Gauss's law and its applications

Why it matters

Gauss's law turns many field problems that would need a messy Coulomb integral into one line of algebra. Fields of coaxial cables, charged wires, spheres and plates — the geometries of real capacitive sensors, cable insulation and shielding — all come straight from it. Its point form, ∇·D = ρv, is also the first of Maxwell's equations.

Key ideas

Electric flux density D. In a linear isotropic medium D = εE = ε₀εrE, in C/m². D depends only on free charge (it is the same in free space and in a dielectric for a given free-charge arrangement with symmetric geometry), which is why Gauss's law is cleanest when written with D.

Electric flux. The flux through a surface is Ψ = ∫ D·dS (unit: coulomb). Only the component of D normal to the surface contributes. If you work with E instead, the flux of E through a closed surface is Q/ε₀ (in V·m) — be clear which one a question uses.

Gauss's law. The net outward flux of D through any closed surface equals the free charge enclosed: ∮ D·dS = Q_enc. It holds for every closed surface and every charge distribution; charges outside the surface contribute zero net flux (their field lines enter and leave). It only gives E directly when symmetry allows you to choose a Gaussian surface on which:

• D is everywhere either normal to the surface with constant magnitude, or tangential to it (contributing zero flux).

The three usable symmetries are spherical (point charge, uniformly charged sphere or shell), cylindrical (infinite line, coaxial cylinders) and planar (infinite sheet, slab).

Point form. Applying the divergence theorem to a tiny volume gives ∇·D = ρv: the divergence of D at a point equals the free volume-charge density there. Where there is no charge, D has zero divergence — field lines neither start nor end.

Conductors. Inside a conductor in electrostatic equilibrium E = 0, so any Gaussian surface inside the material encloses zero net charge; all excess charge sits on the surface. A cavity with no charge inside it is field-free (electrostatic shielding).

Limits. Gauss's law is always true, but for an irregular distribution (a finite line, a disc off its axis, a square plate) the field magnitude is not constant over any simple surface, so you cannot pull |D| outside the integral; use Coulomb's law instead.

Formulas

• Ψ = ∮ D·dS = Q_enc — Gauss's law (integral form); Ψ in C, D in C/m², dS in m².
• ∇·D = ρv — point form; ρv in C/m³.
• D = εE = ε₀εr E — constitutive relation; ε₀ = 8.854 × 10⁻¹² F/m.
• E = Q / (4πεr²) a_r — point charge, or outside a spherically symmetric charge Q (r ≥ radius).
• E = ρv r / (3ε) a_r (r ≤ a) and E = ρv a³ / (3εr²) a_r (r ≥ a) — uniformly charged sphere of radius a.
• E = ρL / (2περ) a_ρ — infinite line charge; ρL in C/m.
• E = ρS / (2ε) a_n — infinite sheet of charge; ρS in C/m².
• E = 0 for ρ < a, E = ρS a / (ερ) a_ρ for ρ > a — infinitely long cylindrical shell of radius a with surface charge ρS.
• ∮ E·dS = Q_enc / ε₀ — the same law in terms of E, in vacuum (V·m).

Worked examples

Example 1 (standard). Find D and E at 0.5 m from a 5 μC point charge in vacuum.

1. Gaussian surface: a sphere of radius r = 0.5 m centred on the charge; D is radial and constant over it.
2. ∮ D·dS = D × 4πr² = Q, so D = Q / (4πr²) = 5 × 10⁻⁶ / (4π × 0.25) = 1.59 μC/m².
3. E = D / ε₀ = 1.5915 × 10⁻⁶ / 8.854 × 10⁻¹² = 1.80 × 10⁵ V/m, radially outward.

Example 2 (GATE level). A sphere of radius a = 0.1 m carries a uniform volume charge ρv = 2 μC/m³ in free space. Find E at r = 0.05 m and r = 0.2 m.

1. Inside (r < a): charge enclosed = ρv × (4/3)πr³. Gauss: ε₀E × 4πr² = ρv (4/3)πr³, so E = ρv r / (3ε₀).
2. E(0.05) = 2 × 10⁻⁶ × 0.05 / (3 × 8.854 × 10⁻¹²) = 3.76 kV/m.
3. Outside (r > a): all charge Q = ρv (4/3)πa³ = 8.38 nC is enclosed, so E = ρv a³ / (3ε₀r²).
4. E(0.2) = 2 × 10⁻⁶ × 10⁻³ / (3 × 8.854 × 10⁻¹² × 0.04) = 1.88 kV/m.
5. Check: at r = a both formulas give ρv a/(3ε₀) = 7.53 kV/m, so E is continuous at the surface (there is no surface charge). E grows linearly inside and falls as 1/r² outside, peaking at r = a.

Example 3 (flux by symmetry). A 1 μC point charge sits at the centre of a cube. Find the flux of E through one face.

1. Total flux of E through the closed cube = Q/ε₀. By symmetry the six faces share it equally.
2. Flux per face = Q / (6ε₀) = 10⁻⁶ / (6 × 8.854 × 10⁻¹²) = 1.88 × 10⁴ V·m (in terms of D: 1/6 μC per face).

Common mistakes

• Thinking Gauss's law is "only valid for symmetric charges". It is always valid; symmetry is only needed to extract E from it.
• Including charge outside the Gaussian surface in Q_enc, or forgetting that only enclosed charge counts.
• Using the total charge Q of a sphere for a point inside it — inside a uniform sphere only the fraction (r/a)³ is enclosed.
• Concluding that zero net flux means zero field on the surface: a surface with a charge outside it has zero net flux but a non-zero E everywhere on it.
• Mixing up the flux of D (in C) with the flux of E (in V·m, = Q/ε₀).
• Applying the infinite-line or infinite-sheet formula to a short wire or small plate far from it.

For GATE IN

Common items: E or D inside and outside a charged sphere, shell or cylinder (often asking where E is maximum or the ratio of fields at two radii); flux through a face of a cube or through part of a sphere; recovering ρv from a given D using ∇·D; and the field between coaxial cylinders, which leads straight into capacitance. Practise all three symmetries, and the cylindrical and spherical divergence formulas.

Quick check

1. What is the net flux of D through a closed surface containing +3 μC and −1 μC?
2. Inside a uniformly charged solid sphere, how does E vary with r?
3. D = 2x a_x C/m² — what is ρv?
4. A charge Q is at the centre of a cube; what fraction of its flux leaves one face?
5. What is E inside a charged hollow conducting sphere with no charge in the cavity?

Answers: 1. 2 μC. 2. Linearly (E ∝ r). 3. 2 C/m³. 4. One sixth. 5. Zero.