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SCIENCE · INVESTIGATE AND EXPLAIN

Compare direct shell integration with Gauss symmetry

A spherical shell provides a useful independent benchmark because its field can be found both by a direct source integral and by symmetry with Gauss law. The original shell has radius R=1 m and total charge Q=2 nC. Observers lie on the positive z axis, strictly away from the charged surface. The model does not assign an on-surface self-field.

Learning outcome: Predict the exact field at each allowed observer before evaluating the midpoint integral. Compare 20,40,100 cells at z=0.5,0.75,1.25,2 m. Retain signed errors and explain why simply increasing cells does not validate an omitted surface-charge or external-field assumption.

Starting recap

Review oriented normals, definite integrals, source volume measures and partial derivatives as one-coordinate rates. Potential differences use an explicit reference and field work uses a signed line integral.

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Try

At observer z=0.5 m the exact field is zero despite nonzero individual shell contributions. At z=2 m it is 18/4=4.5 N/C. At the center, opposite source contributions cancel exactly analytically. Comparing 20 and 100 latitude cells near z=0.75 m tests numerical resolution without changing the physical shell.

Open the experiment

Predict

A coarse direct quadrature reports a small nonzero field inside the stipulated uniform shell. What should be checked first?

Experiment

Predict the exact field at each allowed observer before evaluating the midpoint integral. Compare 20,40,100 cells at z=0.5,0.75,1.25,2 m. Retain signed errors and explain why simply increasing cells does not validate an omitted surface-charge or external-field assumption.

Compare direct shell quadrature with a symmetry result

Worked teaching model · synthetic values · Uniform thin spherical shell radius1 m and charge2 nC, observed on positive z axis away from the charged surface. Integrate over u=cos(theta) with dq=Q du/2. Midpoint quadrature is compared with an independent spherical-symmetry result. On-surface self-field is excluded, not assigned an averaged value.

Reset keeps your prediction, captures and practice history.

Latitude-ring contribution after the area substitution

field integrand (N/C)u = cos(theta)-113.491366-144

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: signed source contribution (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Latitude-ring contribution after the area substitution

signed source contribution

u = cos(theta)field integrand (N/C)
-12.938776
-0.9752.950826
-0.952.962986
-0.9252.975256
-0.92.987635
-0.8753.000123
-0.853.012719
-0.8253.025422
-0.83.038231
-0.7753.051143
-0.753.064157
-0.7253.077271
-0.73.090481
-0.6753.103784
-0.653.117177
-0.6253.130655
-0.63.144213
-0.5753.157845
-0.553.171544
-0.5253.185304
-0.53.199115
-0.4753.212969
-0.453.226853
-0.4253.240756
-0.43.254664
-0.3753.268561
-0.353.28243
-0.3253.29625
-0.33.31
-0.2753.323653
-0.253.337182
-0.2253.350555
-0.23.363734
-0.1753.37668
-0.153.389346
-0.1253.40168
-0.13.413623
-0.0753.425108
-0.053.436058
-0.0253.446389
03.456
0.0253.464781
0.053.472603
0.0753.479319
0.13.484762
0.1253.488739
0.153.491026
0.1753.491366
0.23.489459
0.2253.484959
0.253.47746
0.2753.466485
0.33.451477
0.3253.431775
0.353.406594
0.3753.375
0.43.335873
0.4253.28786
0.453.229321
0.4753.158252
0.53.072186
0.5252.968059
0.552.842035
0.5752.689264
0.62.503542
0.6252.27684
0.651.998622
0.6751.654854
0.71.226512
0.7250.6872933
0.750
0.775-0.8893906
0.8-2.061819
0.825-3.643164
0.85-5.838291
0.875-9
0.9-13.78146
0.925-21.51411
0.95-35.30356
0.975-64.03612
1-144

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Actual axial observer coordinate0.75m
Midpoint direct-integral field0.4661142N/C
Independent symmetry field0N/C
Signed quadrature error0.4661142N/C
Cosine-coordinate cell width0.05ratio

Interior cancellation is exact for the full shell but need not be exact on a coarse quadrature grid.

A latitude strip has dq=Q du/2 after the spherical area Jacobian is included. Its axial contribution is KQ(z−Ru)du/[2(R²+z²−2Rzu)^(3/2)]. The independent spherical-symmetry solution is zero inside and KQ/z² outside. Near-surface integrands vary rapidly, so refining a source grid matters. The charged surface itself is excluded; quadrature residue must not be promoted to a physical violation of Gauss law.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Interior cancellation is exact for the full shell but need not be exact on a coarse quadrature grid.

A latitude strip has dq=Q du/2 after the spherical area Jacobian is included. Its axial contribution is KQ(z−Ru)du/[2(R²+z²−2Rzu)^(3/2)]. The independent spherical-symmetry solution is zero inside and KQ/z² outside. Near-surface integrands vary rapidly, so refining a source grid matters. The charged surface itself is excluded; quadrature residue must not be promoted to a physical violation of Gauss law.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the geometry and reference

A spherical shell provides a useful independent benchmark because its field can be found both by a direct source integral and by symmetry with Gauss law. The original shell has radius R=1 m and total charge Q=2 nC. Observers lie on the positive z axis, strictly away from the charged surface. The model does not assign an on-surface self-field.

Derive the integral or gradient

Parameterize a latitude by u=cosθ. Its area measure converts to dq=Q du/2 over u from −1 to 1. A strip contributes axial field KQ(z−Ru)du/[2(R²+z²−2Rzu)^(3/2)]. The signed numerator matters: for an interior observer, different parts of the shell contribute opposite directions and cancel in the exact integral.

Check the inference boundary

Gauss symmetry separately predicts zero field everywhere inside and KQ/z² outside. Midpoint quadrature over u can leave a nonzero interior residue, especially near the surface where the kernel varies rapidly. That residue is a numerical integration error, not evidence of a physical field inside the ideal shell. Keep the benchmark, quadrature result and signed discrepancy distinct.

Worked original case

At observer z=0.5 m the exact field is zero despite nonzero individual shell contributions. At z=2 m it is 18/4=4.5 N/C. At the center, opposite source contributions cancel exactly analytically. Comparing 20 and 100 latitude cells near z=0.75 m tests numerical resolution without changing the physical shell.

Investigation

Predict the exact field at each allowed observer before evaluating the midpoint integral. Compare 20,40,100 cells at z=0.5,0.75,1.25,2 m. Retain signed errors and explain why simply increasing cells does not validate an omitted surface-charge or external-field assumption.

Changed-case practice

Try the changed case before feedback

For the same Q=2 nC shell, find the exact field at z=1.5 m in N/C.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original calculations checked 30 October 2026 against the primary explanation. All source distributions, fields and readings are hypothetical. This is not a physical high-voltage or electrical handling activity, and public numerical feedback does not establish independent mastery. Preserve first predictions, critique, revision and newly assigned work in the separate retained course workspace.

Apply

A coarse direct quadrature reports a small nonzero field inside the stipulated uniform shell. What should be checked first?

Public practice checks a specific claim. It does not establish independent mastery.

Which geometry, calibration or source assumption could fail while an exact arithmetic fit remains possible?

Continue

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