Gauss's Law and Electric Fields

Concise answer

Gauss's law, $\oint \vec{E}\cdot d\vec{A} = Q_{\text{enc}}/\varepsilon_0$ , turns a hard field calculation into one line whenever the charge distribution has spherical, cylindrical, or planar symmetry — and it explains why fields vanish inside conductors.

Definitions

Electric flux: The surface integral of the electric field over a surface; loosely, the number of field lines passing through it.
Gaussian surface: An imaginary closed surface chosen to exploit the symmetry of a charge distribution when applying Gauss's law.
Electrostatic equilibrium: The state of a conductor in which charges have stopped moving, forcing the field inside the conducting material to zero.
Electric potential: The potential energy per unit charge; a scalar field whose spatial rate of change gives the electric field.
Charge density: Charge per unit length ($\lambda$), area ($\sigma$), or volume ($\rho$), used to describe continuous distributions.

Intuition

Flux counts field lines leaving a closed surface. Since field lines start and end only on charges, the net number leaving depends only on the charge inside — not on where the charge sits inside or on any charge outside.

Gauss's law is always true but only sometimes useful. It pays off when symmetry guarantees the field has constant magnitude over the Gaussian surface, so the integral collapses to field times area.

Concept walkthrough

Gauss's law relates the electric flux through any closed surface to the enclosed charge: $\oint \vec{E}\cdot d\vec{A} = Q_{\text{enc}}/\varepsilon_0$ . For the three standard symmetries the resulting fields are worth memorizing: a spherically symmetric charge looks like a point charge from outside ( $E \propto 1/r^2$ ), an infinite line falls off as $1/r$ , and an infinite plane produces a uniform field independent of distance.

Conductors in electrostatic equilibrium follow directly: if the interior field were nonzero, charges would still be moving. So the field inside the conducting material is zero, excess charge sits on the surface, the surface is an equipotential, and the field just outside is perpendicular to the surface with magnitude $\sigma/\varepsilon_0$ .

Field and potential are two descriptions of the same physics. The field is the negative gradient of the potential, so the field points from high potential toward low potential and is strongest where the potential changes fastest. A region of constant potential has zero field — but a point of zero potential can still have a nonzero field.

After this page, you should be able to

State Gauss's law and identify the charge enclosed by a chosen Gaussian surface.
Match spherical, cylindrical, and planar symmetry to the Gaussian surface that makes the flux integral trivial.
Use conductor properties: zero interior field, surface charge, and the field just outside a conductor.
Relate electric field and electric potential, and translate between the two descriptions.

Formulas and assumptions

Gauss's law

\oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}

Variables

E: electric field in newtons per coulomb
dA: outward area element in square meters
Q_enc: charge enclosed by the surface in coulombs
epsilon_0: vacuum permittivity

Assumptions

The surface is closed.
Q_enc counts only charge inside the surface; outside charge affects E locally but not the net flux.

Field outside a spherical charge distribution

E = \dfrac{Q}{4\pi\varepsilon_0 r^2}

Variables

Q: total charge in coulombs
r: distance from the center in meters
E: radial field magnitude in newtons per coulomb

Assumptions

The charge distribution is spherically symmetric.
The field point lies outside the entire distribution.

Field of an infinite line charge

E = \dfrac{\lambda}{2\pi\varepsilon_0 r}

Variables

lambda: charge per unit length in coulombs per meter
r: perpendicular distance from the line in meters

Assumptions

The line is infinite (or the field point is close compared with the line's length).
The field is radial by cylindrical symmetry.

Field of an infinite charged plane

E = \dfrac{\sigma}{2\varepsilon_0}

Variables

sigma: charge per unit area in coulombs per square meter
E: field magnitude, directed away from a positive plane

Assumptions

The plane is infinite (or edge effects are negligible).
The field is uniform and independent of distance from the plane.

Field from potential (one dimension)

E_x = -\dfrac{dV}{dx}

Variables

E_x: field component along x in volts per meter
V: electric potential in volts

Assumptions

The potential is differentiable.
In three dimensions the field is the negative gradient of V.

Worked example

Field of a charged conducting sphere

A solid conducting sphere of radius $0.10\,\text{m}$ carries charge $Q = +5.0\,\text{nC}$ and is in electrostatic equilibrium. Find the electric field at $r = 0.05\,\text{m}$ and at $r = 0.20\,\text{m}$ from the center.

1Inside the conductor ( $r = 0.05\,\text{m} < 0.10\,\text{m}$ ): a Gaussian sphere at this radius encloses no charge, because all excess charge sits on the conductor's surface. So $E = 0$ .
2Outside ( $r = 0.20\,\text{m}$ ): by spherical symmetry the sphere acts like a point charge, so $E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{kQ}{r^2}$ with $k \approx 8.99\times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ .
3Numerator: $kQ = (8.99\times 10^9)(5.0\times 10^{-9}) \approx 45\,\text{N}\cdot\text{m}^2/\text{C}$ .
4Divide by $r^2 = (0.20)^2 = 0.040\,\text{m}^2$ : $E \approx 45/0.040 \approx 1.1\times 10^3\,\text{N/C}$ , pointing radially outward.

$E = 0$ at $r = 0.05\,\text{m}$ (inside the conductor) and $E \approx 1.1\times 10^3\,\text{N/C}$ radially outward at $r = 0.20\,\text{m}$ .

Common traps

Including charge outside the Gaussian surface in $Q_{\text{enc}}$ ; external charge changes the local field but not the net flux.
Applying the point-charge $1/r^2$ formula to line or plane geometries, which fall off as $1/r$ and not at all, respectively.
Concluding the field is zero wherever the potential is zero; the field depends on how fast the potential changes, not on its value.
Forgetting that the field just outside a conductor is perpendicular to the surface and is $\sigma/\varepsilon_0$ , not $\sigma/2\varepsilon_0$ .
Using Gauss's law on a distribution without enough symmetry, where the flux integral cannot be reduced to field times area.

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Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
University Physics Volume 2, Section 6.2: Explaining Gauss's Law — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 6.3: Applying Gauss's Law — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 6.4: Conductors in Electrostatic Equilibrium — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 7.4: Determining Field from Potential — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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Last source check: 2026-08-02
Next scheduled review: 2026-11-02

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Gauss's Law and Electric Fields

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

Gauss's law

Field outside a spherical charge distribution

Field of an infinite line charge

Field of an infinite charged plane

Field from potential (one dimension)

Worked example

Field of a charged conducting sphere

Common traps

Related pages and practice

Sources

Review and maintenance