Gauss’s Law

A Gaussian surface is an imaginary, closed mathematical surface chosen to apply Gauss's Law conveniently.

• It is not a real physical surface — it is purely mathematical​
• Can be any shape (sphere, cylinder, pillbox, cube)
• Chosen such that E is either uniform and parallel to dS, or perpendicular to dS, at all points​
• Must completely enclose the charge distribution of interest

How to Choose the Right Gaussian Surface

"The total electric flux through any closed surface is equal to \[\frac {1}{ε_0}\] times the net charge enclosed by that surface."

Three Forms of the Law

1. Verbal Form:
The net outward electric flux through a closed surface equals the net enclosed charge divided by ε₀.

2. Algebraic Form:

Φ_E = \[\frac {Q_enc}{ε_0}\]

3. Integral Form:

\[\oint\vec{E}\cdot d\vec{S}=\frac{Q_{\mathrm{enc}}}{\varepsilon_0}\]

Variable Legend

Real-Life Analogy

Imagine holding a fishing net around a sprinkler head. The number of water streams passing through the net depends only on how many sprinkler heads are inside — not on the shape or size of the net.

Similarly, the electric flux through any closed surface depends only​ on the charge enclosed — not on the shape of the surface or charges outside it.

Intuitive Meaning

• More charge inside → More field lines → Greater flux
• Zero charge inside → Field lines enter and exit equally → Net flux = 0
• Charges outside the surface do not contribute to the net flux

Application 1 — Electric Field due to an Infinitely Long Straight Wire

Linear charge density = λ (C/m), find E at distance r from wire.
Gaussian Surface: Co-axial cylinder of radius r and length l.

Derivation:

Flux through the two flat ends = 0 (E ⊥ dS on flat faces)​

Φ = E × 2πrl

Enclosed charge: Q_enc = λl

Applying Gauss's Law:

E × 2πrl = \[\frac {λl}{ε_0}\]

E = \[\frac {λ}{2πε_0r}\]

Direction: Radially outward (for +λ), radially inward (for −λ)

Application 2 — Electric Field due to a Uniformly Charged Infinite Plane Sheet

Surface charge density = σ (C/m²), find E at perpendicular distance from sheet.
Gaussian Surface: Pillbox (short cylinder) straddling the sheet.

Derivation:

Flux through curved part = 0; flux through two flat faces:

Φ = 2E A

Enclosed charge: Q_enc = σ A

2EA = \[\frac {σA}{ε_0}\]

E = \[\frac {σ}{2ε_0}\]

Note: E is independent of the distance from the plane sheet — it remains constant everywhere!

Direction: Normal to plane, outward (for +σ); inward (for −σ)

Application 3 — Electric Field due to a Uniformly Charged Thin Spherical Shell

Total charge Q on a spherical shell of radius R, find E at distance r from the centre.
Gaussian Surface: Concentric sphere of radius r.

Case (a): Outside the shell (r > R)

E × 4πr² = \[\frac {Q}{ε_0}\]

E = \[\frac {1}{4πε_0}\] ⋅ \[\frac {Q}{r^2}\] (r>R)

Shell behaves like a point charge at its centre.

Case (b): On the surface (r = R)

\[{E=\frac{1}{4\pi\varepsilon_0}\cdot\frac{Q}{R^2}}\]

Case (c): Inside the shell (r < R)

Enclosed charge = 0, therefore:

E = 0 (r < R)

Key Result: The electric field inside a uniformly charged spherical shell is zero everywhere.

Summary Table

Flux through a sphere enclosing a charge

Data

• Charge at centre: q = 5.0 C.
• Radius of sphere: R = 1.0 m (later doubled).

Result

• Flux through the sphere: ΦE = \[\frac {q}{ε_0}\]
• This is independent of R.
• If R is doubled, E decreases as 1/R2, area increases as R2, and the product remains the same.
• So total flux remains q/ε₀.

Cube in a non‑uniform field

Field

• E_x = αx^1/2, E_y = 0, E_z = 0, with α = 800 N C⁻¹m^−1/2.
• Cube of side a = 0.1 m.

Flux

• Only two faces (perpendicular to x) contribute to flux; others are perpendicular to E.
• Field on left face: E_L = αa^1/2.
• Field on right face: E_R = α(2a)^1/2.
• Flux: Φ = a²(E_R − E_L) = αa^5/2(\[\sqrt 2\] − 1) ≈ 1.05 N m²C⁻¹

Charge enclosed

• From Gauss’s law: q_enclosed = ε₀Φ ≈ 9.27 × 10⁻¹² C

Cylinder in a piecewise-uniform field

Field

• For x > 0: \[\vec E\] = +200 \[\hat i\] N C⁻¹.
• For x < 0: \[\vec E\] = −200 \[\hat i\] N C⁻¹.

Cylinder

• Radius 0.05 m, length 0.20 m, axis along x, centre at origin.
• Faces at x = +0.10 m and x = −0.10 m.

Flux

• Curved surface: E ⟂ dS ⇒ flux = 0.
• Left face (x < 0): Φ_L = +200 × π(0.05)² ≈ 1.57 N m²C⁻¹
• Right face (x > 0): Φ_R = +200 × π(0.05)² ≈ 1.57 N m²C⁻¹
• Net flux: Φ = Φ_L + Φ_R = 3.14 N m²C⁻¹

Charge enclosed

• From Gauss’s law: q_enclosed = ε0Φ ≈ 2.78 × 10−11 C

• Applicable to any closed surface, regardless of shape or size — sphere, cube, irregular shape
• Only enclosed charges contribute to the net flux; external charges do not
• The electric field E at the Gaussian surface is due to all charges (inside and outside), but the net flux depends only on enclosed charge​
• Gauss's Law is valid for both stationary and moving charges​
• It is one of Maxwell's four equations of electromagnetism​
• Gauss's Law can be derived from Coulomb's Law for static charges, and vice versa — both are equivalent​
• If net enclosed charge = 0, net flux = 0 (but E ≠ 0 necessarily)