SCIENCE · INVESTIGATE AND EXPLAIN

Match potential branches without a false infinity reference

Integrate a Gaussian field while preserving a valid finite reference.

Learning outcome: Compare reference radii 1, 2 and 3 m at fixed lambda. Retain center, surface and exterior potentials, then subtract the same endpoint pair in every case. Independently differentiate both branches and verify continuity at the surface.

Starting recap

Keep source normalization, potential reference, signed energy and evidence independence explicit.

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Try

For lambda = 2 nC/m, R = 1 m and rref = 2 m, V(1) = 2k lambda ln 2 ≈ 24.9188 V. V(0) exceeds V(1) by k lambda ≈ 17.9751 V. V(2) = 0 by definition, not because the exterior field vanishes there.

Open the experiment

Predict

Why is zero potential at infinity unavailable for this nonzero infinite line-charge geometry?

Your prediction and reason

Experiment

Compare reference radii 1, 2 and 3 m at fixed lambda. Retain center, surface and exterior potentials, then subtract the same endpoint pair in every case. Independently differentiate both branches and verify continuity at the surface.

Choose a finite reference for an infinite cylinder

Worked teaching model · synthetic values · Infinite uniform insulating cylinder radius 1 m, selected total line charge. Reference potential V(r_ref) = 0 with r_ref≥1 m. Interior and exterior branches joined continuously. A finite reference is essential; no V(infinity) = 0 convention for nonzero infinite line charge.

Total charge per length: 2 nC/m Reference radius: 2 m

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Matched cylinder potential branches

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Matched cylinder potential branches: numerical coordinates

radius (m)	V relative to reference (V)
0	42.8939
0.0666667	42.814
0.133333	42.5743
0.2	42.1749
0.266667	41.6157
0.333333	40.8967
0.4	40.0179
0.466667	38.9793
0.533333	37.781
0.6	36.4229
0.666667	34.905
0.733333	33.2273
0.8	31.3898
0.866667	29.3926
0.933333	27.2356
1	24.9188
1.06667	22.5986
1.13333	20.4191
1.2	18.3643
1.26667	16.4206
1.33333	14.5766
1.4	12.8225
1.46667	11.1501
1.53333	9.55208
1.6	8.02206
1.66667	6.5545
1.73333	5.1445
1.8	3.78773
1.86667	2.48031
1.93333	1.21877
2	0
2.06667	-1.1788
2.13333	-2.32017
2.2	-3.42642
2.26667	-4.49964
2.33333	-5.54175
2.4	-6.5545
2.46667	-7.5395
2.53333	-8.49823
2.6	-9.43205
2.66667	-10.3422
2.73333	-11.2299
2.8	-12.0962
2.86667	-12.9422
2.93333	-13.7687
3	-14.5766
3.06667	-15.3667
3.13333	-16.1399
3.2	-16.8967
3.26667	-17.638
3.33333	-18.3643
3.4	-19.0762
3.46667	-19.7743
3.53333	-20.4591
3.6	-21.1311
3.66667	-21.7907
3.73333	-22.4385
3.8	-23.0748
3.86667	-23.7
3.93333	-24.3146
4	-24.9188

Calculated values under the declared model
Quantity	Value	Unit
Potential at center	42.8939	V
Potential at radius 0.5 m	38.4001	V
Potential at cylinder surface	24.9188	V
Potential at radius 2 m	0	V
Potential at declared reference	0	V
Interior field at radius 0.5 m	17.9751	N/C
Potential difference V(2) minus V(1)	-24.9188	V

Changing the finite reference adds a constant but leaves differences and fields unchanged.

Outside, integrate E = 2k lambda/r to obtain a logarithmic potential difference. Inside the uniform core, E = 2k lambda r/R² with R = 1 m, giving a quadratic branch matched at the surface. The exterior logarithm diverges at infinity for nonzero lambda, so assigning zero there is invalid for this infinite geometry.

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Changing the finite reference adds a constant but leaves differences and fields unchanged.

Observation to attach to a capture

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

Understand

Recognize the logarithmic exterior

For an infinite cylinder with total line charge lambda and radius R, the exterior field is 2k lambda/r. The potential difference between exterior radii r and rref is −2k lambda ln(r/rref). The integral to infinity diverges for nonzero lambda. It is inconsistent to impose the localized-source convention V(infinity) = 0 on this geometry.

Match the interior branch

For a uniform insulating core, Einside = 2k lambda r/R². Integrating inward from the surface gives Vinside(r) = k lambda(1 − r²/R²) + 2k lambda ln(rref/R). The potential and radial derivative match the exterior at R. This is an insulator with distributed charge, not an equipotential conductor.

Move the reference without moving the source

Changing rref adds a common constant to both branches. It cannot alter a measured difference between fixed locations or their fields. An isolated point-source formula kQ/r would have both the wrong geometry and the wrong dimensional inventory: lambda is charge per length, not total charge.

Worked original case

For lambda = 2 nC/m, R = 1 m and rref = 2 m, V(1) = 2k lambda ln 2 ≈ 24.9188 V. V(0) exceeds V(1) by k lambda ≈ 17.9751 V. V(2) = 0 by definition, not because the exterior field vanishes there.

Investigation

Changed-case practice

Try a new practice case

What is [V(2R) − V(R)] divided by k lambda for this uniform infinite cylinder?

Numerical result in the requested units

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 teaching and cases checked 30 October 2026 against the current official framework, Unit 9 and OpenStax electric potential. All records are supplied or generated, not measured by a learner. No physical charging, battery connection, high-voltage equipment or particle experiment is requested. Retain predictions, derivations and critique in the course evidence workspace; public feedback is not independent mastery or performed laboratory time.

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

Which reference, source geometry or evidence-independence assumption limits your inference?

Your explanation and limitations

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