AP Physics C: E&M · Unit 1: Electric Charges, Fields & Gauss's Law
1.6 Gauss's Law
Gauss's Law Basics
Gauss's Law relates electric flux through a closed surface to the enclosed charge: Φ_E = ∮E·dA = Q_enc/ε₀. It works for any closed surface, but only the charge *inside* the surface matters. Charges outside contribute zero net flux, since their field lines enter and exit the surface equally.
Choosing a Gaussian Surface
To use Gauss's Law effectively, pick a Gaussian surface matching the symmetry of the charge distribution: spheres for point/spherical charges, cylinders for line charges, and flat pillboxes for planes. A good surface makes E constant and either parallel or perpendicular to dA, so E can be pulled out of the integral.
Applying Gauss's Law
Once flux is simplified to E·A = Q_enc/ε₀, solve for E directly. This method quickly gives field expressions for spheres (E ∝ 1/r²), infinite lines (E ∝ 1/r), and infinite planes (E = constant), avoiding messy integration required by Coulomb's Law.
Key terms
- Gauss's Law (formula)
- Φ_E = ∮E·dA = Q_enc/ε₀ — flux through a closed surface equals enclosed charge over ε₀.
- Gaussian surface
- An imaginary closed surface chosen to match the symmetry of a charge distribution, simplifying flux calculations.
- Electric flux (Φ_E)
- A measure of the electric field passing through a surface, Φ_E = ∫E·dA.
- Field outside infinite line charge
- E = λ/(2πε₀r), found using a cylindrical Gaussian surface.
- Field from infinite charged plane
- E = σ/(2ε₀), constant regardless of distance from the plane, using a pillbox Gaussian surface.
Try an AP-style question
A closed surface encloses +3 μC and −1 μC. What is the net flux through it? (ε₀ = 8.85 × 10⁻¹²)
- 0
- About 3.4 × 10⁵ N·m²/C
- About 2.3 × 10⁵ N·m²/C
- About 4.5 × 10⁵ N·m²/C
Show the answer
C. Q_enc/ε₀ = 2 × 10⁻⁶ / 8.85 × 10⁻¹².
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