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Parallel-Plate Capacitor Calculator

Find the capacitance of a parallel-plate capacitor.

Result

708.34 pF
0.7083 nF

How it is calculated

For two flat plates close together, the capacitance is C = ε₀ · εr · A ÷ d. A is the plate area in square metres (m²), d is the gap between plates, entered in millimetres and converted to metres by the tool, εr is the relative permittivity of the material between the plates (1 for vacuum or air, higher for plastics, ceramics and so on) and ε₀ = 8.8541878128 × 10⁻¹² F/m. The result is shown in picofarads and in nanofarads. If the dielectric field is empty or not positive, 1 is used. Area and gap must be greater than zero. The formula ignores fringe fields at the edges, so it is accurate when the gap is small compared with the plate size.

Worked example

Plates of 0.01 m² (10 cm × 10 cm), a gap of 0.5 mm and a dielectric with εr = 4. Gap in metres: 0.5 ÷ 1000 = 0.0005 m. C = 8.854 × 10⁻¹² × 4 × 0.01 ÷ 0.0005 = 7.083 × 10⁻¹⁰ F. In picofarads that is 708.34 pF, or 0.7083 nF. Halving the gap would double the capacitance to about 1,416.7 pF.

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Common mistakes

  • Entering the gap in metres or centimetres. The gap field is in millimetres, so 0.5 means half a millimetre.
  • Entering the area in cm² or mm². The area is in m², so 10 cm × 10 cm must be typed as 0.01.
  • Leaving relative permittivity at 1 for a real insulator, which understates the capacitance by the factor εr.
  • Using the formula with a large gap relative to the plates, where edge effects make the real value higher.

About this calculator

This calculator computes the capacitance of a parallel-plate capacitor: C = ε₀·εr·A/d, where ε₀ is the vacuum permittivity constant, εr is the dielectric material's relative permittivity, A is the plate overlap area, and d is the gap between the plates.

Electronics students and circuit designers use it to size capacitors from physical geometry, and to see how inserting a dielectric (like ceramic or mica) or narrowing the plate gap increases capacitance.

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Comments (1)

  • Editorial team

    Area goes in square metres, so convert first: 10 cm² is 0.001 m². The gap is in millimetres. As a sanity check, a 1 m² plate with a 1 mm air gap gives about 8854 pF, or 8.85 nF. The formula is the ideal one and ignores fringe fields, so it holds best when the gap is much smaller than the plate. Leave the dielectric constant at 1 for air. Which dielectric material would you like to see as a preset?

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