Electromagnetism
Parallel-plate capacitor
Uniform field E = V/d between the plates; read C = ε₀A/d and stored energy U = ½CV².
Investigate a parallel-plate capacitor: uniform electric field, plate charge, and capacitance. Ideal for electrostatics and Gauss's law lessons.
Parallel-plate capacitor
Uniform field E = V/d between the plates; read C = ε₀A/d and stored energy U = ½CV².
Parallel plates at voltage V and separation d create a nearly uniform field E = V/d between them (fringing ignored at the edges). Capacitance C = ε₀A/d and stored energy U = ½CV² follow from geometry. Press Play to watch the plate charge indicators pulse; read C and U in the Properties panel.
- E = V/d
- C = ε₀A/d
Investigation brief
Plan the question before you open the lab
The brief mirrors the prerendered page: driving question, competing predictions, variable roles, governing laws, setup, analysis and extension prompts remain visible and in this order.
Driving question
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?
Predictions to weigh
- 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.
Variable roles
What you set:
- Voltage V (V)
What you measure:
- Capacitance C (pF)
- Stored charge Q (nC)
- Stored energy U (nJ)
How the investigation runs
- 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.
Governing equation
Parallel-Plate Capacitance — C = ε₀A/d
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.
What the printable worksheet asks students to work out
- 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.
Where this shows up beyond the lab
- 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.
- AP Physics 2 — Unit 10: Electric Force, Field, and Potential
- AP Physics C: Electricity and Magnetism — Unit 10: Conductors and Capacitors
- AP Physics C: Electricity and Magnetism — Unit 9: Electric Potential
- IB Physics — D.2 Electric and magnetic fields
- General High School Physics — Magnetism & electromagnetism
- NGSS High School Physics — Energy in particle motion and fields
- Welcome to Parallel Plates
- Select the capacitor
- Press Play
- Uniform field between plates
- Open the Properties panel
- You did it!
Open the interactive simulation to build the scene, press Play, and explore with live measurements and a guided tutorial.