Parallel Plate Capacitor

C = ε₀ × εr × A / d

Quick answer

This calculator solves C = ε₀ × εr × A / d for Capacitance. With Relative Permittivity 0.5, Plate Area 10 cm², Plate Separation 10 mm, the result is 0.4427 pF. It solves in 4 directions, so you can enter the values you have and get the one you need.

Calculator

Capacitance (pF)
—

Formula

C = ε₀ × εr × A / d

Description

The capacitance of a parallel plate capacitor is directly proportional to the plate area and the dielectric constant, and inversely proportional to the plate separation. The vacuum permittivity ε₀ = 8.854 × 10⁻¹² F/m is a fundamental physical constant. The relative permittivity εr (dielectric constant) of the material between the plates multiplies the capacitance: air has εr ≈ 1, FR-4 PCB material εr ≈ 4.5, ceramic capacitor dielectrics range from 6 (C0G/NP0) to over 10,000 (Y5V). This formula also applies to parasitic capacitance in PCB traces, connectors, and IC packaging.

Variables

  • C — Capacitance (F)
  • εr — Relative permittivity of the dielectric (dimensionless)
  • A — Overlapping plate area (m²)
  • d — Distance between plates (m)

Practical Notes

MLCC (multilayer ceramic capacitors) achieve high capacitance in small packages by stacking hundreds of thin dielectric layers in parallel. A 100-layer capacitor with each layer 2µm thick and 1mm² area using X7R ceramic (εr ≈ 3000) gives about 133 nF. Parasitic capacitance between PCB traces is typically 0.5-1 pF/cm for adjacent traces on the same layer.

Symbols and units

Symbol Meaning Default unit Accepted units
C Capacitance pF pF, nF, µF, mF, F
er Relative Permittivity — —
A Plate Area cm² mm², cm², m²
d Plate Separation mm mm, cm, m, km

Step by step

  1. 1. What you know: Relative Permittivity 0.5, Plate Area 10 cm², Plate Separation 10 mm
  2. 2. The relation: C = ε₀ × εr × A / d
  3. 3. Substitute: er = 0.5, A = 10 cm², d = 10 mm
  4. 4. Result: Capacitance = 0.4427 pF

Nearest standard component values

Series Nearest value Error
E12 0.47 pF 6.2%
E24 0.43 pF 2.9%

Worked examples

Minimum range

Given: Relative Permittivity 0.025, Plate Area 10 mm², Plate Separation 0.1 mm

Result: Capacitance = 0.02214 pF

Low range

Given: Relative Permittivity 0.25, Plate Area 1 cm², Plate Separation 1 mm

Result: Capacitance = 0.2214 pF

Typical

Given: Relative Permittivity 0.5, Plate Area 10 cm², Plate Separation 10 mm

Result: Capacitance = 0.4427 pF

High range

Given: Relative Permittivity 0.8, Plate Area 100 cm², Plate Separation 10 cm

Result: Capacitance = 0.7083 pF

Maximum range

Given: Relative Permittivity 8, Plate Area 0.1 m², Plate Separation 1 m

Result: Capacitance = 7.083 pF

Solving for each variable

Solve for Capacitance

Relative Permittivity 0.025, Plate Area 10 mm², Plate Separation 0.1 mm → 0.02214 pF

Solve for Relative Permittivity

Capacitance 1 pF, Plate Area 10 mm², Plate Separation 0.1 mm → 1.129

Solve for Plate Area

Capacitance 1 pF, Relative Permittivity 0.025, Plate Separation 0.1 mm → 4.518 cm²

Solve for Plate Separation

Capacitance 1 pF, Relative Permittivity 0.025, Plate Area 10 mm² → 0.002214 mm

How Capacitance changes with Relative Permittivity

Relative Permittivity (input) Capacitance (result)
0.05 0.04427 pF
0.125 0.1107 pF
0.25 0.2214 pF
0.375 0.332 pF
0.5 0.4427 pF
0.75 0.6641 pF
1 0.8854 pF
2 1.771 pF
3.5 3.099 pF
5 4.427 pF

The same result in other units

Unit Value
pF 0.4427
nF 0.0004427

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Frequently asked questions

How do you calculate Parallel Plate Capacitor?

Use C = ε₀ × εr × A / d. With Relative Permittivity 0.5, Plate Area 10 cm², Plate Separation 10 mm, Capacitance works out to 0.4427 pF.

What is the formula for Parallel Plate Capacitor?

The formula is C = ε₀ × εr × A / d. The capacitance of a parallel plate capacitor is directly proportional to the plate area and the dielectric constant, and inversely proportional to the plate separation.

How do you find Relative Permittivity when you know the other values?

Rearrange for Relative Permittivity: with Capacitance 1 pF, Plate Area 10 mm², Plate Separation 0.1 mm, Relative Permittivity is 1.129.

How do you find Plate Area when you know the other values?

Rearrange for Plate Area: with Capacitance 1 pF, Relative Permittivity 0.025, Plate Separation 0.1 mm, Plate Area is 4.518 cm².

What units does Parallel Plate Capacitor use?

Capacitance is shown in pF, nF. Values are held in SI base units internally, so you can mix prefixes without converting by hand.

Related Concepts

All Basic Electronics formulas →
Ohm's Law Electrical Power Voltage Divider Current Divider Thévenin Equivalent Norton Equivalent Temperature Coefficient of Resistance Kirchhoff's Voltage Law (KVL)

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