Applications of Gauss’s law
A long wire and a flat sheet
An infinitely long straight wire carries a uniform charge λ per metre. By symmetry, its field must point straight out from the wire, at right angles to it, and must have the same size at every point a distance r away.
So choose a Gaussian surface that fits: a cylinder of radius r and length l, with the wire along its axis.
On the two flat ends, E runs parallel to the surface, so no flux goes through them. On the curved surface, E is along the normal and has the same size everywhere. The flux there is E times 2πrl.
The charge inside the cylinder is λl. Gauss’s law gives E times 2πrl equals λl over ε₀. So E equals λ over 2πε₀r, pointing outwards if λ is positive.
The field of a line of charge falls as 1 over r, not 1 over r². For λ equals 1 microcoulomb per metre, at 10 centimetres, E is 2 times 9 × 10⁹ times 10⁻⁶, over 0.1: 1.8 × 10⁵ newtons per coulomb.
Now an infinite flat sheet with uniform surface charge density σ. By symmetry, E is at right angles to the sheet, pointing away from it on both sides if σ is positive.
Choose a small cylinder of cross-section A, passing through the sheet, with one end on each side. On its curved side, E is parallel to the surface: no flux. Each end has flux E A, so the total is 2EA.
The charge inside is σA. So 2EA equals σA over ε₀, and E equals σ over 2ε₀. Notice: r does not appear. The field is the same at every distance from the sheet.
For σ equals 1 microcoulomb per square metre, E is 10⁻⁶ over 2 times 8.854 × 10⁻¹²: about 5.6 × 10⁴ newtons per coulomb. For a real, finite sheet, this holds near its middle, away from the edges.
A thin spherical shell
A thin spherical shell of radius R carries a charge q spread evenly over its surface: σ equals q over 4πR². By symmetry, E points radially, and its size depends only on the distance r from the centre.
First, a point outside, with r greater than R. Take a Gaussian sphere of radius r, centred on the shell. The flux through it is E times 4πr², and it encloses the whole charge q.
So E times 4πr² equals q over ε₀, and E equals q over 4πε₀r². Outside, the shell behaves exactly as if all its charge were a point charge at its centre.
Written with σ, since q is 4πR²σ, the field outside is E equals σR² over ε₀r².
Now a point inside, with r less than R. The Gaussian sphere is now inside the shell, and it encloses no charge at all.
So E times 4πr² equals zero, and E equals zero. The field is zero everywhere inside a uniformly charged thin shell.
Draw the graph of E against r. It is zero all the way from the centre up to R. At the shell it jumps up to q over 4πε₀R², and then falls off as 1 over r².
Example: a shell of radius 6 centimetres carries 10 nanocoulombs. At 3 centimetres from the centre, inside, E is zero. At 10 centimetres, E is 9 × 10⁹ times 10⁻⁸ over 0.1 squared: 9 × 10³ newtons per coulomb.
So, with symmetry and Gauss’s law: a wire gives E proportional to 1 over r, a sheet gives a constant E, and a shell gives 1 over r² outside and zero inside.
Try it: Gaussian surfaces
- Wire: drag the cylinder wider. The graph follows 1/r: double r and E halves. At 10 cm, E = 1.8 × 10⁵ N/C.
- Sheet: move the ends of the pillbox. E stays σ/2ε₀ ≈ 5.6 × 10⁴ N/C for σ = 1 μC/m², at every distance.
- Shell: shrink the Gaussian sphere inside the shell. It encloses no charge, so E = 0.
- Shell: move outside. The graph jumps at R and then falls as 1/r².
Field of a long wire
The field 2 cm from a long, uniformly charged straight wire is 9 × 10⁴ N/C. What is the field 6 cm from the wire?
- 1 × 10⁴ N/C
- 3 × 10⁴ N/C
- 9 × 10⁴ N/C
- 2.7 × 10⁵ N/C
Answer
B. For a line charge, E = λ/2πε₀r, so E ∝ 1/r. Three times the distance gives one third of the field: 9 × 10⁴ / 3 = 3 × 10⁴ N/C. (1 × 10⁴ uses the 1/r² law of a point charge.)
Inside and outside a shell
A thin spherical shell of radius R carries charge q. What is the field at a distance R/2 from the centre, and at 2R?
- kq/(R/2)² and kq/(2R)²
- 0 and kq/4R²
- 0 and 0
- kq/R² at both
Answer
B. At R/2 the Gaussian sphere encloses no charge, so E = 0. At 2R it encloses q, so E = kq/(2R)² = kq/4R², as if q were at the centre.
Summary
- Infinite line charge:
E = λ / 2πε₀r - Infinite plane sheet:
E = σ / 2ε₀ - Thin shell, outside (r > R):
E = q / 4πε₀r² - Thin shell, inside (r < R):
E = 0
- Use symmetry to find the direction of E, then choose a Gaussian surface on which E is either along the normal and constant, or parallel to the surface.
- Wire: a coaxial cylinder; only the curved surface has flux.
- Sheet: a small cylinder through the sheet; only the two ends have flux. The field does not depend on distance.
- Shell: a concentric sphere. Outside, the shell acts like a point charge at its centre; inside, the field is zero.