Relate capacitor geometry to charge and voltage

Name the stored quantity carefully

Two conducting plates can carry charges +Q and −Q while the pair remains net neutral. Here Q is the magnitude on one plate. It is not the net charge of the capacitor, and adding the two magnitudes does not give the Q used in the capacitance formula.

Capacitance describes charge magnitude per voltage difference: C = Q/V, measured in farads (F), where 1 F = 1 C/V. The letter C in a formula denotes capacitance; the unit C denotes coulombs. A picofarad (pF) is 10⁻¹² F, and a picocoulomb (pC) is 10⁻¹² coulomb. Voltage is measured between the plates.

For ideal vacuum parallel plates, C = ε₀ A/d. Here ε₀ = 8.854 × 10⁻¹² F/m is a rounded vacuum permittivity, A is overlapping area in square meters, and d is gap in meters. This neglects edge fields. These capacitance relations are developed in OpenStax: capacitors and capacitance.

Convert geometry before calculating

At the reference setting A = 30 cm², d = 3 mm:

• 1 cm² = (0.01 m)² = 10⁻⁴ m², so A = 0.003 m².
• 1 mm = 0.001 m, so d = 0.003 m.
• C = (8.854 × 10⁻¹²)(0.003)/(0.003) = 8.854 pF.

At 6 V, Q = CV = 53.124 pC on each plate in magnitude. Squaring the length conversion for area matters: treating 30 cm² as 0.30 m² would introduce a factor-of-100 error.

Compare two experiments with the same geometry

The connected scenario holds voltage at 6 V using an ideal source; charge can enter or leave each plate through the wires as geometry changes. The isolated scenario holds charge magnitude at 53.124 pC after disconnection from the source at the reference geometry; its voltage can change. The browser compares these conditions side by side, not as successive steps of one switching animation.

At A = 30 cm², compare d = 2 mm with d = 4 mm:

Quantity	Gap 2 mm	Gap 4 mm
Capacitance, both cases	13.281 pF	6.6405 pF
Connected voltage	6 V	6 V
Connected charge magnitude	79.686 pC	39.843 pC
Isolated charge magnitude	53.124 pC	53.124 pC
Isolated voltage	4 V	8 V

Predict every row before capturing the settings. Doubling gap halves capacitance in both scenarios. At fixed voltage, charge halves; at fixed charge, voltage doubles. The geometry formula did not change when the electrical constraint changed.

Now hold gap fixed and double area from 20 to 40 cm². Predict the opposite pattern: capacitance doubles, connected charge doubles, and isolated voltage halves. The plan-view square grows with the square root of area; its side does not double when area doubles.

Interpret the field and the approximation

The uniform interior field magnitude is E = V/d, in V/m, directed from the positive plate toward the negative plate. At fixed area and isolated charge, doubling gap doubles voltage but leaves E unchanged. At fixed voltage, doubling gap halves E. The side-view arrows show direction only; their count and size do not encode field strength or charge magnitude. Read the numerical table for those quantities.

For a square overlap, the model reports d/√A, a dimensionless gap-to-side ratio. At the reference it is about 0.0548; at A = 10 cm² and d = 5 mm it is about 0.1581. Larger ratios make the neglected edge region more significant. This ratio is not a calculated percentage error, and the browser does not solve the finite-plate field. The model also omits dielectrics, leakage and breakdown; capacitance is not a statement of a component’s safe voltage rating.

Transfer to unfamiliar geometry

For vacuum plates with A = 20 cm² and d = 2 mm, calculate capacitance and charge magnitude at 3 V. If that capacitor is isolated and its gap doubles, predict its new voltage. This example uses a different initial voltage from the browser.

Check the capacitance transfer

A = 0.002 m² and d = 0.002 m, so C = 8.854 pF. At 3 V, Q = 26.562 pC. After isolation, doubling gap halves C to 4.427 pF while Q stays fixed, so V doubles to 6 V. The browser's isolated reference charge remains 53.124 pC; do not treat its readout as this different initial condition.