Q.Four different closed surfaces, of different shapes and different sizes, are considered. Each one of the four surfaces encloses one and the same single point charge $+q$ (and no other charge). Consider the electric flux through each surface.

Gauss's Law is a shortcut. Instead of adding up the Coulomb contribution of every charge — a nightmare of vectors and integrals — it lets you read the field straight off the symmetry of a problem. The whole idea rests on one quantity: electric flux.

Flux — field-lines counted through a surface. For a uniform field E crossing a flat area A, the flux is Φ = E·A = EA cosθ, where θ is the angle between the field and the normal to the surface. Picture the field as a bundle of lines; flux is how many pierce the surface. When E is edge-on (θ = 90°) nothing passes through and Φ = 0; when E is ⊥ to the surface (θ = 0) the count is maximal. For a closed surface, Φ is the net number of lines leaving it — lines that enter and exit cancel.

The law itself. Gauss's law states that the net flux out of any closed surface depends only on the charge trapped inside:

Φ = q_enclosed / ε₀.

Two consequences do most of the work. First, only enclosed charge counts — a charge outside sends as many lines in as out, so its net contribution is exactly zero. Second, the surface's shape is irrelevant; move the charge around inside or deform the surface, and Φ never changes.

Why symmetry makes it powerful. By itself Φ = q/ε₀ has E buried in an integral. It becomes a tool only when you pick a Gaussian surface matched to the symmetry — one where E is constant and everywhere either ⊥ to the surface (so Φ = EA) or ∥ to it (contributing nothing). Then E slides out and you solve in one line. This works for exactly three geometries:

1 — Infinite line charge (linear density λ). Use a coaxial cylinder: E = λ / 2πε₀r, falling off as ∝ 1/r.

2 — Infinite sheet (surface density σ). Use a pillbox pierced through the sheet: E = σ / 2ε₀ — uniform and completely independent of distance. The field near a large charged plane simply doesn't weaken as you step back.

3 — Spherical shell / sphere. For a thin shell of charge Q, a Gaussian sphere inside encloses nothing, so E = 0 everywhere within; outside, the charge acts as if concentrated at the centre, E = kQ/r² — indistinguishable from a point charge. For a solid uniformly charged sphere, an interior surface encloses only the charge within radius r, giving E ∝ r (rising linearly from zero at the centre) up to the surface, then 1/r² beyond.

Field just outside a conductor. A charged conductor holds all its charge on the surface with E = 0 inside, so a straddling pillbox gives E = σ / ε₀ just outside — twice the sheet result, because all the flux escapes on the one outer face.

How it's examined. JEE questions test whether you can spot the symmetry, pick the right surface, and recall which result scales as 1/r, which is flat, and which is 1/r². The physics is always the one line Φ = q_enclosed / ε₀, and the skill is knowing that only the enclosed charge — never the far-off one — ever matters.

"Gauss law class 12 physics derivation" and "electric field due to infinite sheet using Gauss law" are heavily searched terms, since this is one of the core results of the Electrostatics chapter in the NCERT/CBSE Class 12 Physics curriculum. Gauss's law applications for spheres, sheets, and line charges are near-guaranteed questions in JEE Main, NEET, and state CETs.