Gauss’s theorem
Electric flux and fields of a wire, plane sheet and spherical shell.
Electric flux and Gauss’s theorem
Electric flux through a small flat area is Φ = E A cos θ, where θ is the angle between the field and the normal to the area. It measures how many field lines pass through the surface. Unit: N m² C⁻¹ (or V m).
Gauss’s theorem: the total electric flux through any closed surface equals the net charge enclosed divided by ε₀: Φ = q_enclosed / ε₀. Charges outside the surface contribute zero net flux. The closed surface used is called a Gaussian surface; we choose its shape to match the symmetry of the charge.
Standard results:
- Infinitely long straight wire with linear charge density λ: E = λ / (2πε₀r), so E ∝ 1/r. Use a cylindrical Gaussian surface.
- Infinite plane sheet with surface charge density σ: E = σ / (2ε₀), the same at all distances (uniform field).
- Thin spherical shell of radius R and charge q: outside (r > R), E = q / (4πε₀r²), as if all charge sat at the centre; inside (r < R), E = 0.
Worked example: a closed surface encloses 8.85 nC. Φ = 8.85 × 10⁻⁹ / 8.854 × 10⁻¹² ≈ 1.0 × 10³ N m² C⁻¹. Its shape and size do not matter.
Key points
- Φ = q_enclosed / ε₀ for any closed surface.
- Wire: E ∝ 1/r. Sheet: E independent of r. Shell: E = 0 inside, kq/r² outside.
- Flux through a closed surface with no net enclosed charge is zero.
- Gauss’s theorem is most useful when the charge distribution is highly symmetric.
Exam tips
- “Flux through one face of a cube with charge q at its centre” = q/(6ε₀).
- Questions often ask how E depends on distance for wire, sheet or shell.
Common mistakes
- Thinking the flux depends on the size of the Gaussian surface.
- Counting charges outside the surface.
- Mixing up the sheet result σ/(2ε₀) with the field near a conductor σ/ε₀.
Practice questions
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