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Capacitance (Parallel Plate)

A capacitor stores charge, and how much it can store depends on how it is built. Bigger metal plates (A) hold more charge, and plates that sit closer together (d) hold more charge because the pull between them is stronger. The symbol epsilon stands for how well the material between the plates, such as air, plastic, or ceramic, allows an electric field. The result is C = epsilon x A / d, measured in farads.

Electrical Electromagnetics Capacitance
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Parallel Plate Capacitance Calculator

C = ε · A / dε = ε₀ · εr
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CAPACITANCE (C)
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C = ε₀ · εr · A · (N − 1) / d
Explanation

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MaterialεrStrength (MV/m)C (1 cm², 1 mm)

Typical room-temperature values; real parts vary with grade, frequency, temperature and thickness.

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Variables

Symbol Quantity Unit
C Capacitance Farads
ε Permittivity F/m
A Plate area m²
d Distance between plates m

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