SCIENCE · INVESTIGATE AND EXPLAIN

Keep line charge and total charge distinct in a coaxial gap

Two long coaxial conductors have inner radius a, outer radius b and active length L.

Learning outcome: Compare lengths 1, 2 and 3 m at fixed Q. Then write a separate fixed-lambda prediction without changing the simulator’s declared control. Integrate gap energy independently and identify which end-field evidence would be needed to assess a finite apparatus.

Starting recap

Declare conductor boundaries, integrate potential or charging work, and distinguish controlled charge from voltage.

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Try

For a = 0.01 m, b = 0.02 m, L = 2 m and Q = 2 nC, lambda = 1 nC/m, C ≈ 160.521 pF, delta V ≈ 12.4594 V and U ≈ 12.4594 nJ. The outer-radius-to-length ratio is 0.01, supporting but not quantifying the neglected end effects.

Open the experiment

Predict

Which expression is the correct dimensionless quantity inside the coaxial potential logarithm?

Your prediction and reason

Experiment

Compare lengths 1, 2 and 3 m at fixed Q. Then write a separate fixed-lambda prediction without changing the simulator’s declared control. Integrate gap energy independently and identify which end-field evidence would be needed to assess a finite apparatus.

Integrate a logarithmic coaxial gap

Worked teaching model · synthetic values · Inner radius 0.01 m, outer selected ratio, opposite total charges ±2 nC over the selected active length. Long coaxial approximation ignores ends; line charge is total charge divided by length. The infinite-source field is used only in this explicit finite active-length approximation.

Outer-to-inner radius ratio: 2 ratio Active cylinder length: 2 m

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Coaxial gap field

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.

Coaxial gap field: numerical coordinates

radius (m)	radial field (N/C)
0.01	1797.51
0.0101667	1768.04
0.0103333	1739.53
0.0105	1711.91
0.0106667	1685.17
0.0108333	1659.24
0.011	1634.1
0.0111667	1609.71
0.0113333	1586.04
0.0115	1563.05
0.0116667	1540.72
0.0118333	1519.02
0.012	1497.93
0.0121667	1477.41
0.0123333	1457.44
0.0125	1438.01
0.0126667	1419.09
0.0128333	1400.66
0.013	1382.7
0.0131667	1365.2
0.0133333	1348.13
0.0135	1331.49
0.0136667	1315.25
0.0138333	1299.41
0.014	1283.94
0.0141667	1268.83
0.0143333	1254.08
0.0145	1239.66
0.0146667	1225.58
0.0148333	1211.8
0.015	1198.34
0.0151667	1185.17
0.0153333	1172.29
0.0155	1159.68
0.0156667	1147.35
0.0158333	1135.27
0.016	1123.44
0.0161667	1111.86
0.0163333	1100.52
0.0165	1089.4
0.0166667	1078.51
0.0168333	1067.83
0.017	1057.36
0.0171667	1047.09
0.0173333	1037.03
0.0175	1027.15
0.0176667	1017.46
0.0178333	1007.95
0.018	998.617
0.0181667	989.455
0.0183333	980.46
0.0185	971.627
0.0186667	962.952
0.0188333	954.43
0.019	946.058
0.0191667	937.831
0.0193333	929.747
0.0195	921.8
0.0196667	913.988
0.0198333	906.308
0.02	898.755

Calculated values under the declared model
Quantity	Value	Unit
Outer radius	0.02	m
Line charge magnitude	1	nC/m
Capacitance	160.521	pF
Potential difference	12.4594	V
Charging energy	12.4594	nJ
Integrated gap field energy	12.4594	nJ
Outer-radius-to-length ratio	0.01	ratio

The logarithm belongs to the radial gap ratio, while total capacitance grows with active length.

E = lambda/(2pi epsilon0 r); integrating across the gap yields deltaV = lambda ln(b/a)/(2pi epsilon0). Multiplying capacitance per length by active length gives C. The field-energy integral uses cylindrical volume element 2pi r L dr. Neither an absolute infinite-source potential nor an arbitrary end correction is assumed.

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

The logarithm belongs to the radial gap ratio, while total capacitance grows with active length.

Observation to attach to a capture

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Understand

Declare the long-cylinder approximation

Two long coaxial conductors have inner radius a, outer radius b and active length L. End effects are neglected, so the field in the modeled central gap is E = lambda/(2pi epsilon0 r), with lambda = Q/L. Q is the charge on the selected active length, not an infinite total charge.

Integrate a finite radial difference

The gap voltage is lambda ln(b/a)/(2pi epsilon0), giving C = 2pi epsilon0 L/ln(b/a). No absolute potential at infinity is needed. The logarithm acts on a dimensionless radius ratio. Using b − a in place of ln(b/a) without an approximation loses both the radial integral and dimensional consistency.

Reconcile energy and length

Integrating epsilon0 E²/2 over cylindrical volume element 2pi r L dr yields L lambda² ln(b/a)/(4pi epsilon0) = Q²/(2C). At fixed total Q, increasing L spreads the charge and lowers voltage and energy. At fixed lambda, total Q increases with L and energy instead increases with L. State the control before giving a scaling prediction.

Worked original case

Investigation

Changed-case practice

Try a new practice case

At fixed radius ratio, triple the active length while holding total Q fixed. What is the new stored energy divided by the original?

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 boundaries

Original cases checked against the official framework on 30 October 2026, with OpenStax verification on 31 October 2026 against the current official framework, Unit 10 and OpenStax capacitance. These are virtual supplied models, not instructions to charge, ground, wire or handle real equipment. Public answers are practice; retained plans and self-reports do not establish independent mastery, actual laboratory participation or AP authorization.

Apply

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Which conductor boundary, geometric approximation or independent observation could change this conclusion?

Your explanation and limitations

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