Integrate a radial density before invoking Gauss symmetry

Normalize the spherical volume element

For rho(r) = rho 0(r/R)^n, a thin spherical shell hasdq = 4pi r²rho(r)dr. Integrating 0..R givesQ = 4pi rho 0 R³/(n+3), so rho 0 = Q(n+3)/(4piR³). The power ofr² comes from geometry; integrating density alone has the wrong units and dependence.

Justify the field factor before extracting it

Spherical source symmetry makes the field radial and of equal magnitude everywhere on a concentric Gaussian sphere. Only then may the surface integral becomeE×4pi r². Gauss law itself is valid without symmetry, but the simplification is not. For r≤R, Qenclosed = Q(r/R)^(n+3), giving E = kQ r^(n+1)/R^(n+3).

Join the interior and exterior branches

For r≥R all source charge is enclosed, soE = kQ/r². The insulating model’s field is continuous at R and zero at the center for the supported nonnegative density powers. Its slope can change at the boundary. A field profile and its derivative are distinct observations; matching one point at the boundary does not establish identical interiors.

Compare a conductor without overgeneralizing

For a spherical conducting shell in electrostatic equilibrium, excess charge lies on its surface and the empty interior has zero field in this isolated symmetric packet. Just outside, the field is kQ/R² and normal to the surface; the ideal surface-charge sheet produces a discontinuity. The displayed surface value uses the exterior limit. A general charged cavity or externally driven conductor requires a separately stated geometry and equilibrium account.

Normalize an area density without inventing a disk field

For a separate charged annulus, let the surface density be sigma(r) = c r, with c = 3 nC/m³, inner radius 1 m and outer radius 2 m. The area element is dA = 2pi r dr, so Q = integral sigma dA = 2pi c(Router³ − Rinner³)/3 = 14pi nC. The extra power of radius comes from area, not from a spherical volume element. Sigma itself has units nC/m², despite c having different units.

Predict the change when only the outer radius becomes 3 m while c and the inner radius remain fixed: Q becomes 52pi nC, not 21pi from simply scaling by radius. Conversely, keeping total Q = 14pi nC with the enlarged annulus would require c = 21/26 nC/m³. This is an area-charge normalization action only; no disk or annulus electric-field formula outside the stated calculus geometries is claimed.

Worked original case

WithQ = 2 nC,R = 1 m andn = 1, rho 0 = Q/pi≈.636620 nC/m³. At radius .5 m, enclosed charge isQ(.5)^4 = .125 nC and Er = kQ(.5)²≈4.49378 N/C. A conducting shell with the same total Q has zero interior field.

Investigation

Compare density powers 0,1,2 at fixed Q and R. Integrate enclosed charge at radii.25,.5,.75 and 1, then derive the interior and exterior fields. Contrast the volume distribution with the same-charge conducting shell. Explain why an exterior measurement alone cannot distinguish these symmetric interior distributions.

Public numerical transfer practice

Try a new practice case

For n = 2,total Q = 2 nC and observation radiusR/2, what charge is enclosed in nC?

Numerical result in the requested units Check this practice result

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Original cases and teaching checked 30 October 2026 against the current official calculus-based electricity and magnetism framework and OpenStax electric fields. All displayed records are generated or supplied, not learner measurements. No charging, mains connection, high-voltage equipment, physical circuit or material handling is requested. Public feedback and retained self-reports are practice evidence, not independent mastery, performed laboratory time or AP authorization.