Physics · Electromagnetism · Electrostatics
Capacitance Calculator
Calculates the capacitance of a parallel-plate capacitor given permittivity, plate area, and separation distance.
Calculator
Formula
C is the capacitance in farads (F); \varepsilon_r is the relative permittivity (dielectric constant) of the material between the plates (dimensionless); \varepsilon_0 = 8.854 \times 10^{-12} \text{ F/m} is the permittivity of free space; A is the area of one plate in square metres (m²); d is the separation distance between the plates in metres (m).
Source: Griffiths, D.J. — Introduction to Electrodynamics, 4th ed., Cambridge University Press (2017), §2.5.
How it works
Capacitance quantifies a capacitor's ability to store electric charge. When two parallel conductive plates are separated by an insulating medium (dielectric), an electric field forms between them when a voltage is applied. Charges of opposite sign accumulate on each plate, and the system stores electric potential energy. The ratio of charge stored to voltage applied is the capacitance C, measured in farads (F). Larger capacitance means more charge stored per volt — critical for timing circuits, energy storage, signal filtering, and power smoothing applications.
The parallel-plate capacitance formula is C = (ε_r · ε₀ · A) / d. Here, ε₀ = 8.854 × 10⁻¹² F/m is the permittivity of free space, a universal physical constant. The relative permittivity ε_r (also called the dielectric constant) characterises how much the insulating material between the plates enhances charge storage relative to vacuum — air has ε_r ≈ 1.0006, glass roughly 4–10, and barium titanate ceramics can exceed 10,000. Plate area A is the overlapping surface of one plate, and d is the gap distance. Capacitance increases linearly with both ε_r and A, but decreases inversely with separation d.
Practical applications of this formula span an enormous range. In PCB design, engineers calculate parasitic capacitance between copper traces. Sensor manufacturers design capacitive proximity and pressure sensors by exploiting how d changes under mechanical load. In power electronics, large capacitors store energy for burst delivery. In RF electronics, tunable capacitors (varactors) vary their plate overlap or gap to shift resonance frequency. Understanding the formula directly informs every one of these design decisions.
Worked example
Suppose you are designing a simple parallel-plate capacitor using a polypropylene dielectric sheet with the following parameters:
- Relative permittivity ε_r = 2.25 (typical for polypropylene)
- Plate area A = 200 cm² = 0.02 m²
- Plate separation d = 0.5 mm = 0.0005 m
Step 1 — Insert values into the formula:
C = (ε_r × ε₀ × A) / d = (2.25 × 8.854 × 10⁻¹² × 0.02) / 0.0005
Step 2 — Calculate the numerator: 2.25 × 8.854 × 10⁻¹² × 0.02 = 3.985 × 10⁻¹³ F·m
Step 3 — Divide by d: 3.985 × 10⁻¹³ / 0.0005 = 7.969 × 10⁻¹⁰ F ≈ 797 pF
This means the capacitor stores approximately 797 picofarads at any applied voltage. If you increased the plate area to 400 cm², capacitance would double to roughly 1594 pF. If instead you halved the gap to 0.25 mm, capacitance would also double — demonstrating the inverse relationship with separation distance. This kind of sensitivity analysis is essential when tolerancing capacitor designs for production.
Limitations & notes
This calculator assumes an ideal parallel-plate geometry with uniform electric field throughout the dielectric. In practice, fringing fields at the plate edges increase effective capacitance slightly — this effect is significant when plate dimensions are comparable to the separation gap. The formula also assumes a uniform, homogeneous dielectric material with a constant relative permittivity; in reality, ε_r varies with temperature, frequency, and applied field strength, particularly in ferroelectric materials like barium titanate. For very thin dielectrics (nanometre scale), quantum tunnelling effects and surface roughness become significant, invalidating classical predictions. Rolled or wound capacitor geometries require corrected formulas accounting for geometry. Additionally, this tool does not model leakage resistance, equivalent series resistance (ESR), or inductance — all of which affect real capacitor performance at high frequencies.
Frequently asked questions
What is the unit of capacitance and what does one farad represent?
Capacitance is measured in farads (F), named after Michael Faraday. One farad means the capacitor stores one coulomb of electric charge for every one volt applied across its plates. In practice, one farad is an extremely large capacitance — most electronic capacitors are rated in picofarads (pF, 10⁻¹² F), nanofarads (nF, 10⁻⁹ F), or microfarads (μF, 10⁻⁶ F). Only supercapacitors and ultracapacitors approach or exceed one farad.
How does the dielectric material affect capacitance?
The dielectric material between the plates increases capacitance by a factor equal to its relative permittivity ε_r compared to vacuum. A higher ε_r means the material polarises more strongly in response to an electric field, partially cancelling the field and allowing more charge to accumulate on the plates for the same voltage. For example, a capacitor with a glass dielectric (ε_r ≈ 7) stores seven times more charge than the same geometry in vacuum.
What happens to capacitance if I double the plate separation?
Doubling the plate separation d halves the capacitance, because C is inversely proportional to d. The electric field between the plates weakens as the plates move apart, reducing the charge that accumulates for a given voltage. This inverse relationship is why capacitive pressure sensors work — mechanical compression of the dielectric reduces d and increases the measurable capacitance.
Can this formula be used for cylindrical or spherical capacitors?
No — this formula applies specifically to parallel-plate capacitors. Cylindrical capacitors (used in coaxial cables) require C = 2πε₀ε_r L / ln(b/a), where L is the length and a, b are inner and outer radii. Spherical capacitors have yet another formula. Each geometry produces a different field distribution, requiring a separate derivation from Gauss's law.
What is the energy stored in a capacitor?
The energy stored in a charged capacitor is given by U = ½CV², where C is the capacitance in farads and V is the voltage in volts. For example, a 100 μF capacitor charged to 50 V stores U = 0.5 × 100×10⁻⁶ × 2500 = 0.125 joules. This stored energy is released when the capacitor discharges through a circuit, making capacitors essential for power delivery in flash photography, defibrillators, and pulsed power systems.
Last updated: 2025-01-15 · Formula verified against primary sources.