# Gauss’s Law Visualizer - Gaussian Surfaces, Flux, E(r) and V(r) Step by Step

> Visualize Gauss’s law for spheres, shells, coaxial cables, planes and plates: drag the Gaussian surface, see enclosed charge and flux, and plot E(r) and V(r).

- URL: https://qilingtools.com/en/physics/gauss-law
- Category: Physics Tools (https://qilingtools.com/en/physics)
- Tags: [Physics](https://qilingtools.com/en/tags/physics), [Education](https://qilingtools.com/en/tags/education)
- Last updated: 2026-10-29
- Price: Free, no sign-up
- Privacy: Runs locally in the browser; files are never uploaded

Pick a symmetric charge distribution and drag the Gaussian surface (a sphere, a cylinder or a pillbox) across the cross-section. The enclosed charges light up and the tool shows the enclosed charge, the flux Φ = Q_enc/ε₀, the field and the potential. E(r) and V(r) charts mark your position, and a step-by-step derivation fills in the symmetry argument, the area, the enclosed charge and Gauss’s law with your current numbers. For a non-uniform sphere with ρ ∝ rⁿ, the enclosed charge is integrated numerically and checked against the closed form.

## How to use

1. Choose a distribution: point charge, charged sphere, conducting sphere, concentric shells, line, cylinder, coaxial cable, plane, slab or parallel plates
2. Enter the charge (nC), charge density and sizes (cm); for a charged sphere you can also set the density exponent n
3. Drag the Gaussian surface on the canvas or use the slider to change r (or x for planar cases) and see which charges are enclosed
4. Read the flux, E, V and the E(r)/V(r) curves, and follow the derivation below

## Features

- 11 spherical, cylindrical and planar distributions, including E = 0 inside conductors and induced surface charges
- Draggable Gaussian surface with enclosed charges highlighted, E arrows on the surface and a background field map
- Flux Φ = Q_enc/ε₀, cross-checked against E·A
- E(r) and V(r) charts, with V from piecewise Simpson integration and a clear choice of zero potential
- Five-step derivation with numbers: symmetry → area → enclosed charge → Gauss’s law → E and Φ
- Non-uniform sphere ρ = ρ₀(r/R)ⁿ: numerical integral compared with the closed form Q(r/R)ⁿ⁺³

## FAQ

### If Gauss’s law always holds, why does it only give the field in a few cases?

∮E·dA = Q_enc/ε₀ is true for every closed surface, but it only tells you the integral of E over that surface. You can solve for E only when the charge has enough symmetry (sphere, infinite cylinder, infinite plane) to choose a surface where E has the same size everywhere and is perpendicular (or parallel) to it, so E comes out of the integral. For other distributions the law is still true, but you need Coulomb superposition or numerics to find the field.

### Why is the field zero inside a conductor, with all the charge on the surface?

In electrostatic equilibrium the free electrons have stopped drifting, which requires E = 0 inside the metal. Any Gaussian surface drawn inside the conductor then has zero flux and so encloses no net charge: any excess charge must sit on the surface. With concentric conducting shells, the inner surface of the shell carries a charge equal and opposite to the inner sphere, and the outer surface carries the total. The tool lists these surface charges.

### Why does the field inside a uniformly charged sphere grow linearly with r?

A Gaussian sphere of radius r encloses only Q(r/R)³ of the charge, while its area is 4πr², so E = kQ(r/R)³/r² = kQr/R³, proportional to r. It peaks at kQ/R² on the surface and then falls as 1/r². The potential at the centre is 1.5 times the surface value, 1.5kQ/R. With a density ρ ∝ rⁿ, the enclosed charge becomes Q(r/R)ⁿ⁺³ and E inside grows as rⁿ⁺¹.

### Why is the field of an infinite plane independent of distance?

Use a pillbox through the plane with two faces of area A; the sides are parallel to E and carry no flux, so 2EA = σA/ε₀ and E = σ/(2ε₀), with no dependence on distance. An infinite plane looks equally “infinite” from any distance. For parallel plates the two fields add to σ/ε₀ between the plates and cancel outside.

### Why can’t the potential of a line charge be zero at infinity?

A line charge’s field E = λ/(2πε₀r) falls off only as 1/r, so integrating it out to infinity diverges logarithmically and infinity can’t serve as the zero of potential. The tool sets V = 0 at the right edge r_max for cylindrical cases and at x = 0 for planar ones. Potential differences don’t depend on that choice; between the conductors of a coaxial cable it is λ/(2πε₀)·ln(b/a).

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