Theory — Capacitance, Charge, and Energy
What a Capacitor Does
A capacitor stores electric charge and electric energy. The simplest type is two parallel conducting plates separated by a small gap. When a battery is connected, it pushes charge onto the plates: one plate gains +Q, the other an equal −Q. The capacitance C measures how much charge the device holds per volt of potential difference across it.
Units: farad (F) = coulomb / volt. Practical capacitors are measured in
microfarads (µF = 10⁻⁶ F), nanofarads (nF = 10⁻⁹ F), and picofarads (pF = 10⁻¹² F).
The Parallel-Plate Capacitor
For two flat plates of area A separated by a distance d, the capacitance depends only on the geometry and on the material filling the gap. A larger plate area holds more charge; a smaller gap pulls the opposite charges closer and increases the capacitance. Inserting an insulating material — a dielectric — multiplies the capacitance by the dielectric constant κ (kappa).
ε₀ = 8.854 × 10⁻¹² F/m (permittivity of free space)
κ = dielectric constant (κ = 1 for vacuum/air, > 1 for insulators)
The Field Between the Plates
The charge on the plates sets up a nearly uniform electric field in the gap, pointing from the positive plate to the negative plate. For a given voltage, a smaller gap means a stronger field.
Energy Stored
Charging a capacitor takes work, because each additional bit of charge must be pushed onto a plate that already repels it. That work is stored as electric potential energy and is released when the capacitor discharges. The three forms below are all equal — use whichever variables you know.
Battery Connected vs. Disconnected
This is the idea students most often get wrong, so the simulation lets you test it directly. While the battery stays connected, the voltage V is held fixed and the charge adjusts. Once you disconnect the battery, the charge Q is trapped and stays fixed, so changing the geometry or dielectric changes the voltage instead.
Battery connected — V fixed
V is held by the battery. Change A, d, or κ → C changes → Q = CV changes to match. Adding a dielectric pulls more charge from the battery.
Battery disconnected — Q fixed
Charge is trapped on the plates. Change A, d, or κ → C changes → V = Q/C changes. Pulling the plates apart raises the voltage; adding a dielectric lowers it.
| Change | Effect on C | If V fixed (connected) | If Q fixed (disconnected) |
|---|---|---|---|
| Increase area A | C increases | Q increases | V decreases |
| Increase gap d | C decreases | Q decreases | V increases |
| Insert dielectric κ | C increases | Q increases | V decreases |
Instructions — Running the Virtual Experiment
Four experiments build from the basic geometry of a capacitor to the subtle behaviour of a disconnected one. Record every reading in your lab notebook.
Simulation — Parallel-Plate Capacitor
Controls
Set separation d
| d (mm) | C measured (pF) | C·d (pF·mm) |
|---|---|---|
| No readings yet — set a separation and click "Record reading". | ||
Team Questions
Example Lab Report
Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.
Capacitor Basics: Capacitance, Charge, and Stored Energy of a Parallel-Plate Capacitor
Physics | Section: [Your Section] | Date: [Date]
Lab Members: [Names of all members present]
Purpose
To investigate how the capacitance of a parallel-plate capacitor depends on plate area, plate separation, and the dielectric in the gap, and to verify the relationships Q = CV and U = ½CV². The lab also tests the difference between a capacitor that remains connected to its battery (constant voltage) and one that has been disconnected (constant charge).
Theory
For two parallel plates of area A separated by a distance d with a dielectric of constant κ in the gap, the capacitance is C = ε₀κA/d, where ε₀ = 8.854 × 10⁻¹² F/m. The charge and energy follow directly from the definition C = Q/V.
Q = C·V (Q ∝ V)
U = ½C·V² (U ∝ V²)
Multiplying C by d removes the separation, leaving C·d = ε₀κA, a constant for fixed area and dielectric. While the battery is connected the voltage is fixed; once disconnected the charge is fixed and V = Q/C responds to any change in geometry or dielectric.
Calculations — Sample: A = 100 cm² = 0.0100 m², d = 5.0 mm = 0.0050 m, κ = 1, V = 10 V
Capacitance: C = ε₀A/d = (8.854 × 10⁻¹²)(0.0100)/(0.0050) = 1.771 × 10⁻¹¹ F = 17.71 pF
Charge: Q = CV = (1.771 × 10⁻¹¹)(10) = 1.771 × 10⁻¹⁰ C = 177.1 pC
Energy: U = ½CV² = ½(1.771 × 10⁻¹¹)(10)² = 0.885 nJ
Field: E = V/d = 10 / 0.0050 = 2000 V/m
Results Table — Capacitance vs. Separation (A = 100 cm², κ = 1; ε₀A = 88.54 pF·mm)
| d (mm) | C measured (pF) | C·d (pF·mm) |
|---|---|---|
| 2 | 44.27 | 88.54 |
| 4 | 22.13 | 88.54 |
| 6 | 14.76 | 88.54 |
| 8 | 11.07 | 88.54 |
| 10 | 8.85 | 88.54 |
Connected-battery checks (A = 100 cm², d = 5 mm): doubling V from 5 V to 10 V doubled Q (88.5 → 177.1 pC) but quadrupled U (0.221 → 0.885 nJ). Raising κ from 1 to 3 tripled both C and Q (17.71 → 53.12 pF; 177.1 → 531.2 pC). Disconnected check: at fixed Q, increasing d raised V proportionally; inserting κ = 2 halved V.
Discussion
The capacitance was inversely proportional to the plate separation: every reading of the product C·d gave the same value, 88.54 pF·mm, which matched the predicted ε₀A = (8.854 × 10⁻¹²)(0.0100) = 8.854 × 10⁻¹⁴ F·m to the full precision of the model. This confirms C ∝ 1/d and recovers ε₀ directly from the geometry. Increasing the plate area or the dielectric constant raised the capacitance in direct proportion, as C = ε₀κA/d requires.
With the battery connected, the voltage stayed fixed and the charge tracked the capacitance: Q = CV rose when C rose. The energy, however, depends on V², so doubling the voltage quadrupled the stored energy. The disconnected capacitor behaved in the opposite way. With the charge trapped, pulling the plates apart lowered C and therefore raised V = Q/C, while inserting a dielectric raised C and lowered V. The stored energy of the disconnected capacitor fell when the dielectric was inserted (U = Q²/2C with C increasing); the "missing" energy is the work the capacitor does pulling the dielectric into the gap.
Conclusion
The experiment confirmed the parallel-plate capacitor relationships. Capacitance obeyed C = ε₀κA/d, with C·d constant and equal to ε₀A; charge obeyed Q = CV; and energy obeyed U = ½CV². The contrasting behaviour of a connected capacitor (constant V, charge adjusts) and a disconnected capacitor (constant Q, voltage adjusts) was demonstrated directly, clarifying which quantity is held fixed in each case.
Practice Questions
Show all work and include units in your answers.