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Capacitance (Parallel Plate)

Capacitance of a parallel-plate capacitor.

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Capacitance Calculator Parallel Plate

C = ε · A / d
C = capacitance (F)  ·  ε = permittivity (F/m)  ·  A = plate area (m²)  ·  d = plate separation (m)
⟹ Solve C, ε, A, d
F
F/m
m
Please fix the errors above.
Solve for:
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Capacitance
C: ε: A: d:
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C = ε A / d  ·  ε0 ≈ 8.854 × 10⁻¹² F/m (vacuum). All units in SI.

Interpretation

Parallel‑plate capacitance depends on the dielectric permittivity, plate area, and separation distance.
Larger area and smaller gap both increase capacitance.
Example: ε=8.85×10⁻¹² F/m, A=0.1m², d=0.001m → C = (8.85e-12 × 0.1) / 0.001 = 8.85×10⁻¹⁰ F.

C = εA / d
Capacitance (Parallel Plate)

Variables

SymbolQuantityUnit
CCapacitanceFarads
εPermittivityF/m
APlate area
dDistance between platesm

What it means

The capacitance of a parallel‑plate capacitor is given by C = εA/d, where ε is the permittivity of the dielectric material between the plates, A is the area of each plate, and d is the separation distance. Permittivity ε = ε₀ * ε_r, where ε₀ is the vacuum permittivity (8.85×10⁻¹² F/m) and ε_r is the relative permittivity (dielectric constant) of the material. This formula shows that capacitance is directly proportional to plate area and inversely proportional to plate separation. Increasing the area or decreasing the distance increases capacitance. The formula also shows that inserting a dielectric (ε_r > 1) multiplies the capacitance. This is the basis for the design of variable capacitors (by changing plate overlap) and for high‑value capacitors using thin dielectric films. Understanding parallel‑plate capacitance is fundamental in circuit design, filter design, and energy storage. It also helps in understanding parasitic capacitance in circuit boards. Example: With ε=8.85×10⁻¹² F/m, A=0.1 m², d=0.001 m, C = (8.85e-12 * 0.1)/0.001 = 8.85×10⁻¹⁰ F (885 pF).

Worked example

Parallel Plate Capacitance – Practical Example

Real‑World
Scenario: Design a 100 pF capacitor using two parallel plates with air (ε₀ = 8.85 pF/m). The plate area is 10 cm² and separation is 1 mm. Verify the capacitance.
ParameterValue
ε₀8.85 pF/m
A10 cm² = 10×10⁻⁴ m²
d1 mm = 10⁻³ m
FormulaC = ε₀·A / d
1Substitute:C = 8.85e-12 · (10e-4) / 1e-3
2Simplify:C = 8.85e-12 · 0.001 / 0.001 = 8.85e-12 F
3Convert to pF:C = 8.85 pF
Final Design C ≈ 8.85 pF (10 cm², 1 mm gap) ✓ Suitable for RF
Why: The capacitance is proportional to area and inversely proportional to distance – small gaps give higher capacitance.

Common mistakes

  • Permittivity ε: Use the absolute permittivity of the dielectric (ε = ε₀·ε_r).
  • Area A: The overlap area of the plates – not the total surface area.
  • Distance d: The separation between plates – in metres.
  • Edge effects: The formula assumes infinite plates – for finite plates, fringe fields cause deviations.
  • Dielectric breakdown: If the voltage is too high, the dielectric may break down – the formula gives capacitance only.

Applications

Parallel‑plate capacitance is determined by the dielectric permittivity, plate area, and separation distance; larger area and smaller gap increase capacitance. This formula is essential for designing capacitors, touch sensors, and many electronic components. Engineers use it to calculate the capacitance of printed circuit board traces, to design capacitive sensing interfaces, and to model the behaviour of dielectric materials. In power electronics, it helps determine the capacitance required for filtering and energy storage. By understanding the dependence on geometry and material, professionals can optimise capacitor performance and size. This fundamental equation is also used in the design of MEMS capacitive sensors and in the study of dielectric breakdown.

  • Capacitor design for power supplies and filters
  • Capacitive touch sensing and proximity detection
  • PCB trace capacitance modelling for high‑speed signals
  • Dielectric material characterisation and selection
  • MEMS capacitive sensors and actuators

Frequently Asked Questions

Q01What is the capacitance of a parallel-plate capacitor?
A01

The capacitance is C = ε·A / d, where ε is the permittivity of the dielectric, A is the plate area, and d is the separation between plates.

Q02What is the permittivity ε in the formula?
A02

ε is the absolute permittivity of the dielectric material between the plates. For vacuum, ε = ε₀ = 8.85 × 10⁻¹² F/m. For other materials, ε = ε₀·ε_r.

Q03How does capacitance change with plate area and separation?
A03

Capacitance is directly proportional to plate area (A) and inversely proportional to plate separation (d). Increasing area or decreasing distance increases capacitance.

Q04What is the effect of inserting a dielectric between the plates?
A04

Inserting a dielectric increases capacitance by a factor equal to the dielectric constant (ε_r). It also increases the breakdown voltage, allowing higher voltages.

Q05What are typical applications of parallel plate capacitors?
A05

They are used in electronic circuits for filtering, tuning, energy storage, and coupling/decoupling. Also used in sensors (capacitive touch, humidity).

Q06What is the energy stored in a capacitor?
A06

The energy stored is E = ½ C V². This energy is stored in the electric field between the plates.

Q07What are the limitations of the parallel plate capacitance formula?
A07

The formula assumes infinite plates (neglects fringing fields) and uniform electric field. For finite plates, fringing effects cause slight deviations, especially for small gaps.

Q08How does temperature affect capacitance?
A08

Temperature can change dielectric constant and plate dimensions. Some capacitors have specific temperature coefficients (like C0G, X7R) that describe capacitance change with temperature.

Q09How do you measure capacitance in practice?
A09

Capacitance is measured using LCR meters, capacitance bridges, or by measuring the time constant in an RC circuit (charging/discharging curves).

Q10What are the common mistakes when using the parallel plate capacitance formula?
A10

Common errors include: 1) using the wrong permittivity (e.g., vacuum instead of dielectric), 2) using the wrong units for area and distance, 3) forgetting the factor ε_r, 4) applying the formula to non-parallel plates, and 5) ignoring fringing effects.