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SCIENCE · INVESTIGATE AND EXPLAIN

Choose reference levels before taking an infinite-source limit

A potential reference valid for a finite localized source need not survive an infinite-source limit. For a uniform finite rod of half-length L and fixed λ=1 nC/m, the perpendicular-bisector potential with zero at infinity is the integral Kλ ds/√(s²+y²). It evaluates to 2Kλ asinh(L/y)=18 asinh(L/y) V.

Learning outcome: Compare L=1,3,5 m for the potentials at y=1 and2 m and their difference. Retain the increasing absolute potential separately from the converging difference. Explain why declaring V at infinity zero for the infinite line would contradict the integral.

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

For any finite L, the potential difference at y=1 m relative to itself is exactly zero. The infinite-line difference from y0=1 m to y=2 m is −18ln2≈−12.476649 V. For finite L the corresponding expression is 18[asinh(L/2)−asinh(L)], which approaches that result as L grows at fixed λ.

Open the experiment

Predict

Which reference statement is valid for an ideal infinite charged line?

Experiment

Compare L=1,3,5 m for the potentials at y=1 and2 m and their difference. Retain the increasing absolute potential separately from the converging difference. Explain why declaring V at infinity zero for the infinite line would contradict the integral.

Separate finite-source reference from infinite-line differences

Worked teaching model · synthetic values · Finite uniform rod with fixed lambda=1 nC/m, half-length L. V(y)=18 asinh(L/y) V for y>0 and reference zero at spatial infinity for each finite L. Compare V(y)−V(1). Infinite-line total potential with infinity reference is not finite; only a difference relative to finite y0 is used.

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Reference-subtracted finite and infinite-line comparisons

potential difference from distance 1 (V)perpendicular distance (m)0312.12-16.86731

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: finite-rod potential difference (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Reference-subtracted finite and infinite-line comparisons

finite-rod potential difference

perpendicular distance (m)potential difference from distance 1 (V)
0.512.12
0.5312511.04452
0.562510.03235
0.593759.076714
0.6258.171908
0.656257.313044
0.68756.495921
0.718755.716896
0.754.97279
0.781254.26081
0.81253.578495
0.843752.923656
0.8752.294346
0.906251.68882
0.93751.105508
0.968750.5429956
10
1.03125-0.5246433
1.0625-1.031996
1.09375-1.523029
1.125-1.99863
1.15625-2.459619
1.1875-2.906748
1.21875-3.340711
1.25-3.762154
1.28125-4.171672
1.3125-4.569821
1.34375-4.957116
1.375-5.334039
1.40625-5.701041
1.4375-6.058542
1.46875-6.406938
1.5-6.746598
1.53125-7.077871
1.5625-7.401085
1.59375-7.716549
1.625-8.024555
1.65625-8.325379
1.6875-8.619281
1.71875-8.906509
1.75-9.187298
1.78125-9.461868
1.8125-9.730432
1.84375-9.993191
1.875-10.25034
1.90625-10.50205
1.9375-10.74851
1.96875-10.98987
2-11.2263
2.03125-11.45795
2.0625-11.68496
2.09375-11.90747
2.125-12.12562
2.15625-12.33952
2.1875-12.54932
2.21875-12.75511
2.25-12.95702
2.28125-13.15514
2.3125-13.3496
2.34375-13.54048
2.375-13.72788
2.40625-13.9119
2.4375-14.09262
2.46875-14.27013
2.5-14.44452
2.53125-14.61586
2.5625-14.78423
2.59375-14.94971
2.625-15.11236
2.65625-15.27226
2.6875-15.42948
2.71875-15.58408
2.75-15.73611
2.78125-15.88566
2.8125-16.03276
2.84375-16.17749
2.875-16.31989
2.90625-16.46003
2.9375-16.59794
2.96875-16.73369
3-16.86731

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
Finite rod total charge6nC
Finite rod potential at distance 132.73204V
Finite rod potential at selected distance21.50574V
Finite rod potential difference-11.2263V
Infinite-line finite-reference difference-12.47665V
Finite minus infinite difference1.250351V

An infinite line has a finite potential difference between finite radii, not a finite potential with zero at infinity.

Integrating Kλ ds/√(s²+y²) along a finite rod gives 2Kλ asinh(L/y). At fixed λ it diverges logarithmically as L grows. Subtracting the potential at y0=1 before taking the limit gives −2Kλ ln(y/y0), a finite difference. A logarithm must have a dimensionless argument. Gauge choice removes an additive constant, not the physical radial field.

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

An infinite line has a finite potential difference between finite radii, not a finite potential with zero at infinity.

Integrating Kλ ds/√(s²+y²) along a finite rod gives 2Kλ asinh(L/y). At fixed λ it diverges logarithmically as L grows. Subtracting the potential at y0=1 before taking the limit gives −2Kλ ln(y/y0), a finite difference. A logarithm must have a dimensionless argument. Gauge choice removes an additive constant, not the physical radial field.

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

Understand

Define the geometry and reference

A potential reference valid for a finite localized source need not survive an infinite-source limit. For a uniform finite rod of half-length L and fixed λ=1 nC/m, the perpendicular-bisector potential with zero at infinity is the integral Kλ ds/√(s²+y²). It evaluates to 2Kλ asinh(L/y)=18 asinh(L/y) V.

Derive the integral or gradient

At fixed λ this absolute potential grows logarithmically as L increases. An infinite line therefore cannot be assigned a finite potential with zero at infinity. Instead subtract the potential at a finite reference radius y0 before taking the limit. The difference tends to −2Kλ ln(y/y0); its logarithm has a dimensionless argument.

Check the inference boundary

This reference issue does not make the electric field undefined. Differentiating the finite-rod potential yields its known finite-line field, and the infinite-line difference yields E=2Kλ/y. A gauge shift removes a constant but cannot remove a radial field or cure an invalid divergent reference. Holding total Q fixed while increasing L would be a different limiting procedure.

Worked original case

For any finite L, the potential difference at y=1 m relative to itself is exactly zero. The infinite-line difference from y0=1 m to y=2 m is −18ln2≈−12.476649 V. For finite L the corresponding expression is 18[asinh(L/2)−asinh(L)], which approaches that result as L grows at fixed λ.

Investigation

Compare L=1,3,5 m for the potentials at y=1 and2 m and their difference. Retain the increasing absolute potential separately from the converging difference. Explain why declaring V at infinity zero for the infinite line would contradict the integral.

Changed-case practice

Try the changed case before feedback

For the stated infinite-line comparison, find V(0.5 m)−V(1 m) in volts.

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

Which reference statement is valid for an ideal infinite charged line?

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

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