Application of Gauss' Law

Follow this procedure for every application:

1. Identify the symmetry of the charge distribution (spherical / cylindrical / planar)
2. Choose the Gaussian surface that matches the symmetry
3. Evaluate the flux Φ = E ⋅ A (since \[\vec E\] ∥ d\[\vec A\] on the chosen surface)
4. Find the enclosed charge Q_enc
5. Apply Gauss's Law E ⋅ A = Q_enc/ε₀ and solve for E

Setup

• Infinitely long, straight wire with uniform linear charge density λ (C/m)
• Find E at a perpendicular distance r from the wire​

Gaussian Surface

• Coaxial cylinder of radius rr and length l
• Flux through flat ends = 0 (since \[\vec E\] ⊥ d\[\vec A\] at ends)
• Flux through curved surface = E ⋅ (2πrl)

Derivation Outline

Qenc = λl

Φ = E ⋅ 2πrl = \[\frac {λl}{ε_0}\]

✦ Result

E = \[\frac {λ}{2πε_0r}\]

Key Points

• E ∝ \[\frac {1}{r}\] (decreases as distance increases)
• Direction: radially outward for positive λ; radially inward for negative λ
• λ = linear charge density; SI unit = C/m

Setup

• Uniformly charged thin insulating infinite plane sheet
• Surface charge density σ (C/m²); same on both sides​

Gaussian Surface

• Rectangular pillbox (cylinder) straddling the sheet
• Flux through curved side wall = 0 (since \[\vec E\] || sheet surface)
• Flux through two flat faces (each area A) = 2EA

Derivation Outline

Q_enc = σA

2EA = \[\frac {σA}{ε_0}\]

✦ Result

E = \[\frac {σ}{2ε_0}\]

Key Points

• E is uniform and independent of distance from the sheet​
• Direction: perpendicular to the sheet (outward for +σ, inward for −σ)
• Valid only for an ideal infinite insulating sheet​

Setup

• A single conducting plate with surface charge density σ on its outer surface only​

Key Difference from Insulating Sheet

The pillbox Gaussian surface here has flux only through one face (the field is zero inside a conductor):

EA = \[\frac {σA}{ε_0}\]

✦ Result

E = \[\frac {σ}{ε_0]\]

"Why is the conducting plate formula different from the insulating sheet?"

• Insulating sheet: charge is on both surfaces → flux exits from both faces → divide by 2 → E = σ/2ε₀
• Conducting plate: charge resides only on outer surface; field is zero inside the conductor → flux exits from one face only → E = σ/ε₀

Setup

• Thin shell of radius R, total charge Q, surface charge density σ

Gaussian Surface

• Concentric spherical surface of radius r

Sub-case A: Outside the Shell (r > R)

E ⋅ 4πr² = \[\frac {Q}{ε_0}\]

E = \[\frac {Q}{4πε_0r^2}\] = \[\frac {σR^2}{ε_0r^2}\]

The shell behaves as a point charge at its centre for external points.​

Sub-case B: On the Surface (r = R)

E = \[\frac {Q}{4πε_0R^2}\] = \[\frac {σ}{ε_0}\]

Sub-case C: Inside the Shell (r < R)

The Gaussian surface encloses zero charge → Q_enc = 0

E = 0

Analogy: Think of the shell like a hollow coconut. Anywhere outside, the whole coconut "feels" like one concentrated point. But inside the hollow shell, there is complete shielding — no electric field at all.

Radius R, total charge Q, volume charge density ρ

Outside (r > R)

E = \[\frac {Q}{4πε_0r^2}\]

Inside (r < R)

Q_enc = \[\frac {4}{3}\]πr³ ρ = Q ⋅ \[\frac {r^3}{R^3}\]

E = \[\frac {Qr}{4πε_0R^3}\] ∝ r