Gauss’s Law
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 8 practice questions.
Gauss’s law, ∮E · dA = Qenc/ε0, is always true, but it’s only a shortcut for finding E when symmetry makes E constant over the Gaussian surface. Then the integral becomes E × (area), and you can solve for E.
1. Three symmetries
2. A routine that works
1. Pick a Gaussian surface that matches the symmetry and passes through the point.
2. Argue that E is constant (and perpendicular) on it, so ∮E · dA = E × area.
3. Find Qenc, using the density if the charge is spread out. Solve for E.
Inside a uniform solid sphere: Qenc = Q(r/R)³, giving E = kQr/R³.
Inside a thin shell: Qenc = 0, so E = 0.
Non-uniform ρ(r): Qenc = ∫ρ 4πr² dr.
3. Example, then try it
A very long line of charge has λ = 2.0 nC/m. Find the field 0.10 m away.
E = λ/(2πε0r) = 2kλ/r = 2(9.0 × 10⁹)(2.0 × 10⁻⁹) ÷ 0.10 = 360 N/C, pointing away from the line.
A solid insulating sphere of radius 0.10 m carries 5.0 nC spread uniformly through its volume. Find E at r = 0.050 m, at the surface, and at r = 0.20 m.
Inside: Qenc = Q(r/R)³, and E(4πr²) = Qenc/ε0 gives E = kQr/R³.
r = 0.050 m: E = (9.0 × 10⁹)(5.0 × 10⁻⁹)(0.050) ÷ (0.10)³ = 2250 N/C.
Surface: E = kQ/R² = 45 ÷ 0.010 = 4500 N/C, the maximum.
r = 0.20 m: E = kQ/r² = 45 ÷ 0.040 ≈ 1100 N/C (1125), a quarter of the surface value.
Answer: 2250 N/C, 4500 N/C, about 1100 N/C.
4. Common mistakes
How does the electric field near a very large, uniformly charged flat sheet depend on the distance from it?
Gauss’s law with a pillbox gives E = σ/(2ε0), with no r in it. For a sheet much larger than the distance, the field is uniform.
Ready to practice?
8 questions on Gauss’s Law, each with a hint and an explanation for every wrong answer.