Fields from Gauss's law.

E against r for a thin shell, a solid (insulating or conducting) sphere, a long line and a large sheet, with the matching Gaussian surface drawn at the probe radius.

For a uniformly charged solid sphere of radius 10 cm, find both distances where E is half its surface value (5 cm and 14.1 cm)

Using kq/r² inside a charged shell, or making a sheet's field fall with distance.

• Infinitely long straight wire with uniform λ: by symmetry the field is radial and depends only on r. A coaxial cylinder of length l gives E × 2πrl = λl/ε₀, so E = λ/(2πε₀r), falling as 1/r.
• Infinite plane sheet with uniform σ: the field is normal to the sheet on both sides. A box straddling the sheet gives 2EA = σA/ε₀, so E = σ/(2ε₀), the same at every distance from the sheet.
• For a large finite sheet, E = σ/(2ε₀) holds well in the middle region away from the edges; for a long finite wire, λ/(2πε₀r) holds near its middle.
• Uniformly charged thin spherical shell, outside (r > R): E = q/(4πε₀r²), as if the whole charge q = 4πR²σ were at the centre.
• Inside the shell (r < R): the Gaussian sphere encloses no charge, so E = 0 everywhere inside. Experiments confirming this confirm the 1/r² law.
• A solid sphere with uniform volume charge also has, outside it, the field of a point charge at its centre.
• In each case the field on the Gaussian surface comes from the whole distribution, even though only the enclosed part appears in the law; the answers rely on the symmetry of an infinite wire or sheet.

Teaching this idea? Put the simulation on your own page or LMS with this code.