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SCIENCE · INVESTIGATE AND EXPLAIN

Challenge an exact two-gap capacitance fit

For a vacuum parallel-plate source, geometric C = epsilon0 A/d.

Learning outcome: Vary gain and parasitic offset separately, retaining which fit parameter changes. Predict a third-gap indication and explain why it does not resolve common gain. Then calculate the supplied finite-bound area interval and name an additional uncertainty that it omits.

Starting recap

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

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Try

For true A = 0.01 m², g = 1.05 and Cp = 20 pF, the indications at 1 and 2 mm are approximately 112.969 and 66.4845 pF. The unit-gain fit returns Afit = 0.0105 m² and Cp = 20 pF. The independent supplied optical area is 0.01 m², revealing a 5% conditional discrepancy.

Open the experiment

Predict

Which operation removes an additive stray capacitance but not a common multiplicative gain?

Experiment

Vary gain and parasitic offset separately, retaining which fit parameter changes. Predict a third-gap indication and explain why it does not resolve common gain. Then calculate the supplied finite-bound area interval and name an additional uncertainty that it omits.

Challenge a parasitic-capacitance inference

Worked teaching model · synthetic values · True plate area 0.01 m², vacuum gaps 1 and 2 mm. Indication = gain times geometric capacitance plus additive parasitic term. Fit assumes unit gain and solves area plus offset. Independent optical area stipulated; no real metrology or material identification claim.

Reset keeps your prediction, captures and practice history.

Shared-gain capacitance indications

indicated capacitance (pF)gap (mm)14112.9690

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.

Shared-gain capacitance indications: numerical coordinates
gap (mm)indicated capacitance (pF)
1112.969
1.05108.542
1.1104.517
1.15100.843
1.297.4741
1.2594.3752
1.391.5146
1.3588.8659
1.486.4064
1.4584.1165
1.581.9793
1.5579.98
1.678.1056
1.6576.3448
1.774.6876
1.7573.1251
1.871.6494
1.8570.2535
1.968.931
1.9567.6764
266.4845
2.0565.3507
2.164.2709
2.1563.2414
2.262.2586
2.2561.3195
2.360.4213
2.3559.5613
2.458.7371
2.4557.9465
2.557.1876
2.5556.4584
2.655.7573
2.6555.0826
2.754.433
2.7553.8069
2.853.2032
2.8552.6207
2.952.0583
2.9551.5149
350.9897
3.0550.4816
3.149.99
3.1549.514
3.249.0528
3.2548.6058
3.348.1724
3.3547.7519
3.447.3438
3.4546.9475
3.546.5626
3.5546.1884
3.645.8247
3.6545.471
3.745.1267
3.7544.7917
3.844.4655
3.8544.1478
3.943.8382
3.9543.5364
443.2422
Calculated values under the declared model
QuantityValueUnit
Indication at gap 1 mm112.969pF
Indication at gap 2 mm66.4845pF
Fitted plate area0.0105m²
Fitted additive capacitance20pF
Stipulated independent optical area0.01m²
Relative area inference error0.05ratio
Third-gap indication at 3 mm50.9897pF

Two-gap subtraction removes an additive parasitic term but leaves common gain hidden in fitted area.

The known inverse-gap form yields Afit = gain times true area. A third same-instrument gap can still fit the biased area exactly. An independently justified area measurement tests that gain hypothesis only if dielectric composition, plate parallelism and gap calibration are also controlled. This packet does not uniquely identify a real material permittivity.

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

Two-gap subtraction removes an additive parasitic term but leaves common gain hidden in fitted area.

The known inverse-gap form yields Afit = gain times true area. A third same-instrument gap can still fit the biased area exactly. An independently justified area measurement tests that gain hypothesis only if dielectric composition, plate parallelism and gap calibration are also controlled. This packet does not uniquely identify a real material permittivity.

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

Understand

Write the indicated rather than ideal quantity

For a vacuum parallel-plate source, geometric C = epsilon0 A/d. The synthetic instrument reports Cm = g epsilon0 A/d + Cp, with common gain g and additive parasitic capacitance Cp. A model that assumes g = 1 can solve area and offset from two gap readings, but that does not prove the assumed gain.

Eliminate one nuisance parameter

Subtracting two readings removes Cp and gives Afit = (C1 − C2)/[epsilon0(1/d1 − 1/d2)] = gA. Substituting back returns Cp for this exact packet. A third same-instrument gap still follows the same inverse-gap curve with biased Afit, so exact held-out agreement tests the curve shape but not common gain independently.

Choose an independent constraint

A separately justified optical area record can expose the fitted-area bias, conditional on known permittivity, parallel plates and calibrated gaps. A wrong gap zero, fringing, material polarization or time-dependent stray capacitance could create other discrepancies. Distinguish rejecting one joint model from uniquely diagnosing a gain error.

Bound a two-reading inference

For a separate supplied pair of finite reading bounds, let C1 be 105 to 107 pF and C2 be 60 to 62 pF at gaps 1 and 2 mm. The independent rectangular bound on the difference is 43 to 47 pF, not 45 pF exactly. Dividing by epsilon0×500 per metre gives an inferred area interval of approximately 0.009713 to 0.010616 m². These are stipulated hard bounds, not confidence intervals; common gain remains outside that two-reading bound unless explicitly bounded too.

Worked original case

For true A = 0.01 m², g = 1.05 and Cp = 20 pF, the indications at 1 and 2 mm are approximately 112.969 and 66.4845 pF. The unit-gain fit returns Afit = 0.0105 m² and Cp = 20 pF. The independent supplied optical area is 0.01 m², revealing a 5% conditional discrepancy.

Investigation

Vary gain and parasitic offset separately, retaining which fit parameter changes. Predict a third-gap indication and explain why it does not resolve common gain. Then calculate the supplied finite-bound area interval and name an additional uncertainty that it omits.

Changed-case practice

Try a new practice case

If true area is 0.01 m² and common gain is 0.95, what area in m² does the unit-gain fit infer?

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

Which operation removes an additive stray capacitance but not a common multiplicative gain?

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

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

Continue

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