Physics
Gauss's law calculator: flux, enclosed charge and symmetric fields
Pick a relation and clear the quantity you are after. Alongside the theorem itself you get the three fields Gauss settles in one line: an infinite line, an infinite plane, and a charged sphere from inside and out.
When Gauss earns its keep, and when it is only an elegant shortcut
Gauss's law says the flux of the electric field through a closed surface equals the enclosed charge divided by ε₀. It is always true, for any surface and any distribution: it is neither an approximation nor a special case. But always true is not the same as always useful, and that gap is exactly where exam mistakes collect.
The theorem becomes a calculating tool only when symmetry lets you take E outside the integral. You need a surface on which the field has constant magnitude and a known direction relative to the normal: a concentric sphere for a spherical distribution, a coaxial cylinder for a line, a box straddling the plane. Without that symmetry the theorem is still valid but settles nothing, because the unknown stays inside the integral.
The three classic geometries give three different fall-offs, and they are worth holding in mind as a single picture: a point charge and a sphere go as 1/r², a line as 1/r, a plane not at all. The reason is purely geometric — the area of the Gaussian surface grows as r², as r, or stays constant. Recognising which of the three you are looking at is half the work.
Common mistakes
- Counting charges outside the surface: they do contribute to the field at every point, but their net flux through a closed surface is zero. Only the enclosed charge enters the theorem.
- Using the interior formula for a point outside the sphere, or the other way round: inside, the field grows with r; outside, it falls as 1/r². The two agree only at the surface.
- Confusing the field of a single plane, σ/2ε₀, with the field between two opposite plates, σ/ε₀. The factor of two comes from the two sheets' fields adding in the gap.
- Giving the infinite line a total charge: over an infinite length only the linear density means anything. If the problem states a charge, it also states the length it is spread over.
Frequently asked questions
What does Gauss's law state?
That the flux of the electric field through any closed surface equals the enclosed charge divided by ε₀. It does not depend on the shape of the surface, on where the charges sit inside it, or on any charges outside.
When should I use Gauss's law rather than Coulomb's law?
When the distribution has spherical, cylindrical or planar symmetry. In those cases you can choose a surface on which the field is constant in magnitude and perpendicular, and the integral collapses to E·A. Without symmetry the theorem is still true, but integrating Coulomb's law directly is easier.
Why doesn't the field of an infinite plane depend on distance?
Because as you move away you see contributions from an ever larger patch of the plane, and the growth in area exactly offsets the weakening of each element's contribution. It holds as long as the distance stays small compared with the real plane's size.
What is the field inside a charged conducting sphere?
Zero. In a conductor at equilibrium all the charge sits on the surface, so a Gaussian surface drawn inside encloses none of it. The case handled here, where the field grows linearly with r, is an insulating sphere with charge spread uniformly through its volume.
What unit is electric flux measured in?
Newton square metres per coulomb (N·m²/C), equivalently volt metres (V·m). You get it by multiplying a field in N/C by an area in m².
How this calculation works
Flux through a flat surface: Φ = E·A·cos θ, with θ the angle between the field and the surface normal. Gauss's law: Φ = Q₀/ε₀ through any closed surface, with ε₀ = 8.854 × 10⁻¹² F/m. Infinite straight line of density λ: E = λ/(2πε₀r), pointing radially. Infinite plane of density σ: E = σ/(2ε₀), uniform and perpendicular to the plane. Between two flat plates with opposite charges: E = σ/ε₀. Sphere of charge Q, point outside at distance r: E = kQ/r² with k = 1/(4πε₀). Uniform insulating sphere of radius R, point inside at r ≤ R: E = kQr/R³, which meets kQ/R² at the surface.