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

How to use Gauss’s Law Visualizer & Field Calculator

1. 1Choose a distribution: point charge, charged sphere, conducting sphere, concentric shells, line, cylinder, coaxial cable, plane, slab or parallel plates
2. 2Enter the charge (nC), charge density and sizes (cm); for a charged sphere you can also set the density exponent n
3. 3Drag 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. 4Read the flux, E, V and the E(r)/V(r) curves, and follow the derivation below

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).