State Gauss's theorem. Use it to derive the electric field due to a uniformly charged thin spherical shell at a point outside it and at a point inside it.

1. 1 · State the theorem. The net electric flux through a closed surface is 1/ε₀ times the charge enclosed: φ = q/ε₀.
2. 2 · Draw the figure. [Shell of radius R with + signs; a dashed Gaussian sphere of radius r > R through P; a dashed sphere of radius r < R through P′; E arrows radially out]
3. 3 · Outside point. On a Gaussian sphere of radius r, E is radial and equal all over, so the flux is E × 4πr² = Q/ε₀.
4. 4 · Result outside. E = Q/(4πε₀r²): the shell acts as if all its charge were at the centre.
5. 5 · Inside point. A Gaussian sphere of radius r < R encloses no charge, so E × 4πr² = 0 and E = 0.
6. 6 · Draw the graph. [E against r: zero up to R, a jump to Q/(4πε₀R²) at R, then falling as 1/r²]
7. 7 · Real life. A closed metal shell shields what is inside it, which is why instruments are kept in metal boxes.
8. 8 · Close in one line. Outside the shell E falls as 1/r²; inside it E is zero.

Scoring tip: Say why the flux is E × 4πr²: E is the same everywhere on the sphere and parallel to the area vector. That reason earns the method mark.