Spherical charge distributions

Take a spherically symmetric distribution and suppose that at some point P the field had a sideways component, tilted off the radius. Now rotate the whole distribution about the axis through P and the centre. The source is mapped exactly onto itself, so the field it makes must be unchanged — yet the tilted component would have swung to a new direction. The only escape is that there was no such component: E points along the radius. Running the same rotation between two points at equal r shows that the magnitude cannot depend on which way you look either, so the whole field is carried by one unknown function E(r) of one variable. This is an argument about the source, not about any surface you later draw, and it is what licenses pulling E outside the flux integral.

Draw an imaginary sphere of radius r about the same centre. At every point of it the outward area element dA is radial, so E ⋅ dA = E(r) dA with no cosine left to carry. E(r) is the same number everywhere on that surface, so it comes out of the integral: Φ = ∮E ⋅ dA = E(r) ∮dA = E(r) × 4πr². Setting this equal to q_enc/ε₀ gives E(r) = q_enc(r)/(4πε₀r²). Two habits follow. Choose r to be exactly where you want the answer — the surface is imaginary and costs nothing to place. And keep the sign: a negative q_enc returns a negative E(r), which means the field points inward, towards the centre.

For a uniform solid sphere of total charge Q and radius R the density is ρ = 3Q/(4πR³), so a sphere of radius r ≤ R encloses q_enc = ρ × (4/3)πr³ = Q(r/R)³. Put Q = 4.0 nC and R = 5.0 cm. At r = 2.0 cm, q_enc = 4.0 × 0.4³ = 0.26 nC and E = 5.8 × 10³ N C⁻¹, while at the surface E = 1.44 × 10⁴ N C⁻¹ — exactly 0.4 of the surface value, because the r³ upstairs beats the r² downstairs by one power of r. When ρ is not uniform, do the integral instead: q_enc(r) = ∫₀ʳ ρ(r′)4πr′² dr′, slicing the sphere into shells of volume 4πr′² dr′. A density rising as ρ ∝ r gives q_enc ∝ r⁴ and therefore E ∝ r², a field that starts far flatter than the uniform case.

Once r ≥ R the enclosed charge is the whole of Q no matter how it is arranged inside, so E = Q/(4πε₀r²) exactly. Not approximately, and not only far away: one millimetre above the surface of a uniformly charged sphere the field is already the point-charge value. From outside, the internal layering is invisible. Run the same argument inwards through a hollow shell and q_enc = 0, so E = 0 everywhere in the cavity — not merely at the centre, where cancellation in pairs would be obvious, but at every point of it. Geometry is the reason: a narrow cone drawn from an off-centre point cuts a patch of shell whose area grows as the square of its distance, while the field of each patch falls as the inverse square, so the near patch and the far patch cancel term by term. Newton needed the same theorem for gravity.

In electrostatic equilibrium the field inside conducting material is zero, or the free electrons would still be moving. Put a Gaussian sphere anywhere within the metal and its flux is therefore zero, so it encloses no net charge: any excess charge has to live on a surface. For an isolated solid conducting sphere of radius R = 5.0 cm carrying Q = 4.0 nC, that surface charge is σ = Q/(4πR²) = 1.27 × 10⁻⁷ C m⁻², and just outside the field is σ/ε₀ = 1.44 × 10⁴ N C⁻¹ — the same number as Q/(4πε₀R²), as it must be. Across the layer E leaps from 0 to σ/ε₀ over no thickness at all. Contrast the uniformly charged insulating sphere, whose charge is spread through a volume: there E climbs smoothly to Q/(4πε₀R²) at r = R and falls smoothly away beyond it, with no jump anywhere.

Nothing new is needed for nested geometries — at each radius, ask only what is inside. Fix a point charge q = +3.0 nC at the centre of an uncharged thick conducting shell of inner radius 2.0 cm and outer radius 3.0 cm. At r = 1.0 cm only q is enclosed, so E = 2.7 × 10⁵ N C⁻¹ outward. At r = 2.5 cm we are inside metal, so E = 0, and Gauss's law then forces q_enc = 0 there — which means −3.0 nC has been drawn onto the inner surface. The shell was neutral, so +3.0 nC is left on the outer surface. At r = 5.0 cm the enclosure is +3.0 − 3.0 + 3.0 = +3.0 nC and E = 1.1 × 10⁴ N C⁻¹, precisely what the bare point charge would have given. The shell rearranges its own charge; it does not hide the charge inside it from the world outside.