Read geometry and stored energy from a capacitor model

Establish the model and evidence

A capacitor separates equal amounts of positive and negative charge. The combined device may have zero net charge while maintaining a substantial field and stored energy between its plates. The quantity Q in Q=CV means the magnitude on one plate, not the net charge of the two-plate system. Capacitance C measures charge separated per volt.

For ideal parallel plates in vacuum, C=ε₀A/d, where A is the overlapping area of one plate and d the gap. Increasing area creates more capacity for separated charge at a fixed voltage; increasing the gap reduces capacitance. The relationship assumes a gap small compared with plate dimensions and ignores edge fields. It must not be applied to arbitrary shapes without a suitable model.

With the plates connected to a fixed voltage source, Q changes when geometry changes. The nearly uniform gap field has magnitude E=V/d and points from the positive plate to the negative plate. The field is perpendicular to each conducting plate. This is a region-specific model: there are edge corrections near the boundaries and no claim that the field is uniform everywhere around the device.

The stored energy is U=½CV²=½QV. During charging the voltage grows from zero to its final value, so multiplying the final voltage by the final charge overcounts the field energy by a factor of two. On a Q versus V graph, the triangle area gives energy; the slope gives capacitance. The plot uses a family of hypothetical voltages for the selected geometry, while the ledger evaluates the specified 12 V endpoint.

Worked evidence

For A=200 cm²=0.0200 m² and d=2 mm=0.002 m, C=88.54 pF. At 12 V, Q=1.06248 nC, E=6000 V/m and U=6.37488 nJ. The two plates together still have net charge zero.

Predict and investigate

At fixed 12 V, doubling area changes which quantities: capacitance, plate charge, field magnitude or energy?

Compare area 200 and 400 cm² at gap 2 mm, then compare gaps 2 and 4 mm at fixed area. Retain the numerical ratios and explain which boundary condition keeps voltage fixed.

Transfer to changed conditions

A 60 pF capacitor is held at 8 V. Find one-plate charge and stored energy; then triple voltage with unchanged geometry.

Compare the worked transfer

Initially Q=480 pC and U=1.92 nJ. At 24 V, Q=1440 pC and U=17.28 nJ: charge triples, energy increases by nine, capacitance stays 60 pF.

Explain the boundary

Explain why equal opposite charges do not cancel the electric field inside the gap, and identify where the ideal uniform-field assumption fails.

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. These cases, diagrams and numerical records are original. All experiments are bounded browser calculations or synthetic data, not measured laboratory evidence. Do not construct or handle charged capacitors, high-voltage sources or particle-beam equipment for these activities.