Gauss's law applications — sphere, cylinder, plane symmetry selection

Question

How do we apply Gauss’s law to find the electric field for charge distributions with spherical, cylindrical, or planar symmetry? How do we choose the right Gaussian surface?

(CBSE 12, JEE Main — Gauss’s law derivations for the three standard geometries are must-know for boards and competitive exams)

Solution — Step by Step

\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_0}

The total electric flux through any closed surface equals the net charge enclosed divided by $\varepsilon_0$ .

The power of Gauss’s law: if we choose a surface where $\vec{E}$ is constant in magnitude and either parallel or perpendicular to $d\vec{A}$ at every point, the integral simplifies to a simple multiplication. This is why symmetry is everything.

Charge distribution: Point charge, uniformly charged solid sphere, hollow sphere.

Gaussian surface: Concentric sphere of radius $r$ .

On this surface, $\vec{E}$ is radially outward (or inward) and has constant magnitude. So:

\oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2

Solid sphere (radius $R$ , total charge $Q$ ):

Outside ( $r > R$ ): $E = \frac{Q}{4\pi\varepsilon_0 r^2}$ (same as point charge)
Inside ( $r < R$ ): $q_{enc} = Q \cdot \frac{r^3}{R^3}$ , so $E = \frac{Qr}{4\pi\varepsilon_0 R^3}$ (linearly increasing)

Hollow sphere:

Outside: same as above
Inside: $q_{enc} = 0$ , so $E = 0$

Charge distribution: Infinite line charge (linear charge density $\lambda$ ), uniformly charged long cylinder.

Gaussian surface: Coaxial cylinder of radius $r$ and length $L$ .

Flux through curved surface: $E \cdot 2\pi r L$ (field is radial). Flux through flat ends: 0 (field is perpendicular to the normal of the flat faces).

E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}

E = \frac{\lambda}{2\pi\varepsilon_0 r}

The field falls as $1/r$ — not $1/r^2$ as for a point charge.

Charge distribution: Infinite plane sheet (surface charge density $\sigma$ ).

Gaussian surface: Cylinder (pill-box) with flat faces parallel to the sheet, straddling it.

Flux through both flat faces: $2EA$ (field is perpendicular to the sheet on both sides). Flux through curved surface: 0 (field is parallel to it).

2EA = \frac{\sigma A}{\varepsilon_0}

E = \frac{\sigma}{2\varepsilon_0}

The field is uniform — independent of distance from the sheet. This is a unique property of infinite plane charge distributions.

flowchart TD
    A["Charge distribution"] --> B{"What symmetry?"}
    B -->|"Spherical (point, sphere)"| C["Gaussian surface: Sphere<br/>∮E·dA = E × 4πr²"]
    B -->|"Cylindrical (line, cylinder)"| D["Gaussian surface: Cylinder<br/>∮E·dA = E × 2πrL"]
    B -->|"Planar (infinite sheet)"| E["Gaussian surface: Pill-box<br/>∮E·dA = 2EA"]
    C --> F["E = Q/(4πε₀r²)"]
    D --> G["E = λ/(2πε₀r)"]
    E --> H["E = σ/(2ε₀)"]

Why This Works

Gauss’s law is always true, but it is useful for calculation only when symmetry allows us to pull $E$ out of the integral. The three standard symmetries — spherical, cylindrical, and planar — each have a natural Gaussian surface where the field is constant. Without symmetry, the integral cannot be simplified, and we must use Coulomb’s law directly (or integration).

The decreasing power of $r$ in each case reflects the dimensionality: 3D spreading (sphere, $1/r^2$ ), 2D spreading (cylinder, $1/r$ ), no spreading (plane, constant).

Common Mistake

Students try to apply Gauss’s law to finite charge distributions — like a short charged rod or a small disc. Gauss’s law is valid for any closed surface, but the symmetry-based simplification works ONLY for infinite (or effectively infinite) distributions. For a finite rod, the field is not purely radial and not constant on any simple Gaussian surface. In such cases, direct integration using Coulomb’s law is the correct approach.

For the charged sphere, memorise these two results: outside it behaves like a point charge, and inside a hollow sphere $E = 0$ (shielding). These are directly tested as 1-mark MCQs in JEE Main. Inside a solid sphere, $E \propto r$ — this linear variation is often tested in graph-based questions.

Question

Solution — Step by Step

Why This Works

Common Mistake

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