Capacitance Calculator
Calculate the capacitance of a parallel plate capacitor from plate area, separation, and dielectric constant using C = ε₀εᵣA/d.
About this calculator
A parallel-plate capacitor stores electrical energy as an electric field between two conductive plates separated by an insulating (dielectric) gap. Its capacitance -- how much charge it can hold per volt applied -- follows directly from plate geometry and the material between the plates: C = ε₀εᵣA/d, where ε₀ is the permittivity of free space (8.854×10⁻¹² F/m), εᵣ is the dielectric material's relative permittivity (1 for vacuum/air, roughly 3.5 for paper, 5-10 for common glass), A is the overlapping plate area, and d is the plate separation. This calculator plugs your plate area, separation, and dielectric constant directly into that formula and reports capacitance in whichever metric prefix (pF, nF, μF, mF, or F) best fits the result's magnitude, since real capacitors span many orders of magnitude from picofarads to farads.
It also reports the electric field strength and stored charge at a reference voltage of 1 volt, which scale linearly with any actual applied voltage -- double the voltage and both double along with it. The underlying model assumes an idealized uniform field between infinite parallel plates and ignores fringing-field effects at the plate edges, which is the standard textbook approximation and holds well when plate separation is small relative to plate dimensions, but will understate real capacitance somewhat for wide gaps or small plates.
Inputs
Results
Capacitance
88.54
How to Use This Calculator
- Enter Plate Area, Plate Separation, and Dielectric Constant (εᵣ).
- Review the Capacitance result.
- Use Unit and Electric Field at 1 V (V/m) to inform your decision.
- Use the chart to visualize the results and explore different scenarios by adjusting inputs.
How the result changes with Plate Separation
| Plate Separation | Capacitance |
|---|---|
| 0 | 177.08 |
| 0 | 118.05 |
| 0 | 59.03 |
| 0 | 35.42 |
What each input means
- Plate Area
- Area of each capacitor plate in square meters. 0.01 m² = 100 cm².
- Plate Separation
- Distance between the two plates. Smaller separation gives larger capacitance.
- Dielectric Constant (εᵣ)
- Relative permittivity of the material between plates. Vacuum = 1, paper ≈ 3.5, glass ≈ 5–10.
How this is calculated
Worked example, using the default values
- Identify Input ParametersPlate Area = 0.01, Plate Separation = 0.001, Dielectric Constant (εᵣ) = 1 = 3 input(s) provided
- Calculate CapacitanceCapacitance = displayCapacitance88.54 = 88.54
- Calculate UnitUnit = displayUnit.trim()pF = pF
- Calculate Electric Field at 1 VElectric Field at 1 V1000 = 1000
Engine last updated . Checked against 1 independently-derived test — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
Why does capacitance increase as plate separation decreases?
A smaller gap means the electric field between the plates is stronger for the same applied voltage, which lets the plates hold more charge before that voltage limit is reached. Since capacitance is defined as charge per volt, a stronger field per volt of separation directly increases capacitance -- this is why real capacitors use very thin dielectric layers to pack more capacitance into a small physical size.
What does the dielectric constant actually represent?
It measures how much an insulating material between the plates amplifies capacitance compared to a vacuum gap of the same size. A dielectric material's molecules polarize in the electric field, partially canceling it and letting the plates store more charge at the same voltage -- vacuum and air sit at essentially 1, while materials like glass or ceramic can reach into the hundreds or thousands for specialty capacitor dielectrics.
Is the electric field between the plates really uniform?
In the idealized model used here, yes -- the field is treated as uniform between infinite parallel plates, giving the simple E = V/d relationship. Real, finite-sized plates have some field bulging (fringing) near their edges that this calculator does not model, which makes the idealized formula most accurate when the plate separation is small compared to the plates' width and length.
How do I find the field or charge at a voltage other than 1 volt?
Both electric field and charge scale linearly with applied voltage in this model, so multiply the calculator's reported Electric Field at 1 V or the underlying charge figure by your actual voltage to get the real value. For example, at 12 volts the field and charge are both 12 times the calculator's 1-volt reference values.
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