Electric flux, Gauss's theorem and its uses

Electric flux and Gauss's theorem

Electric flux through a surface measures how many field lines pass through it. For a small area, draw the area vector Δs at right angles to the surface. The flux is Δφ = E·Δs = EΔs cos θ, where θ is the angle between E and Δs. Flux is a scalar, with unit N m² C⁻¹. For a closed surface, Δs points outward. Gauss's theorem: the net flux through any closed surface equals the total charge inside it divided by ε₀, φ = q/ε₀. The closed surface is a Gaussian surface: an imaginary surface chosen to suit the symmetry. Charges outside it give no net flux. For a point charge, take a Gaussian sphere of radius r round q. E is the same all over it and along Δs, so E × 4πr² = q/ε₀. This gives E = q/(4πε₀r²): Coulomb's law again.

Derivation: field of a long line charge

A long thin wire carries a uniform charge σₗ per unit length. Find E at a point P, a distance r from the wire.

1. 1
   Symmetry: E points radially away from the wire and has the same size at all points a distance r from it.
2. 2
   Gaussian surface: a closed cylinder of radius r and length l, with the wire along its axis.
3. 3
   Flat ends: E runs along them, at 90° to their area vectors, so their flux is zero.
4. 4
   Curved surface: E is along the area vector everywhere, so the flux is E × 2πrl.
5. 5
   Charge enclosed = σₗl. Gauss's theorem: E × 2πrl = σₗl/ε₀.
6. 6★
   So, E = σₗ/(2πε₀r). The field falls as 1/r.

★ key step

Derivation: field of a charged spherical shell

A thin spherical shell of radius R carries a charge Q spread evenly over its surface.

1. 1
   Outside (r > R): draw a Gaussian sphere of radius r through P, with the same centre O as the shell.
2. 2
   By symmetry E is radial and the same all over this sphere. So the flux is E × 4πr².
3. 3
   It encloses the whole charge Q, so E × 4πr² = Q/ε₀ and E = Q/(4πε₀r²).
4. 4
   Outside, the shell acts as if all its charge were at the centre. A charged solid metal sphere gives the same result, as its charge sits on its surface.
5. 5
   Inside (r < R): a Gaussian sphere through a point P′ inside encloses no charge, so E × 4πr² = 0.
6. 6★
   So, E = 0 at every point inside the shell.

★ key step

Derivation: field of a plane sheet of charge

A large thin sheet carries a uniform charge σ per unit area. Find E at a point P in front of it.

1. 1
   Symmetry: E is at right angles to the sheet, pointing away from it on both sides (for σ > 0).
2. 2
   Gaussian surface: a cylinder with its axis along E. One flat end passes through P; the other is at P′, the same distance behind the sheet.
3. 3
   Curved surface: E runs along it, so its flux is zero.
4. 4
   Each flat end (area Δs) has flux EΔs, so the two ends give 2EΔs.
5. 5
   Charge enclosed = σΔs. Gauss's theorem: 2EΔs = σΔs/ε₀.
6. 6★
   So, E = σ/(2ε₀). The field does not depend on the distance from the sheet.

★ key step

How to draw the Gaussian surfaces and the graph

• Point charge: put +q at the centre of a dashed circle, the Gaussian sphere. Draw radial E arrows crossing it, and one small patch Δs with its area vector pointing out.
• Long wire: a vertical line with + signs; a dashed cylinder round it; label r and l; E arrows pointing out from the curved surface.
• Shell: draw a circle of radius R with + signs on it. Add a dashed circle outside it through P (r > R) and one inside it through P′ (r < R).
• Sheet: draw a large plane with + signs and a small dashed cylinder through it, with its ends at P and P′. Draw E arrows pointing away on both sides.
• E against r for a shell: E = 0 from r = 0 to R; a jump to Q/(4πε₀R²) at r = R; then a curve falling as 1/r².

Worked example: flux and three fields