What is the capacitance?
Pick how you know the capacitor, then type its charge and voltage or its size. The capacitance calculator gives C in a readable unit, with the charge and the energy it stores at a voltage.
- Capacitance
- 8.85419 nF
The capacitance is 8.85419 nF.
- Capacitance (F)
- 8.85419 × 10⁻⁹
- Capacitance (µF)
- 0.00885419
- Capacitance (pF)
- 8,854.19
- Charge (C)
- 3.54168 × 10⁻⁷
- Energy stored (J)
- 0.00000708335
Capacitance: 8.85419 nF. The capacitance is 8.85419 nF.
How to calculate
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.
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.
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.
- 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
Each example is checked against the calculator on every build.
- From Parallel plates, Plate area (A) 1,550 in², Plate separation (d) 0.03937 in, Between the conductors Vacuum, κ = 1, Voltage (V) 3,000 V gives Capacitance 8.85419 nF, Capacitance (F) 0, Charge (C) 0.000027, Energy stored (J) 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)
- From Charge and voltage, Charge (Q) 0.8, Charge unit nC, Voltage (V) 40 V gives Capacitance 20 pF, Capacitance (F) 0, Capacitance (pF) 20, Energy stored (J) 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)
- From Parallel plates, Plate area (A) 1,550 in², Plate separation (d) 0.03937 in, Between the conductors PTFE, 2.1 gives Capacitance 18.5938 nF, Capacitance (F) 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)
- From Sphere, Inner radius (R₁) 1.969 in, Outer radius (R₂) 3.937 in, Between the conductors Vacuum, κ = 1 gives Capacitance 11.1265 pF, Capacitance (F) 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
- From Sphere, Inner radius (R₁) 250,826,772 in, Between the conductors Vacuum, κ = 1 gives Capacitance 708.869 µF, Capacitance (F) 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)
- From Cylinder, Inner radius (R₁) 0.01969 in, Outer radius (R₂) 0.07874 in, Length (L) 39.37 in, Between the conductors Vacuum, κ = 1 gives Capacitance 40.1304 pF, Capacitance (F) 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
How it works
Pick one of four ways in. ε₀ = 8.8541878188 × 10⁻¹² F/m (CODATA 2022), and κ is the dielectric constant of the material that fills the gap.
- Charge and voltage: C = Q ÷ V.
- Parallel plates: C = κε₀A ÷ d, with A the area of one plate and d the gap.
- Sphere: two concentric spheres, C = 4πκε₀ × R₁R₂ ÷ (R₂ − R₁), with R₁ the inner radius and R₂ the inner radius of the outer shell. Leave R₂ empty for a sphere on its own: C = 4πκε₀R₁.
- Cylinder: two coaxial cylinders of length L, C = 2πκε₀L ÷ ln(R₂ ÷ R₁).
With a voltage V, the page also gives the charge Q = C × V and the stored energy U = ½ × Q × V = ½CV². In the charge and voltage method, Q is the charge you type.
Dielectric constants (OpenStax Table 8.1, room temperature): vacuum 1; dry air at 1 atm 1.00059; PTFE 2.1; paraffin 2.3; silicone oil 2.5; polystyrene 2.56; nylon 3.4; paper 3.7; fused quartz 3.78; concrete 4.5; diamond 5.5; mica 6.0; neoprene rubber 6.7; water 80; strontium titanate 310. Or type your own κ.
Units (exact): 1 mC = 10⁻³ C, 1 µC = 10⁻⁶ C, 1 nC = 10⁻⁹ C, 1 pC = 10⁻¹² C; 1 mil = 0.0254 mm.
Rules
- Charge from 10⁻¹² to 10¹² in its unit; voltage from 10⁻⁹ V to 10⁹ V; plate area from 10⁻¹⁸ m² to 10¹² m²; gap, radii and length from 10⁻¹² m to 10⁹ m; κ from 1 to 10⁷.
- The charge and voltage method needs the voltage. A cylinder needs both radii. The outer radius must be larger than the inner radius.
- A capacitance too small or too large to show as a number has no answer.
Output format
The headline is C to 6 significant figures in the largest of F, mF, µF, nF, pF and fF that keeps the number at 1 or more (fF below 1 fF), rounded half up. Numbers in text are written with thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ. The other results show to 6 significant figures. Typed values are read back as exact decimals, so C = Q ÷ V and the parallel-plate C are exact until rounded once; the sphere and cylinder use π and a logarithm, so they are 64-bit floats, and the headline rounds the float's shortest decimal.
Worked examples by hand
Parallel plates. A = 1.00 m², d = 1.00 mm, vacuum: C = 8.8541878188 × 10⁻¹² × 1 ÷ 0.001 = 8.8541878188 × 10⁻⁹ F, shown as 8.85419 nF. At 3,000 V: Q = 8.8541878188 × 10⁻⁹ × 3,000 = 2.65626 × 10⁻⁵ C (26.6 µC) and U = ½ × 2.65626 × 10⁻⁵ × 3,000 = 0.0398438 J.
Charge and voltage. Q = 0.8 nC at V = 40 V: C = 0.8 × 10⁻⁹ ÷ 40 = 2 × 10⁻¹¹ F = 20 pF; U = ½ × 0.8 × 10⁻⁹ × 40 = 1.6 × 10⁻⁸ J.
With a dielectric. The same plates with PTFE (κ = 2.1): C = 2.1 × 8.8541878188 nF = 18.5938 nF.
Concentric spheres. R₁ = 5 cm, R₂ = 10 cm, vacuum: C = 4π × 8.8541878188 × 10⁻¹² × 0.05 × 0.1 ÷ 0.05 = 1.11265 × 10⁻¹¹ F = 11.1265 pF.
An isolated sphere. R = 6,371 km (the Earth): C = 4π × 8.8541878188 × 10⁻¹² × 6,371,000 = 7.08869 × 10⁻⁴ F = 708.869 µF.
A coaxial pair. R₁ = 0.5 mm, R₂ = 2 mm, L = 1 m, vacuum: C = 2π × 8.8541878188 × 10⁻¹² × 1 ÷ ln 4 = 4.01304 × 10⁻¹¹ F = 40.1304 pF.
Other questions people ask
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).