Planar symmetry and sheets of charge

Put a uniform surface charge σ on the plane z = 0 and list what leaves the source unchanged: any slide in x, any slide in y, any rotation about the z axis, and a reflection in the plane itself. The field has to inherit every one of them. Translation in the plane means E cannot depend on x or y. Rotation about the normal kills any in-plane component, because such a component would have to point somewhere and no direction in the plane is special. So E = E(z)ẑ. The reflection maps z to −z while leaving σ alone, so the field on one side is the mirror of the field on the other: E(−z) = −E(z). Three sentences of symmetry have cut an unknown vector function of three variables down to one unknown number at each distance, and that is the entire reason Gauss's law is about to work here.

Take a closed cylinder — a pillbox — with its axis along z, flat caps of area A at z = +h and z = −h, and its curved wall cutting through the sheet. On the wall E is parallel to the surface, so E ⋅ dA = 0 there. On each cap E is normal to the surface and constant over it, and the mirror symmetry makes the outward flux equal on the two caps, so ∮E ⋅ dA = 2EA. The charge inside is whatever σ covers: q(enc) = σA. Gauss's law gives 2EA = σA/ε₀, hence E = σ/(2ε₀). Two features deserve attention. A cancels, so the answer cannot depend on the size of the box you imagined — a sign the calculation is honest. And h never entered at all, which is the claim that the field does not weaken with distance. With σ = 2.0 nC m⁻², E = 2.0 × 10⁻⁹ / (2 × 8.854 × 10⁻¹²) = 113 N C⁻¹, and a probe at 5 mm and a probe at 5 m read the same.

A field independent of distance looks like a violation of the inverse-square law, and it is not. Stack rings and integrate to get the axial field of a uniformly charged disc of radius R: E = σ/(2ε₀) [1 − z/√(z² + R²)]. Every ring obeys 1/r², but as you retreat the rings that dominate sit further out, and their circumference grows just fast enough to pay for the weakening. Two limits make the trade explicit. Hold z and let R → ∞: the bracket goes to 1 and E → σ/(2ε₀), so the compensation is exact only when the sheet never runs out. Hold R and let z ≫ R instead: 1 − z/√(z² + R²) ≈ R²/(2z²), so E ≈ σR²/(4ε₀z²) = Q/(4πε₀z²) with Q = σπR². Far enough away the disc is a point charge again, exactly as it should be.

Sheets superpose like anything else: add the vectors. Put σ₁ on a sheet at the left and σ₂ on one at the right, and take x to the right. Each sheet contributes magnitude |σ|/(2ε₀) everywhere, directed away from itself when σ is positive. Left of both, the two contributions point left, giving (σ₁ + σ₂)/(2ε₀) leftward. Between them the left sheet pushes right and the right sheet pushes left, so they subtract: the x component is (σ₁ − σ₂)/(2ε₀). Right of both they add again. Set σ₁ = +σ and σ₂ = −σ and the arithmetic collapses to the capacitor: zero outside, σ/ε₀ in the gap, uniform. With σ = 35 nC m⁻² that is 35 × 10⁻⁹ / (8.854 × 10⁻¹²) = 4.0 × 10³ N C⁻¹, and the gap width never appears. Set σ₁ = σ₂ = +σ instead and the picture inverts: zero between the sheets, σ/ε₀ outside.

Give the charge a thickness. A slab of uniform ρ filling |z| ≤ d has the same planar symmetry, so E = E(z)ẑ with E(−z) = −E(z), and a pillbox of cap area A reaching from −z to +z inside the slab encloses ρ(2zA). Gauss's law gives 2E(z)A = 2ρzA/ε₀, so E(z) = ρz/ε₀: the field climbs linearly from zero at the mid-plane. Push the caps outside and the enclosed charge stops growing at ρ(2dA), so E = ρd/ε₀ everywhere beyond, which is σ/(2ε₀) for the equivalent sheet σ = 2ρd. With ρ = 3.2 μC m⁻³ and d = 3.0 mm that outside field is 1.08 × 10³ N C⁻¹, matching the inside value at the face — a volume charge has no surface density there, so nothing jumps. Now shrink d towards zero holding σ = 2ρd fixed. The ramp steepens and narrows until it is a step, and the step is σ/ε₀. That is the general boundary condition: crossing any surface charge the normal component of E jumps by σ/ε₀, while the tangential component passes through unchanged.

The infinite sheet is a model, and a demanding one. On the axis of a disc of radius R the exact field is σ/(2ε₀) [1 − z/√(z² + R²)], so the fractional shortfall is z/√(z² + R²), which is close to z/R whenever z ≪ R. Landing within 1% of σ/(2ε₀) therefore means staying inside z ≈ 0.01R; at z = 0.1R you are already 10% low. For a plate 12 cm across, R = 6 cm and 1% means under a millimetre — not a few centimetres. The same condition in different costume governs the parallel-plate capacitor: the uniform σ/ε₀ field holds while the gap is small compared with the plate dimensions, and the fringing field at the rim is exactly what the model discards. Off the axis it is worse, because near an edge the field is neither uniform nor perpendicular. Planar symmetry is exact only for an infinite plane; for anything real it is a statement about being close.