# What is the capacitance?

Capacitance calculator: C = Q ÷ V from charge and voltage, or the capacitance of parallel plates, a sphere or a cylinder with a dielectric, with the charge and energy at a voltage.

- Page: https://www.acalculator.org/physics/capacitance-calculator
- JSON spec: https://www.acalculator.org/physics/capacitance-calculator.json
- Version: 7eb50c66310b

## Default answer

Example with the default inputs (From Parallel plates, Plate area (A) 1,550.0031000062 in², Plate separation (d) 0.0393700787401575 in, Between the conductors Vacuum, κ = 1, Voltage (V) 40 V): The capacitance is 8.85419 nF.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| by | From | Work out C from the charge and voltage, or from the shape of the capacitor. |
| q | Charge (Q) | The charge on each plate, in the unit chosen below. |
| qu | Charge unit | The unit of the charge you type. |
| A | Plate area (A) | The area of one plate that faces the other. |
| d | Plate separation (d) | The distance between the plates. |
| r1 | Inner radius (R₁) | The radius of the inner sphere or cylinder. |
| r2 | Outer radius (R₂) | The inner radius of the outer shell. Leave it empty for a sphere on its own. |
| L | Length (L) | The length of the two cylinders. |
| die | Between the conductors | The material that fills the gap, which sets the dielectric constant κ (OpenStax Table 8.1). |
| k | Dielectric constant (κ) | The dielectric constant of the material in the gap; 1 for a vacuum. |
| v | Voltage (V) | The potential difference between the plates. Optional for a shape: gives the charge and energy. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| capacitance | Capacitance | C to 6 significant figures, in F, mF, µF, nF, pF or fF: the largest unit that keeps it at 1 or more. |
| farads | Capacitance (F) | C in farads. |
| microfarads | Capacitance (µF) | C in microfarads: 1 µF = 10⁻⁶ F. |
| picofarads | Capacitance (pF) | C in picofarads: 1 pF = 10⁻¹² F. |
| chargeC | Charge (C) | Q = C × V, in coulombs. |
| energy | Energy stored (J) | U = ½ × C × V² = ½ × Q × V, in joules. |

## Method

C = Q ÷ V; parallel plates C = κε₀A ÷ d; sphere C = 4πκε₀R₁R₂ ÷ (R₂ − R₁), or 4πκε₀R₁ alone; cylinder C = 2πκε₀L ÷ ln(R₂ ÷ R₁); Q = CV; U = ½CV².

## Assumptions

- Ideal capacitors: the plates are large next to their gap (no edge field), and the cylinders are long next to their radii.
- A dielectric fills the whole gap and multiplies the vacuum capacitance by κ; ε₀ = 8.8541878188 × 10⁻¹² F/m (CODATA 2022).

## Worked examples

1. by = plates, A = 1, d = 0.001, die = vacuum, v = 3,000 gives capacitance = 8.85419 nF, farads = 0, chargeC = 0.000027, energy = 0.039844. Source: OpenStax, University Physics Volume 2, §8.1 Capacitors and Capacitance (Equations 8.1, 8.3, 8.4 and 8.6; Examples 8.1 and 8.3), https://openstax.org/books/university-physics-volume-2/pages/8-1-capacitors-and-capacitance (retrieved 2026-10-18), Example 8.1: C = ε₀A ÷ d = 8.85 nF and Q = CV = 26.6 µC (with ε₀ = 8.85 × 10⁻¹²); NIST, CODATA 2022 recommended value of the vacuum electric permittivity ε₀ = 8.8541878188(14) × 10⁻¹² F/m, https://physics.nist.gov/cgi-bin/cuu/Value?ep0 (retrieved 2026-10-18); OpenStax, University Physics Volume 2, §8.3 Energy Stored in a Capacitor (Equation 8.10, U = Q² ÷ 2C = ½QV = ½CV²), https://openstax.org/books/university-physics-volume-2/pages/8-3-energy-stored-in-a-capacitor (retrieved 2026-10-18).
2. by = qv, q = 0.8, qu = nC, v = 40 gives capacitance = 20 pF, farads = 0, picofarads = 20, energy = 0. Source: OpenStax, University Physics Volume 2, §8.4 Capacitor with a Dielectric (Equation 8.11, C = κC₀; Example 8.10: 20.0 pF at 40.0 V holds 0.8 nC and 16.0 nJ), https://openstax.org/books/university-physics-volume-2/pages/8-4-capacitor-with-a-dielectric (retrieved 2026-10-18); OpenStax, University Physics Volume 2, §8.3 Energy Stored in a Capacitor (Equation 8.10, U = Q² ÷ 2C = ½QV = ½CV²), https://openstax.org/books/university-physics-volume-2/pages/8-3-energy-stored-in-a-capacitor (retrieved 2026-10-18).
3. by = plates, A = 1, d = 0.001, die = ptfe gives capacitance = 18.5938 nF, farads = 0. Source: OpenStax, University Physics Volume 2, §8.4 Capacitor with a Dielectric (Equation 8.11, C = κC₀; Example 8.10: 20.0 pF at 40.0 V holds 0.8 nC and 16.0 nJ), https://openstax.org/books/university-physics-volume-2/pages/8-4-capacitor-with-a-dielectric (retrieved 2026-10-18) (C = κC₀) and Table 8.1 (κ = 2.1).
4. by = sphere, r1 = 0.05, r2 = 0.1, die = vacuum gives capacitance = 11.1265 pF, farads = 0. Source: OpenStax, University Physics Volume 2, §8.1 Capacitors and Capacitance (Equations 8.1, 8.3, 8.4 and 8.6; Examples 8.1 and 8.3), https://openstax.org/books/university-physics-volume-2/pages/8-1-capacitors-and-capacitance (retrieved 2026-10-18), Equation 8.4.
5. by = sphere, r1 = 6,371,000, die = vacuum gives capacitance = 708.869 µF, farads = 0.000709. Source: OpenStax, University Physics Volume 2, §8.1 Capacitors and Capacitance (Equations 8.1, 8.3, 8.4 and 8.6; Examples 8.1 and 8.3), https://openstax.org/books/university-physics-volume-2/pages/8-1-capacitors-and-capacitance (retrieved 2026-10-18), Example 8.3 (C = 4πε₀R).
6. by = cylinder, r1 = 0.0005, r2 = 0.002, L = 1, die = vacuum gives capacitance = 40.1304 pF, farads = 0. Source: OpenStax, University Physics Volume 2, §8.1 Capacitors and Capacitance (Equations 8.1, 8.3, 8.4 and 8.6; Examples 8.1 and 8.3), https://openstax.org/books/university-physics-volume-2/pages/8-1-capacitors-and-capacitance (retrieved 2026-10-18), Equation 8.6.

