How this instrument works
Gauss's law states that the total electric flux leaving a closed surface equals the charge enclosed divided by the permittivity of free space: Φ = Q_enc ⁄ ε₀. Flux counts field lines crossing the surface, weighted by area and by the angle between the field and the surface's outward normal, and the formula folds all of that geometric bookkeeping into one ratio. The result Gauss proved with the divergence theorem is that the surface's shape, its size, and how the charge sits inside it are all irrelevant — only the enclosed charge decides the total.
The constant ε₀, vacuum permittivity, sets the strength of the electric force in SI units and is fixed at 8.8541878176 × 10⁻¹² farads per metre. Gauss's law is not a separate assumption bolted onto Coulomb's law; it follows directly from the inverse-square shape of the Coulomb field, and the two describe the same static charges from different angles. What makes Gauss's law the practical shortcut is symmetry: pick a Gaussian surface where the field has constant magnitude and points straight through it — a sphere around a point charge, a cylinder around a wire, a slab beside a charged plane — and the field can be pulled out of the flux integral and solved for algebraically, no integration of Coulomb's law required.
An electrical engineer checking that a metal enclosure blocks outside interference leans on the same charge-in, flux-out bookkeeping: a hollow conductor's interior carries zero enclosed charge, so the flux through any surface drawn inside it is zero, which is the entire reason a Faraday cage works. The recurring mistake is assuming the surface has to be centred on the charge, or shaped to match it, for the law to apply. It does not. Move the enclosed charge anywhere inside the boundary, or swap a sphere for an irregular blob of twice the volume, and the total flux is unchanged — what you lose without that symmetry is the ability to say anything about the field's direction or strength at a single point, which is why this instrument reports flux rather than field.
- Enter the net charge inside the boundary in the Enclosed charge field — positive or negative, in coulombs, with millicoulomb, microcoulomb, or nanocoulomb options in the unit menu.
- Do not worry about the surface itself: Gauss's law gives the same answer whether you imagine a sphere, a cube, or an irregular closed shape, as long as the same charge sits inside it.
- Read the result in Electric flux, N·m² ⁄ C — the total field lines crossing the surface outward, independent of the surface's size or distance from the charge.
- For a symmetric distribution — a point charge, an infinite line, an infinite sheet — divide the flux by the surface area to recover the electric field's magnitude directly.
Worked example — flux from 5 nC of enclosed charge
A small charged bead carries 5 nC of charge, or 5 × 10⁻⁹ C, and a closed surface is drawn around it — say a sphere, though a cube would give the same reading. Gauss's law gives the flux in one step: Φ = Q_enc ⁄ ε₀ = (5 × 10⁻⁹) ⁄ (8.8541878176 × 10⁻¹²) ≈ 564.70 N·m²/C. That is the total electric flux crossing the surface outward, in newton-metres squared per coulomb, and it holds whether the sphere has a radius of one millimetre or one metre.
Push the same bead off-centre inside a much larger, oddly shaped enclosure, or swap the sphere for a cube of twice the volume, and the reading stays exactly 564.70 N·m²/C — only the enclosed charge changes the total. Double the charge to 10 nC instead and the flux doubles too, to about 1129.41 N·m²/C, because Φ scales linearly with Q_enc; nothing about the geometry enters the calculation at all.
Questions
Does the shape or position of the surface change the flux?
No. Gauss's law depends only on the charge enclosed, never on the surface's shape, its size, or where the charge sits inside it. A tight sphere and a lumpy, off-centre blob drawn around the same 5 nC both read exactly 564.70 N·m²/C, because the law follows from the same inverse-square field that makes Coulomb's law shape-independent to begin with.
What happens if the surface encloses no net charge?
The flux comes out to zero, even if strong fields pass straight through the surface. A dipole sitting entirely inside a Gaussian surface, or a metal enclosure shielding an outside field, both give Φ = 0 — either the positive and negative contributions cancel, or there is simply no enclosed charge, which is why a closed conductor blocks fields from reaching its interior.
Can Gauss's law give me the electric field directly, not just the flux?
Only when the charge distribution has enough symmetry to make the field constant over a chosen surface — spherical, cylindrical, or planar symmetry. Then the field can be pulled out of the flux integral and solved for on its own, for example E = Φ ⁄ A around a point charge. Without that symmetry the law still holds exactly, but it no longer isolates the field by itself.
Why does a negative enclosed charge give negative flux?
Because flux measures field lines crossing the surface outward, and field lines point toward negative charge rather than away from it. Enter −5 nC and the instrument returns −564.70 N·m²/C — the same magnitude as the positive case, but every field line runs inward instead of outward through the boundary.
Does doubling the enclosed charge double the flux?
Yes. Flux scales linearly with enclosed charge: 5 nC produces about 564.70 N·m²/C, and 10 nC produces exactly twice that, about 1129.41 N·m²/C, however that charge happens to be distributed inside the surface. Halving the charge halves the flux the same way.
Is vacuum permittivity the same thing as the Coulomb constant?
Related but not identical. The Coulomb constant is k = 1 ⁄ (4πε₀), about 8.988 × 10⁹ N·m²/C², and it appears in the point-charge force law. Vacuum permittivity, ε₀ = 8.8541878176 × 10⁻¹² F/m, is the more fundamental SI constant, and Gauss's law is written directly in terms of it rather than k.