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Parallel-Plate Capacitance, Area & Gap

Solve ideal parallel-plate capacitance, plate area, separation or absolute permittivity with selectable units.

Formula & worked example
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How to use this calculator

Enter the known values in the units shown. Results update as you type.
Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
Use the worked example to check the method. Reset restores the starting fields.

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Formula and method

An area of 1,000 mm², a gap of 1 mm and permittivity of 8.854 pF/m give 8.854 pF of capacitance.

C = εA/d

Worked example

Enter these known values and leave the other values blank.

Overlapping plate area: 1000 mm²
Plate separation: 1 mm
Absolute permittivity: 8.854 pF /m

Capacitance (C): 8.854 pF

Assumptions and limitations

This ideal model assumes parallel plates, uniform dielectric and negligible fringing. All quantities must be positive.
Permittivity is absolute ε in farads per metre, not the dimensionless relative permittivity εᵣ. The displayed default 8.854 pF/m is a rounded vacuum value.
Area means the overlap of one opposing plate face, not the sum of both faces.

Common questions

How do I use relative permittivity?

Multiply relative permittivity by vacuum permittivity to obtain absolute permittivity in F/m, then enter that value.

References

The calculation equations, inverse formulas, units, and input rules were imported from this source. Bookify provides the interface and equation solver.

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