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SCIENCE · INVESTIGATE AND EXPLAIN

Normalize a line density before integrating its field

Compare an analytic field with bounded numerical sums at fixed total charge.

Learning outcome: At observer z = 2, compare beta = 0,1,2 while keeping total charge 2 nC fixed. Derive lambda 0 anew and compare exact,8-cell and 64-cell fields. Then increase observation distance and compare with the leading point-charge limit kQ/z². Explain why one small numerical discrepancy is not a general quadrature error bound.

Starting recap

Choose source and observer coordinates, integrate normalized charge, and justify symmetry before simplifying field or flux.

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Try

For total charge 2 nC on 0–1 m, beta = 1 gives lambda 0 = 4/3 nC/m. At observer z = 2 m the exact field is k lambda 0[3(1−1/2)−ln 2]≈9.66884 N/C. Eight midpoint cells give about 9.63396 N/C; sixty-four give 9.66829 N/C.

Open the experiment

Predict

Why must lambda 0 change when beta changes while total charge is held fixed?

Experiment

At observer z = 2, compare beta = 0,1,2 while keeping total charge 2 nC fixed. Derive lambda 0 anew and compare exact,8-cell and 64-cell fields. Then increase observation distance and compare with the leading point-charge limit kQ/z². Explain why one small numerical discrepancy is not a general quadrature error bound.

Integrate a normalized nonuniform line

Worked teaching model · synthetic values · Source 0≤x≤1 m, total charge 2 nC, lambda=lambda0(1+beta x/L), L=1 m. Observer beyond end. Exact analytic integral compared with midpoint sums, not physical measurement.

Reset keeps your prediction, captures and practice history.

Field beyond the source end

Ex (N/C)observer x (m)1.5526.77960

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Field beyond the source end: numerical coordinates
observer x (m)Ex (N/C)
1.526.7796
1.5583322.9358
1.6166719.9032
1.67517.4609
1.7333315.4604
1.7916713.7979
1.8512.3992
1.9083311.2098
1.9666710.1889
2.0259.30529
2.083338.53491
2.141677.85876
2.27.26177
2.258336.7318
2.316676.25901
2.3755.83531
2.433335.45401
2.491675.10956
2.554.79728
2.608334.51323
2.666674.25406
2.7254.01691
2.783333.79932
2.841673.59918
2.93.41464
2.958333.2441
3.016673.08617
3.0752.93963
3.133332.80339
3.191672.6765
3.252.55812
3.308332.44749
3.366672.34395
3.4252.24691
3.483332.15581
3.541672.07019
3.61.98961
3.658331.91368
3.716671.84205
3.7751.77439
3.833331.71041
3.891671.64986
3.951.59249
4.008331.53807
4.066671.48642
4.1251.43734
4.183331.39066
4.241671.34623
4.31.30391
4.358331.26357
4.416671.22507
4.4751.18832
4.533331.15321
4.591671.11963
4.651.08751
4.708331.05676
4.766671.02729
4.8250.999052
4.883330.971964
4.941670.945966
50.921002
Calculated values under the declared model
QuantityValueUnit
Normalized base line density1.33333nC/m
Exact axial field9.66884N/C
Eight-cell midpoint field9.63396N/C
Sixty-four-cell midpoint field9.66829N/C
Coarse signed error-0.0348825N/C
Fine signed error-0.000548501N/C

Normalize the charge distribution before comparing its integral with numerical sums.

Integrate the signed source density divided by (observer−source)². A finite-cell sum is a quadrature approximation; agreement on one grid is not an error proof or measured evidence.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Normalize the charge distribution before comparing its integral with numerical sums.

Integrate the signed source density divided by (observer−source)². A finite-cell sum is a quadrature approximation; agreement on one grid is not an error proof or measured evidence.

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

Understand

Start with charge and units

For a line, dq = lambda(x)dx; lambda has units C/m. The assigned density is lambda 0(1+beta x/L), with L = 1 m. Integrating from 0 to L gives Q = lambda 0 L(1+beta/2). Holding lambda 0 fixed while changing beta would add charge, a different experiment from redistribution at fixed Q.

Keep source and observation coordinates separate

Call the source coordinate x and the observer z>L. Every source displacement is positive along the line, so dE = k lambda(x)dx/(z−x)². The source coordinate is the integration variable; z is held fixed during that integral. Crossing into the ideal line or approaching its endpoint is excluded from the controls.

Derive the endpoint expression

Using numerical metre coordinates with L = 1, write 1+beta x = (1+beta z)−beta(z−x). Integrating gives E = k lambda 0[(1+beta z)(1/(z−1)−1/z)−beta ln(z/(z−1))]. The logarithm is a ratio of distances, hence dimensionless. Restoring L requires keeping the factors beta/L, not treating a dimensional coordinate as an unlabelled pure number.

Audit approximation and limiting behavior

A midpoint sum approximates each small source cell by its center. Refinement reduces the displayed discrepancy in these supported cases, but neither cell count proves accuracy for arbitrary shapes or near-singular observers. Far from a bounded source, the leading contribution depends on total Q; changing beta affects smaller geometric corrections. Derive that limit before trusting a plotted flat-looking tail.

Worked original case

For total charge 2 nC on 0–1 m, beta = 1 gives lambda 0 = 4/3 nC/m. At observer z = 2 m the exact field is k lambda 0[3(1−1/2)−ln 2]≈9.66884 N/C. Eight midpoint cells give about 9.63396 N/C; sixty-four give 9.66829 N/C.

Investigation

At observer z = 2, compare beta = 0,1,2 while keeping total charge 2 nC fixed. Derive lambda 0 anew and compare exact,8-cell and 64-cell fields. Then increase observation distance and compare with the leading point-charge limit kQ/z². Explain why one small numerical discrepancy is not a general quadrature error bound.

Public numerical transfer practice

Try a new practice case

For a uniform line on 0–1 m viewed at z = 3 m, what is E/[kQ/(1 m)²]?

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.

Apply

Why must lambda 0 change when beta changes while total charge is held fixed?

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

Which source, boundary or independence assumption would have to change before the conclusion changes?

Continue

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