Electric flux and Gauss’s law

A field crosses an empty box

Place an imaginary closed box in a prescribed uniform electric field. Field enters some faces and leaves others. Does an empty interior require the field to vanish? No. The contributions to a signed surface integral can cancel while the field remains nonzero everywhere on the box.

Use positive scales q₀ for charge and L₀ for length. Let E₀ = kₑq₀/L₀², with kₑ = 1/(4πε₀). Our first field is E/E₀ = (2, −3, 5). The box is centered at the origin with full side lengths 2L₀, 4L₀ and 6L₀ along x, y and z. Its faces and interior contain no charge in this prescribed local field model.

The box is a Gaussian surface: a mathematical boundary used to ask a field question. It is not a charged object, a conducting shell or a wall that blocks the field. Specifying a uniform field in a neighborhood of this box does not require us to model the distant equipment that produces it.

This lesson connects signed flux to net enclosed charge, then asks when that integral constraint determines a local field. Keep those two tasks distinct as you follow the box, sphere, cube, line and plane examples. Each charged-source family is a separate prescribed stationary vacuum model unless an external contribution is explicitly added.

Orient the area before measuring flux

An oriented small surface patch has vector area dA = n̂ dA, where dA is its positive area and n̂ is a chosen unit normal. Electric flux through an oriented surface S is Φ_E = ∫_S E·n̂ dA. It is a signed scalar, with SI units N m²/C. It is not the electric field vector and does not measure a flow of electric charge.

For a flat patch in a uniform field, E and its normal are constant, so Φ_E = E·n̂ A = EA cos θ. The angle θ is measured between the field and the normal. A field tangent to the patch gives zero flux even when the field magnitude is nonzero. For a curved surface or a varying field, use the integral rather than inserting one convenient value.

Our open rectangle has unit tangents u = (4/5, −3/5, 0) and v = (0, 0, −1), with sides L₀ and 5L₀. Choose n̂ = u × v = (3/5, 4/5, 0). Then E·n̂/E₀ = −6/5 and Φ_E/(E₀L₀²) = −6. Reversing v reverses the normal and gives +6; it changes neither the field nor the physical rectangle.

A perspective sketch of a patch or box is schematic. The specified vectors, lengths and dot products determine the physical angle and area; apparent screen angles and lengths do not. A chosen number of drawn field lines likewise has no fixed physical flux unless an explicit drawing convention assigns it.

Close the surface and sum every outward face

A closed surface uses outward normals consistently. Positive contributions count field directed outward, and negative contributions count field directed inward. Add every face. Omitting the entering contributions would turn a signed integral into a different quantity.

For our box, the x-face area is 24L₀², the y-face area is 12L₀² and the z-face area is 8L₀². Multiplying each area by its signed normal field gives the following ledger. Opposite faces have equal areas and opposite normals in the same uniform field.

The closed flux is zero, although E/E₀ = (2, −3, 5) is nonzero. This example needs no cancellation of the field itself. It is cancellation of six scalar surface contributions.

An open patch does not enclose a volume, so there is no charge enclosed by that patch to insert into the closed-surface law. A nonzero open-patch flux in the same empty neighborhood is entirely consistent with zero closed-box flux.

What Gauss’s law fixes

For a closed Gaussian surface S bounding a volume V, Gauss’s law in vacuum SI notation is ∮_S E·n̂ dA = Q_enclosed/ε₀. Here Q_enclosed is the signed net charge inside V, and E is the total field on the boundary. For a regular volume density, Q_enclosed = ∫_V ρ dV. Use the charge support and the chosen volume to decide which charge is included.

One way to see the normalization is to put a point charge q at the center of a sphere of radius r > 0. Its radial field has constant normal component kₑq/r², so the flux is 4πr²kₑq/r² = q/ε₀. The radius cancels. In the electrostatic point-source construction, deformation or a solid-angle argument extends this result to enclosing surfaces without placing the charge on their boundary.

A charge outside a closed surface contributes zero net flux through it, yet generally changes the field at points on that surface. Superposition applies to those local field contributions too. Do not delete an external field from E merely because its closed integral vanishes.

Zero net flux means zero signed net enclosed charge. It does not establish that no charges are inside: equal positive and negative charges can have zero net charge. It also does not establish that E vanishes, as the uniform-field box already shows.

We keep ideal point or surface charges off an elementary Gaussian boundary whenever its ordinary field integral would be undefined or ambiguous. A surface carrying an ideal sheet charge needs one-sided limits or a more careful convention; this lesson does not assign a half-charge rule at that boundary.

Keep field, flux and charge units distinct

Our field scale is E₀ = q₀/(4πε₀L₀²), with units N/C; E/E₀ is dimensionless. A convenient flux unit is E₀L₀². A different convenient unit is q₀/ε₀ = 4πE₀L₀². Thus an enclosed charge q₀ gives flux 4π in the first normalization and 1 in the second. Losing that factor of 4π changes the prediction.

A flux divided by an area is an average normal field component. It equals a field component at every point only if additional information establishes that the normal component is constant there. Flux has no direction vector; a negative flux indicates the chosen net orientation of the field crossing.

Symmetry must justify extracting a field

Gauss’s law is an integral constraint, not a general recipe E = Q_enclosed/(ε₀A). To extract an unknown field from an integral, identify which components and values are fixed by the symmetries of the complete physical configuration, including external fields and relevant boundary conditions.

For an isolated spherically symmetric prescribed charge distribution with the corresponding electrostatic field, rotational symmetry makes the field radial and makes its radial component depend only on radius. A concentric Gaussian sphere then has one constant normal component. Its area is 4πr², so E_r(r)4πr² = Q_enclosed(r)/ε₀ for r > 0.

Choosing a spherical Gaussian surface does not impose spherical symmetry on a nonsymmetric field. Nor does geometric symmetry of the surface by itself justify ignoring an external field. State why each contributing part of a chosen cylinder, pillbox or sphere has constant normal component or zero flux before taking a field outside the integral.

The following sphere and shell are assigned stationary charge distributions. Their fields are not claims about how freely moving charge distributes itself in a conductor. Material equilibrium and its boundary conditions belong to the later conductor lesson.

Same total charge, different interiors

Consider two separate sources centered at the origin. A uniform solid sphere has radius 2L₀ and total charge q₀, with density 3q₀/[4π(2L₀)³]. A thin uniformly charged spherical shell has the same radius and total charge, with surface density q₀/[4π(2L₀)²]. Let R = r/L₀ denote the radius of a concentric Gaussian sphere.

For the solid sphere, an interior Gaussian volume contains only the fraction R³/8 of the total charge. For 0 < R ≤ 2, Gauss’s law and spherical symmetry give E_r/E₀ = (R³/8)/R² = R/8. For R ≥ 2, all charge is enclosed and E_r/E₀ = 1/R². These branches agree at R = 2.

For the thin shell, a concentric sphere with R < 2 encloses no charge. Spherical symmetry then gives zero interior field. For R > 2, the field again has E_r/E₀ = 1/R². The conclusion inside this particular shell needs symmetry; an arbitrary charge-free region need not have zero field.

At the center, the full vector field of either positive spherical model is zero. There is no outward radial direction at that one point. The tabulated solid radial value there denotes its continuous limit; do not divide a zero-radius flux by a zero sphere area to derive it.

The table gives signed outward radial components for positive total charge. Reversing the assigned source charge reverses nonzero components and fluxes; field magnitudes remain nonnegative. Scaling the entire source charge to zero gives the zero field in these source-only families.