SCIENCE · INVESTIGATE AND EXPLAIN

Test symmetry of the charge-weighted integrand

A symmetric support does not guarantee a symmetric charge distribution. On −1≤s≤1 m, prescribe λ(s)=1 nC/m times (1+αs/(1 m)). The gradient term integrates to zero in total charge, so Q remains 2 nC while its distribution changes. The registered |α|≤0.75 keeps every density positive.

Learning outcome: Compare α=−0.5,0,+0.5 at fixed y=1 m. Predict the sign of Ex before calculation, retain an odd/even integrand sketch and verify that both total charge and Ey remain fixed. Explain why a centered uniform approximation discards a real component.

Starting recap

Review signed vector components, definite integrals, radians, density units and elementary derivatives. Distinguish observer coordinates from source integration coordinates.

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Try

At α=0.5 and y=1 m, Ex=−9[asinh(1)−1/√2]≈−1.568401 N/C, while Ey=9√2≈12.727922 N/C. The field tilts toward negative x because the positive-charge density is greater on the positive-x side.

Open the experiment

Predict

At positive α, which component behavior is correct for the stated bisector observer?

Your prediction and reason

Experiment

Compare α=−0.5,0,+0.5 at fixed y=1 m. Predict the sign of Ex before calculation, retain an odd/even integrand sketch and verify that both total charge and Ey remain fixed. Explain why a centered uniform approximation discards a real component.

Integrate a density gradient that defeats geometric symmetry

Worked teaching model · synthetic values · Rod −1≤s≤1 m, λ(s)=1 nC/m ×(1+αs/(1 m)), so total charge remains 2 nC and all densities are positive. Observer at (0,y). Geometry is symmetric but charge need not be. Rounded K=9×10^9; Cartesian vector components are signed.

Density asymmetry coefficient: 0.5 ratio Perpendicular observer distance: 1 m

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Charge gradient on symmetric geometry

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

Equivalent numerical coordinates for Charge gradient on symmetric geometry

prescribed line density

source coordinate (m)	density (nC/m)
-1	0.5
-0.975	0.5125
-0.95	0.525
-0.925	0.5375
-0.9	0.55
-0.875	0.5625
-0.85	0.575
-0.825	0.5875
-0.8	0.6
-0.775	0.6125
-0.75	0.625
-0.725	0.6375
-0.7	0.65
-0.675	0.6625
-0.65	0.675
-0.625	0.6875
-0.6	0.7
-0.575	0.7125
-0.55	0.725
-0.525	0.7375
-0.5	0.75
-0.475	0.7625
-0.45	0.775
-0.425	0.7875
-0.4	0.8
-0.375	0.8125
-0.35	0.825
-0.325	0.8375
-0.3	0.85
-0.275	0.8625
-0.25	0.875
-0.225	0.8875
-0.2	0.9
-0.175	0.9125
-0.15	0.925
-0.125	0.9375
-0.1	0.95
-0.075	0.9625
-0.05	0.975
-0.025	0.9875
0	1
0.025	1.0125
0.05	1.025
0.075	1.0375
0.1	1.05
0.125	1.0625
0.15	1.075
0.175	1.0875
0.2	1.1
0.225	1.1125
0.25	1.125
0.275	1.1375
0.3	1.15
0.325	1.1625
0.35	1.175
0.375	1.1875
0.4	1.2
0.425	1.2125
0.45	1.225
0.475	1.2375
0.5	1.25
0.525	1.2625
0.55	1.275
0.575	1.2875
0.6	1.3
0.625	1.3125
0.65	1.325
0.675	1.3375
0.7	1.35
0.725	1.3625
0.75	1.375
0.775	1.3875
0.8	1.4
0.825	1.4125
0.85	1.425
0.875	1.4375
0.9	1.45
0.925	1.4625
0.95	1.475
0.975	1.4875
1	1.5

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
Quantity	Value	Unit
East field from odd density contribution	-1.568401	N/C
North field from even density contribution	12.72792	N/C
Total source charge	2	nC
Field angle from east	97.02487	degrees

Only the charge-weighted integrand symmetry justifies cancellation.

The constant density makes Ex odd and cancels; the αs term turns −s times density into an even negative term. Its primitive uses asinh(s/y)−s/√(s²+y²). For Ey, the gradient part remains odd and cancels. Therefore Ey stays fixed while Ex changes sign with α. Replacing this distribution by a centered uniform rod loses a real component.

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

Only the charge-weighted integrand symmetry justifies cancellation.

Observation to attach to a capture

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

Understand

Define the source and observer

Derive before approximating

At P=(0,y), the field kernel is Kλ(s)(−s,y)/(s²+y²)^(3/2). For Ex, the uniform term is odd and cancels, while the gradient creates an even term proportional to −αs². For Ey, the uniform term is even and the gradient contribution is odd. Thus Ey is unchanged but Ex is generally nonzero. Cancellation must be justified for the whole integrand.

Check evidence and the domain

Using metre numerical coordinates, the horizontal integral becomes −18α[asinh(1/y)−1/√(1+y²)] N/C. One can verify the primitive by differentiating asinh(s/y)−s/√(s²+y²). The vertical result remains 18/[y√(1+y²)] N/C. Reversing α reverses Ex without reversing Ey or total charge.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

At α=0 and y=0.5 m, find Ey in N/C.

Numerical result in the requested units

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 derivations and numerical packets checked 30 October 2026 against the primary explanation. The electrostatic field construction also uses continuous source integration. All records are synthetic; no electrical apparatus, high-voltage handling or practical certification is requested. Public feedback is not independent mastery; retain first predictions, alternative models and criterion-linked critique in the separate course evidence workspace.

Apply

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?

Your explanation and limitations

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