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?

2 · Set Up

  1. Open the parallel-plates preset and press Reset. Plate area = 0.05 m², separation = 2 mm.
  2. Enable the charge and stored-energy readouts.
  3. 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

  1. 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.
  2. 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

  1. 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.
  2. 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.

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