vec{D} = \epsilon \vec{E}$. The capacitance increases by a factor of $\kappa$: $C = \kappa \epsilon_0 A/d$.

Limitations of Gauss's law

Non-symmetric charge distributions

Gauss's law is always true, but it's only useful for finding $\vec{E}$ when symmetry lets you simplify the surface integral. For an arbitrary charge distribution (say, a bent wire or a finite disk), you can't pull $E$ out of the integral, and the law gives you a single equation relating the flux to the charge without telling you the field at any particular point. In those cases, direct integration via Coulomb's law or numerical methods is the way to go.

Electric fields in matter

The basic form $\nabla \cdot \vec{E} = \rho/\epsilon_0$ includes all charges (free and bound). In linear, isotropic dielectrics, you can work with $\vec{D}$ and free charges only, which keeps things manageable.

For nonlinear, anisotropic, or inhomogeneous materials, $\vec{D}$ and $\vec{E}$ are no longer simply proportional ($\vec{D} = \overleftrightarrow{\epsilon} \cdot \vec{E}$ with a tensor permittivity, or $\vec{D}$ depends on $\vec{E}$ nonlinearly). Gauss's law still holds, but extracting $\vec{E}$ from it requires additional constitutive relations and often numerical techniques.

Gauss's law vs. Coulomb's law

Both laws describe how charges produce electric fields, and in fact Gauss's law can be derived from Coulomb's law (plus superposition) and vice versa. They're mathematically equivalent for electrostatics.

Similarities and differences

• Coulomb's law gives you the field from a specified charge distribution by direct vector summation or integration. It works for any geometry.
• Gauss's law gives you a constraint (total flux = enclosed charge/$\epsilon_0$) that, with sufficient symmetry, determines $\vec{E}$ algebraically.
• Coulomb's law is inherently an inverse-square law. Gauss's law encodes the same $1/r^2$ dependence but expresses it through flux, which is why it generalizes naturally to one of Maxwell's equations.

When to use each law

Use Coulomb's law when:

• You have a small number of discrete charges.
• The geometry lacks the symmetry needed for Gauss's law.
• You need the field at a specific point from a known charge configuration.

Use Gauss's law when:

• The charge distribution has spherical, cylindrical, or planar symmetry.
• You want the field over an entire region rather than at a single point.
• You're working with conductors in electrostatic equilibrium.

In many problems, especially at this level, you'll use both: Gauss's law to find the field in a symmetric region, and Coulomb-type integration (or the method of images, multipole expansions, etc.) for everything else.