Gauss's Law — Physica

Equivalent forms

\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}

\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}

Turns a hard integral into a trivial calculation whenever the charge distribution has enough symmetry — spherical, cylindrical, or planar.

Symbol	Name	SI unit	Dimension	Range
$Φ_E$	Electric fluxoutput Total electric flux through the closed surface	V·m	M·L³·T⁻³·I⁻¹	0 – 1000000
$Q_enc$	Enclosed charge Total charge enclosed by the Gaussian surface	C	I·T	-0.00001 – 0.00001
$ε₀$	Vacuum permittivity Permittivity of free space	F/m	I²·T⁴·M⁻¹·L⁻³	8.854e-12 – 8.854e-12

Unit systems

Symbol	SI	CGS	Imperial
$Φ_E$	V·m	statV·cm	V·m
$Q$	C	statC	C

Where it holds

Universally valid in classical electrostatics. Most useful for symmetric charge distributions (spherical, cylindrical, planar). For arbitrary distributions, Coulomb's law or numerical methods are more practical.

Dimensional analysis

$[\Phi _E] = [Q]/[\varepsilon _{0}] \to C/(F/m) = C\cdot m/F = V\cdot m \checkmark$

Discovery

Carl Friedrich Gauss · 1835

Gauss formulated this as a mathematical theorem relating surface integrals to volume integrals. Its power in electrostatics was recognized when Maxwell incorporated it as his first equation.

Try this

Why is the electric field zero inside a hollow metal sphere, even if it's highly charged?

A uniformly charged sphere of radius 0.1 m carries total charge +2 μC. Find the electric field at r = 0.2 m from the center.

Research status: stable

Real-world applications

Designing Faraday cages for electromagnetic shielding
Calculating capacitor fields (parallel plate, cylindrical, spherical)
Electrostatic shielding in sensitive electronics
Understanding charge distribution on conductors

Common misconceptions

Gauss's law is always true but not always useful — it only simplifies calculations when sufficient symmetry exists
The flux depends only on enclosed charge, not on charges outside the surface
E on the Gaussian surface may have contributions from external charges, but they cancel in the flux integral

Experimental verification

Faraday's ice-pail experiment (1843) demonstrated that the induced charge on a hollow conductor equals the enclosed charge. Modern verification via high-precision Cavendish experiments confirming the inverse-square law exponent.

Derivation

Start with Coulomb's law for a point charge.

Integrate

E\cdot dA

over a sphere of radius r centered on q:

\Phi = E(4\pi r^{2}) = (q/4\pi \varepsilon _{0}r^{2})(4\pi r^{2}) = q/\varepsilon _{0}

.

Generalize via superposition: any enclosed charge contributes

q/\varepsilon _{0}

; external charges contribute zero net flux.

Limiting cases

Q_enc = 0

⟶

\Phi _E = 0

No enclosed charge means zero net flux — field lines enter and exit in equal number.

Surface inside a conductor⟶

E = 0

everywhere on surfaceCharges reside on the conductor surface; any interior Gaussian surface encloses zero charge.

What if…

What if you double the Gaussian surface radius?

For a sphere, E drops by $4\times$ (inverse-square). For a cylinder, E drops by $2\times$ (inverse-linear). The flux stays the same — only the surface area changes.

What if there's zero enclosed charge?

Net flux is zero, but E is not necessarily zero — external charges create fields that enter and exit the surface in equal amounts.

What if the charge distribution lacks symmetry?

Gauss's law is still true $(\Phi = Q_enc/\varepsilon _{0})$ but you can't extract E from the integral. Use Coulomb's law or numerical methods instead.

1

Field outside a charged sphere

Given ·

Q:: 0.000002
R:: 0.1
r:: 0.2

Find · E at

r = 0.2\,\mathrm{m}

Steps

Choose a spherical Gaussian surface at $r = 0.2\,\mathrm{m} (r$ > R)
By symmetry, E is radial and constant on the surface: $\oint E\cdot dA = E \times 4\pi r^{2}$
Enclosed charge: $Q_enc = 2 \mu C = 2\times 10^{-6} C$
Apply Gauss's law: $E \times 4\pi r^{2} = Q/\varepsilon _{0}$
Solve: $E = Q / (4\pi \varepsilon _{0}r^{2}) = 8.9875\times 10^{9} \times 2\times 10^{-6} / 0.04 = 4.49\times 10^{5} V/m$

Result ·

E = Q / (4\pi \varepsilon _{0}r^{2}) = 8.9875\times 10^{9} \times 2\times 10^{-6} / 0.04 = 4.49\times 10^{5} V/m

2

Field of an infinite line charge

Given ·

lambda:: 5e-9
r:: 0.02

Find · E at

r = 2\,\mathrm{cm}

from the line

Steps

Choose a cylindrical Gaussian surface of radius $r = 0.02\,\mathrm{m}$ and length L
By symmetry, E is radial: flux through curved surface $= E \times 2\pi rL$
End caps contribute zero flux (E ⊥ dA)
Enclosed charge: $Q_enc = \lambda L$
Gauss's law: $E \times 2\pi rL = \lambda L/\varepsilon _{0} \to E = \lambda /(2\pi \varepsilon _{0}r) = 4494\,\mathrm{V/m}$

Result ·

E = \lambda / (2\pi \varepsilon _{0}r) = 5\times 10^{-9} / (2\pi \times 8.854\times 10^{-12} \times 0.02) = 4494\,\mathrm{V/m}

Coulomb's Law→Electric Field of a Point Charge→Gauss's Law for Magnetism→