Gauss's Law Applications

Integral Form: ∮ E⃗·dA⃗ = Q_enclosed/ε₀

The electric flux through a closed surface equals the enclosed charge divided by ε₀.

Differential Form: ∇·E⃗ = ρ/ε₀

The divergence of the electric field at a point equals the charge density at that point divided by ε₀.

What they tell us: • Integral form: Relates field over a surface to enclosed charge (useful for calculations) • Differential form: Local relationship between field and charge density • Both are equivalent formulations via the divergence theorem • Electric field lines originate from positive charges and terminate on negative charges

Step 1: Choose Gaussian surface Use a cylindrical surface of radius r and length L, coaxial with the line charge.

Step 2: Apply symmetry By symmetry, E is radial and constant at distance r. E⃗ ⊥ end caps (no flux through ends) E⃗ ∥ dA⃗ on curved surface

Step 3: Calculate flux Φ = ∮ E⃗·dA⃗ = E(2πrL) (only curved surface contributes)

Step 4: Find enclosed charge Q_enclosed = λL

Step 5: Apply Gauss's Law E(2πrL) = λL/ε₀ E = λ/(2πε₀r)

Direction: Radially outward from line (if λ > 0)

The field decreases as 1/r for an infinite line charge.

Use spherical Gaussian surfaces of radius r.

OUTSIDE (r > R): Q_enclosed = ρ(4πR³/3) = total charge Φ = E(4πr²)

E(4πr²) = ρ(4πR³/3)/ε₀ E = ρR³/(3ε₀r²) = Q/(4πε₀r²)

Same as point charge! (r > R)

INSIDE (r < R): Q_enclosed = ρ(4πr³/3) Φ = E(4πr²)

E(4πr²) = ρ(4πr³/3)/ε₀ E = ρr/(3ε₀)

Key results: • Inside: E ∝ r (linear increase from center) • Outside: E ∝ 1/r² (like point charge) • At r = 0: E = 0 (by symmetry) • At r = R: Both formulas give same value (continuous) • Maximum field at surface r = R

CONDUCTING PLANE:

Step 1: Choose Gaussian surface Cylindrical "pillbox" with area A, one end inside conductor, one outside.

Step 2: Apply conductor properties • E = 0 inside conductor • E ⊥ surface outside

Step 3: Calculate flux Φ = EA (only outer end cap contributes)

Step 4: Enclosed charge Q_enclosed = σA

Step 5: Gauss's Law EA = σA/ε₀ E = σ/ε₀

NON-CONDUCTING SHEET: E = σ/(2ε₀) on each side

Key difference: • Conductor: E = σ/ε₀ (factor of 2 larger!) • Non-conductor: E = σ/(2ε₀)

Why? For conductor, all charge is on one surface. For non-conducting sheet, charge distributed throughout, field on both sides.

Use spherical Gaussian surfaces:

INSIDE (r < R): Q_enclosed = 0 (all charge on shell at r = R) ∮ E⃗·dA⃗ = 0 E = 0 for r < R

OUTSIDE (r > R): Q_enclosed = Q E(4πr²) = Q/ε₀ E = Q/(4πε₀r²) = kQ/r² for r > R

AT SURFACE (r = R): E = kQ/R² (discontinuous jump)

Sketch of E(r): • E = 0 for r < R (flat line at zero) • Jump discontinuity at r = R • E = kQ/r² for r > R (1/r² decay)

Physical insight: • Inside a uniformly charged shell, field is exactly zero! • This is why Faraday cages work • Outside, shell acts like point charge at center • Surface discontinuity reflects surface charge density