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SCIENCE · INVESTIGATE AND EXPLAIN

Separate a plate capacitance from a stray contribution

Capacitance is the proportionality C=Q/ΔV for the chosen linear device. An ideal capacitor has equal opposite charges on its plates, so the usual stored charge Q is a plate-charge magnitude, not the net charge of the entire two-plate object. For broad vacuum plates with negligible edge fields, C=ε0A/d. Convert area into square metres and separation into metres before using that equation.

Learning outcome: Compare gaps 1,2,3 mm while keeping the stray contribution fixed. Plot charge against voltage and retain slope and intercept separately. Explain why doubling the gap does not halve the total measured slope when a nonzero fixed stray capacitance remains.

Starting recap

Use signed charge, common coordinate axes, SI units, vector components and algebraic ratios. Separate a source field from probe force and a fitted parameter from independently tested evidence.

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Try

At gap 2 mm, C_plate=ε0(.02)/.002=88.541878128 pF. Adding 20 pF stray gives slope 108.541878128 pF. At 4 V the indicated charge is 439.167512512 pC, including the 5 pC readout offset; plate energy alone is about 7.08335×10⁻¹⁰ J.

Open the experiment

Predict

A constant charge-readout offset changes which feature of a linear Q–V fit?

Experiment

Compare gaps 1,2,3 mm while keeping the stray contribution fixed. Plot charge against voltage and retain slope and intercept separately. Explain why doubling the gap does not halve the total measured slope when a nonzero fixed stray capacitance remains.

Separate plate geometry from stray parallel capacitance

Worked teaching model · synthetic values · Ideal vacuum parallel plates area .02 m², epsilon0=8.8541878128e−12 F/m, negligible edge fields. A fixed independent stray capacitance is in parallel. Read charge at 2,4,6 V; charge sensor also adds a known 5 pC offset. No practical capacitor handling.

Reset keeps your prediction, captures and practice history.

Charge versus maintained potential difference

indicated charge (pC)voltage (V)06656.2510

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

Equivalent numerical coordinates for Charge versus maintained potential difference

plate plus stray plus offset

voltage (V)indicated charge (pC)
05
0.1521.2813
0.337.5626
0.4553.8438
0.670.1251
0.7586.4064
0.9102.688
1.05118.969
1.2135.25
1.35151.532
1.5167.813
1.65184.094
1.8200.375
1.95216.657
2.1232.938
2.25249.219
2.4265.501
2.55281.782
2.7298.063
2.85314.344
3330.626
3.15346.907
3.3363.188
3.45379.469
3.6395.751
3.75412.032
3.9428.313
4.05444.595
4.2460.876
4.35477.157
4.5493.438
4.65509.72
4.8526.001
4.95542.282
5.1558.564
5.25574.845
5.4591.126
5.55607.407
5.7623.689
5.85639.97
6656.251

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
Geometric plate capacitance88.5419pF
Measured charge-voltage slope108.542pF
Known charge-reading intercept5pC
Total indicated charge at 4 V439.168pC
Plate field at 4 V2000V/m
Plate energy at 4 V7.08335e-10J
Stray energy at 4 V1.6e-10J

The fitted slope includes both capacitances; a charge offset changes the intercept instead.

C=epsilon0 A/d and Q=CV use SI area and separation. Opposite plates carry +Q and −Q, so the capacitor remains neutral overall. A dielectric changes polarization and effective permittivity; here vacuum is stipulated. A slope alone cannot distinguish larger plate area, smaller gap, dielectric or stray parallel capacitance without independent geometry/material evidence. Energy is one half CV², not QV for gradual charging.

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

The fitted slope includes both capacitances; a charge offset changes the intercept instead.

C=epsilon0 A/d and Q=CV use SI area and separation. Opposite plates carry +Q and −Q, so the capacitor remains neutral overall. A dielectric changes polarization and effective permittivity; here vacuum is stipulated. A slope alone cannot distinguish larger plate area, smaller gap, dielectric or stray parallel capacitance without independent geometry/material evidence. Energy is one half CV², not QV for gradual charging.

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

Understand

Define the source and measurement boundary

Capacitance is the proportionality C=Q/ΔV for the chosen linear device. An ideal capacitor has equal opposite charges on its plates, so the usual stored charge Q is a plate-charge magnitude, not the net charge of the entire two-plate object. For broad vacuum plates with negligible edge fields, C=ε0A/d. Convert area into square metres and separation into metres before using that equation.

Derive the quantitative comparison

The supplied plate area is .02 m², but a fixed stray capacitance is also connected across the same potential difference. Their charges add, so the measured Q–V slope is C_plate+C_stray. A separate charge-readout offset of 5 pC changes the intercept, not that slope. The model makes the stray contribution explicit; a single measured slope without geometry information would not uniquely reveal which physical contribution produced it.

Test an alternative before claiming identification

For a linear capacitor the charging voltage rises from zero to its final value. The work is therefore Q times the average charging voltage, giving U=QV/2=CV²/2, not QV. A dielectric polarizes and increases effective capacitance under the ideal linear model; changing separation, area or material can each change the slope. The present packet uses vacuum, and does not claim that nominal material constants or a calculated field establish a real breakdown or safety rating. Series and parallel networks and changing-dielectric energy accounts continue in the next category.

Worked original case

At gap 2 mm, C_plate=ε0(.02)/.002=88.541878128 pF. Adding 20 pF stray gives slope 108.541878128 pF. At 4 V the indicated charge is 439.167512512 pC, including the 5 pC readout offset; plate energy alone is about 7.08335×10⁻¹⁰ J.

Investigation

Compare gaps 1,2,3 mm while keeping the stray contribution fixed. Plot charge against voltage and retain slope and intercept separately. Explain why doubling the gap does not halve the total measured slope when a nonzero fixed stray capacitance remains.

Changed-case practice

Try the changed case before feedback

A charge–voltage fit has slope 120 pF and known stray parallel capacitance 30 pF. Find the inferred plate capacitance in pF.

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 evidence boundary

Original teaching, numerical cases and diagrams checked 30 October 2026 against the relevant primary textbook explanation. No textbook exercise, image or passage is reproduced. These are bounded synthetic records, not observed apparatus. No charging, wiring, capacitor handling, medical application or safety certification is requested. Preserve first predictions, contrasting settings and critiques in the course evidence workspace; public feedback is not independent mastery or performed laboratory time.

Apply

A constant charge-readout offset changes which feature of a linear Q–V fit?

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

Which independent observation could reject this model rather than merely refit its parameters?

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