Electric flux and Gauss's theorem
Electric flux counts the field lines crossing a surface. Gauss's theorem: the flux through any closed surface is q/ε₀, where q is the charge enclosed.
Electric flux
Electric flux through a surface measures how many field lines cross it. Think of rain falling through a window: tilt the window and fewer drops pass. For a small flat area ΔS in a field E, the flux is ΔΦ = E·ΔS = EΔS cos θ. Here θ is the angle between E and the normal to the surface. Flux is a scalar. It is largest when the surface faces the field (θ = 0°) and zero when the field runs along the surface (θ = 90°). For a closed surface, the normal points outward. So outgoing lines give positive flux and incoming lines give negative flux.
Gauss's theorem
Gauss's theorem: the total electric flux through any closed surface is 1/ε₀ times the net charge enclosed by it: Φ = q/ε₀. The closed surface is called a Gaussian surface. It is imaginary, and it should not pass through a point charge. Charges outside the surface give zero net flux: every line that enters also leaves. But E in the theorem is the total field of all the charges, inside and outside. The theorem follows from the inverse-square nature of Coulomb's law. It is most useful when the charge has symmetry: a line, a plane or a sphere.
Check Gauss's theorem for a point charge
- 1Draw a sphere of radius r centred on a point charge q.
- 2At every point on the sphere, E = q/(4πε₀r²) and points along the outward normal (θ = 0°).
- 3Flux = E × total area = q/(4πε₀r²) × 4πr².
- 4★So Φ = q/ε₀. The radius cancels, so every sphere gives the same flux, as Gauss's theorem says.
★ key step
Worked example: flux through a flat surface
A uniform field of 2 × 10³ N C⁻¹ points along the x-axis. Find the flux through a rectangle 20 cm × 10 cm when its normal (a) is along x, (b) makes 60° with x, (c) is along y.
- Area ΔS = 0.20 m × 0.10 m = 0.02 m².
- (a) θ = 0°: Φ = EΔS cos 0° = 2 × 10³ × 0.02 × 1 = 40 N m² C⁻¹.
- (b) θ = 60°: Φ = 40 × cos 60° = 40 × 0.5 = 20 N m² C⁻¹.
- (c) θ = 90°: Φ = 40 × cos 90° = 0.
(a) 40 N m² C⁻¹; (b) 20 N m² C⁻¹; (c) zero
Worked example: charge inside a cube
A point charge of 8.85 μC sits at the centre of a cube. Find the flux through (a) the whole cube and (b) one face. (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
- (a) By Gauss's theorem, Φ = q/ε₀ = 8.85 × 10⁻⁶ ÷ 8.85 × 10⁻¹² = 1 × 10⁶ N m² C⁻¹.
- (b) The charge is at the centre, so by symmetry all six faces get equal flux.
- Flux through one face = 1 × 10⁶ ÷ 6 ≈ 1.67 × 10⁵ N m² C⁻¹.
(a) 1 × 10⁶ N m² C⁻¹; (b) about 1.7 × 10⁵ N m² C⁻¹ (ε₀ rounded to 8.85 × 10⁻¹²)
Points to remember
- The net flux through a closed surface with no charge inside is zero.
- A closed surface around an electric dipole has zero net flux, since +q and −q cancel.
- Changing the size or shape of the closed surface does not change the flux, if the enclosed charge stays the same.
- Charges outside change the field at points on the surface, but not the total flux.
Flux is like counting rain falling through a window: tilt it, and fewer drops pass. Installers use the same idea when they tilt a rooftop solar panel in Rajasthan to face the sun and catch the most light.
1 marks: “What is the net electric flux through a closed surface that encloses an electric dipole?” Zero, because the net charge enclosed (+q − q) is zero.
2 marks: “State Gauss's theorem. A charge q is at the centre of a cube. Find the flux through one face.” Flux through any closed surface = q(enclosed)/ε₀. By symmetry, one face gets q/(6ε₀).
3 marks: “Define electric flux and give its SI unit. A closed surface encloses +3 μC and −1 μC, while +5 μC lies outside it. Find the net flux through the surface.” Φ = E·ΔS, unit N m² C⁻¹. Only the enclosed charge counts: q = +2 μC, so Φ = 2 × 10⁻⁶ ÷ 8.854 × 10⁻¹² ≈ 2.26 × 10⁵ N m² C⁻¹.
Gauss: what is inside counts, what is outside does not (Φ = q(inside)/ε₀).