Derive a slab field and challenge a symmetry shortcut

State the fixed inventory

The infinite slab fills−a≤x≤a and carries fixed total charge per area sigma. Therefore its uniform volume density isrho = sigma/(2a). Changing thickness at fixed rho would change the total charge per area; this workbench instead holdssigma fixed so geometry and inventory are not silently mixed.

Choose a Gaussian pillbox with a reason

A symmetric pillbox extends from−x tox with equal face areaA. Planar symmetry makes the field normal to the slab and constant on each face. Reflection symmetry givesE(−x) = −E(x), so the two outward fluxes add to 2E(x)A for positive x. Side walls contribute no flux because their normals are parallel to the slab.

Integrate the enclosed charge

For 0≤x≤a, enclosed charge isrho(2xA), soE = rho x/epsilon 0. Forx≥a, all slab thickness is enclosed and E = rho a/epsilon 0 = sigma/(2epsilon 0). The signed formula isE(x) = rho clamp(x,−a,a)/epsilon 0. At fixed sigma, narrowing the slab steepens the interior slope; its sheet limit retains the same exterior magnitude.

Falsify the shortcut rather than Gauss law

For a finite charged patch, the field generally varies over a pillbox face and has edge components. Gauss law remains true, but replacing its integral by one guessedE times area is unjustified. A displaced point charge inside a spherical surface likewise gives the same total flux as a centered charge but a different local field pattern. These counterexamples separate a valid balance law from an invalid symmetry assumption.

Place the result in the field framework

Gauss law is one of Maxwell’s equations, relating electric flux to charge. The present static calculation does not cover time-dependent induction or magnetic fields; those require additional equations and later course evidence. No successful static fit certifies a practical device, an actual conductor’s boundary conditions or a performed field measurement.

Worked original case

For half-thickness 1 m and total charge per area 2 nC/m², rho = 1 nC/m³. Atx = .5 m, Ex = rho x/epsilon 0≈56.4705 N/C. Outside the slab on the positive side, Ex = sigma/(2epsilon 0)≈112.941 N/C, independent of further distance in the infinite-plane approximation.

Investigation

Compare half-thicknesses.5,1,2 m at fixed charge per area. Predict the changing interior slope and unchanged exterior magnitude. Reverse the observation coordinate to check sign. Then replace the verbal infinite-plane source with a finite patch and identify precisely which uniformity step can no longer be assumed from Gauss law alone.

Public numerical transfer practice

Try a new practice case

For half-thickness 2 m at coordinatex = 1 m, what is the field divided by its positive exterior value?

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and evidence boundary

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.