Reconcile six signed fluxes without guessing a local field

Define the oriented area element

For a flat patch in a uniform field, flux isE dot A, where the area vector is perpendicular to the patch. For a closed surface use outward normals consistently. Positive flux is outward; negative flux is inward. Flux units areN m²/C, not the N/C units of field strength.

Integrate rather than multiply incompatible averages

For a nonuniform field, divide the surface into small patches and sumE dot dA, yielding a surface integral. On this box each face has a particularly simple component: the x-normal faces have area 1 m², while the y-normal faces have area 2 m². The values of Ex onx = 0 andx = 2 differ; do not multiply a single arbitrary field value by total surface area.

Integrate a field that varies across one face

For a separate original open patch at x = 1 m, let 0≤y≤2 m and 0≤z≤1 m, with positive-x orientation. The supplied field is E = (c y²,0,0), where c = 3 N/(C m²). This is a mathematical spatial flux packet, not a claim that the displayed vector function is a standalone electrostatic field everywhere: its curl is generally nonzero. The surface integral still has the stated mathematical meaning. Its normal component changes over the patch, so flux is the double integral of c y² dy dz, giving c(2 m)³/3×(1 m) = 8 N m²/C. Evaluating the field only at the midpoint y = 1 and multiplying by area 2 m² would give 6, not 8.

Predict the changed patch with upper y limit 1 m before checking its flux 1 N m²/C. The area halves but the flux becomes one eighth, because the removed part carried a larger normal field. Reversing the patch orientation reverses the sign without changing the field. This open patch alone does not enclose a charge; do not apply the closed-surface charge equation to its single flux.

Apply the closed-surface balance

The net outward integral equals enclosed charge divided byepsilon 0. Here the x contribution is 2a and the transverse terms cancel, soQenclosed = 2a epsilon 0. This infers a charge total consistent with the assigned static field. It does not identify a unique pointwise charge arrangement or calibrate a real instrument.

Separate local strength from total flux

IncreasingB changes the local magnitude and two face fluxes without changing their signed sum. Witha = 0, the box can contain a nonzero uniform transverse field and still have zero net flux. Conversely, equal enclosed charge gives equal total flux for different enclosing surfaces, but not equal local field or equal flux through each small patch.

Worked original case

For E = (2x,3,0) N/C in the 2×1×1 m box, x faces contribute 0 and 4 N m²/C. The y faces contribute−6 and+6, while z faces contribute 0. Net outward flux is 4, requiring enclosed charge epsilon 0×4≈35.4168 pC.

Investigation

Hold field gradient 2 and compare transverse fields 0,2,4. Predict the individual face changes and unchanged total. Then set the gradient to 0 with transverse field 4: record nonzero field with zero net flux. Redraw each outward normal and independently integrate the actual face areas before using Gauss law.

Public numerical transfer practice

Try a new practice case

For the same box with gradienta = 3 N/(C m) and transverse fieldB = 4 N/C, what is the net outward flux in N m²/C?

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.