Parallel-Plate Capacitor: Charge and Energy Storage
1 · Predict
A parallel-plate capacitor has a fixed geometry (area and gap). If you increase the voltage across it, does the stored energy grow in direct proportion, or faster?
- Stored energy grows faster than voltage — it depends on voltage squared.
- Stored energy is directly proportional to voltage.
- Stored energy doesn't depend on voltage.
2 · Set Up
- Open the parallel-plates preset and press Reset. Plate area = 0.05 m², separation = 2 mm.
- Enable the charge and stored-energy readouts.
- Set the plate voltage for each trial and record the stored charge and energy.
3 · Collect Data
| Voltage V (V) | Capacitance C (pF) | Stored charge Q (nC) | Stored energy U (nJ) |
|---|---|---|---|
| 6 | |||
| 12 | |||
| 18 |
Plot stored energy U (y-axis) against voltage V (x-axis) for your three trials. Is the curve straight or does it bend upward?
4 · Analyze
- For one trial, compute C = ε₀A/d using A = 0.05 m², d = 0.002 m (same in every trial), then Q = CV and U = ½CV². Compare all three to the table.
- Explain, using U = ½CV², why doubling the voltage quadruples the stored energy rather than doubling it — even though the stored charge only doubles.
5 · Extend
- The capacitance C = ε₀A/d depends only on the plates' geometry, not on the voltage you apply. Explain why a physically bigger capacitor (larger A or smaller d) can store more charge at the same voltage.
- A medical defibrillator charges a capacitor to a high voltage, then discharges it through the patient in a fraction of a second. Using U = ½CV², explain why doctors use a high voltage rather than a high capacitance to deliver the needed energy quickly.
The Physics Behind This Experiment
Parallel-Plate Capacitance
Two parallel conducting plates separated by a gap store charge in proportion to their area and inversely to their separation: C = ε₀A/d. The stored energy is U = ½CV² = ½QV.