## FAQ

### How do you calculate capacitance?

Divide the charge on one plate by the voltage between the plates: C = Q ÷ V. A capacitor that holds 0.8 nC at 40 V has C = 0.8 × 10⁻⁹ ÷ 40 = 2 × 10⁻¹¹ F, which is 20 pF.

### What is the formula for a parallel-plate capacitor?

C = κε₀A ÷ d, with A the area of one plate, d the gap, ε₀ = 8.854 × 10⁻¹² F/m and κ the dielectric constant of the material in the gap (1 for a vacuum). Plates of 1 m² that are 1 mm apart give 8.85 nF.

### How does a dielectric change capacitance?

A dielectric that fills the gap multiplies the capacitance by its dielectric constant κ: C = κC₀. PTFE (κ = 2.1) between the 1 m², 1 mm plates raises 8.85 nF to 18.59 nF.

### How much energy does a capacitor store?

U = ½CV² = ½QV. A 20 pF capacitor at 40 V stores ½ × 20 × 10⁻¹² × 40² = 1.6 × 10⁻⁸ J, or 16 nJ.

### What is the capacitance of a sphere?

An isolated sphere of radius R has C = 4πε₀R. A sphere the size of the Earth (R = 6,371 km) has only about 709 µF. Two concentric spheres have C = 4πε₀R₁R₂ ÷ (R₂ − R₁).

### What is a farad?

One farad stores one coulomb per volt. It is a very large unit, so capacitors are usually rated in microfarads (µF, 10⁻⁶ F), nanofarads (nF, 10⁻⁹ F) or picofarads (pF, 10⁻¹² F).

## Sources

- OpenStax, University Physics Volume 2, §8.1 Capacitors and Capacitance: Equation 8.1, C = Q ÷ V; 8.3, C = ε₀A ÷ d; 8.4, spherical capacitor; 8.6, cylindrical capacitor; Example 8.1 (8.85 nF, 26.6 µC); Example 8.3 (isolated sphere, 4πε₀R), retrieved 2026-10-18. https://openstax.org/books/university-physics-volume-2/pages/8-1-capacitors-and-capacitance
- OpenStax, University Physics Volume 2, §8.3 Energy Stored in a Capacitor: Equation 8.10, U = Q² ÷ 2C = ½QV = ½CV², retrieved 2026-10-18. https://openstax.org/books/university-physics-volume-2/pages/8-3-energy-stored-in-a-capacitor
- OpenStax, University Physics Volume 2, §8.4 Capacitor with a Dielectric: Equation 8.11, C = κC₀; Example 8.10 (20.0 pF at 40.0 V, 0.8 nC, 16.0 nJ), retrieved 2026-10-18. https://openstax.org/books/university-physics-volume-2/pages/8-4-capacitor-with-a-dielectric
- OpenStax, University Physics Volume 2, §8.5 Molecular Model of a Dielectric: Table 8.1, dielectric constants at room temperature, retrieved 2026-10-18. https://openstax.org/books/university-physics-volume-2/pages/8-5-molecular-model-of-a-dielectric
- NIST, CODATA 2022 recommended values: vacuum electric permittivity ε₀ = 8.8541878188(14) × 10⁻¹² F/m, retrieved 2026-10-18. https://physics.nist.gov/cgi-bin/cuu/Value?ep0
- NIST Special Publication 811 (2008), Appendix B.8: 1 in = 0.0254 m, 1 ft = 0.3048 m, 1 mil = 0.001 in, retrieved 2026-10-18. https://www.nist.gov/pml/special-publication-811