Capacitance Calculator
Two ways to find capacitance: solve the charge relation C = Q / V for any unknown, or design a parallel plate capacitor from its plate area, gap, and dielectric material. Either way you get the capacitance in readable units plus the charge, the energy stored, and the electric field between the plates.
Formula
Worked example
Parallel plate: two plates of 100 cm² (0.01 m²) separated by 1 mm of air (κ ≈ 1.0006). C = 1.0006 × 8.854e-12 × 0.01 / 0.001 = 8.86e-11 F = 88.6 pF. At 12 V they hold Q = C·V = 1.06 nC and store E = ½·C·V² = 6.38 nJ, with a field of 12000 V/m.
What capacitance means
Capacitance is the ratio of the electric charge stored on a conductor to the voltage across it, written C = Q / V. Its SI unit is the farad (F), defined as one coulomb of charge per volt. A larger capacitance means the component can hold more charge at the same voltage, much as a wider bucket holds more water at the same depth. Because one farad is enormous in practical terms, everyday capacitors are rated in microfarads (1 µF = 10⁻⁶ F), nanofarads (1 nF = 10⁻⁹ F), or picofarads (1 pF = 10⁻¹² F). This calculator offers two routes: solve the charge relation directly, or design a parallel plate capacitor from its physical dimensions.
Solving C = Q / V for any unknown
The single relation C = Q / V can be rearranged to find whichever quantity you do not know. To find the charge a capacitor holds, multiply its capacitance by the applied voltage, Q = C · V. To find the voltage that a known charge produces on a known capacitor, divide charge by capacitance, V = Q / C. Pick what you want to solve for and supply the other two, so you never have to manipulate the algebra by hand. Keep your units consistent, coulombs, volts, and farads, and convert prefixes such as µF or nC to base SI units before entering them, or the result will be off by powers of ten.
Designing a parallel plate capacitor
For two flat plates the capacitance follows from geometry: C = κ · ε0 · A / d. Here A is the overlapping plate area, d is the gap between the plates, ε0 is the permittivity of free space (8.854 × 10⁻¹² F/m), and κ (kappa) is the dielectric constant of the insulator filling the gap. Capacitance rises with a larger area or a higher-κ dielectric, and falls as the plates move apart. Choosing a dielectric such as mica (κ ≈ 5.4) or a high-k ceramic (κ ≈ 100) multiplies the capacitance over an air gap of the same size. Pick a material from the list or enter a custom κ, set the area and gap in your preferred units, and the calculator converts everything to SI before applying the formula.
Energy, charge, and field strength
A charged capacitor stores energy in the electric field between its plates, and that energy equals E = ½·C·V². Equivalent forms are E = ½·Q·V and E = Q² / (2C). The factor of one half appears because the voltage climbs from zero to its final value as the capacitor charges, so the average voltage doing work is half the final value. Energy scales with the square of voltage: doubling the voltage stores four times the energy. In plate mode, supplying an applied voltage also reports the charge stored (Q = C·V) and the electric field between the plates (E = V / d). Keep that field well under the dielectric strength of your material, since exceeding it causes breakdown, and remember that large charged capacitors can be dangerous even after a circuit is switched off.
Dielectric constants of common materials
| Material | Dielectric constant (κ) | Typical use |
|---|---|---|
| Vacuum | 1 | Reference value |
| Air | 1.0006 | Variable / tuning capacitors |
| Teflon (PTFE) | 2.1 | RF and high-frequency parts |
| Paper | 2.3 | Older film capacitors |
| Polystyrene | 2.5 | Precision film capacitors |
| Mica | 5.4 | Stable RF and timing |
| Glass | 5.5 | High-voltage capacitors |
| Ceramic (high-k) | 100+ | Compact bypass / decoupling |
Relative permittivity κ used in C = κ·ε0·A / d. Values are approximate and vary with frequency and temperature.
Frequently asked questions
What is the formula for capacitance?
Capacitance is charge divided by voltage, C = Q / V, measured in farads (F). Rearranged, the charge stored is Q = C · V and the voltage is V = Q / C. For a parallel plate capacitor you can also compute it from geometry: C = κ · ε0 · A / d, where A is the plate area, d the gap, ε0 the permittivity of free space, and κ the dielectric constant.
How does the dielectric affect capacitance?
The dielectric constant κ multiplies the capacitance directly. Replacing the air gap with mica (κ ≈ 5.4) makes a capacitor about 5.4 times larger for the same plate area and spacing, and a high-k ceramic (κ ≈ 100) increases it roughly a hundredfold. That is why high-value capacitors use special dielectrics rather than just larger plates.
How much energy does a capacitor store?
The energy stored in a capacitor is E = ½·C·V² joules, where C is capacitance in farads and V is the voltage in volts. You can also write it as E = ½·Q·V or E = Q² / (2C). Because energy depends on the square of voltage, doubling the voltage stores four times as much energy.
Why is one farad considered so large?
A one-farad capacitor stores one coulomb of charge at one volt, which is an enormous amount of charge for a small component. Most circuits need far less, so real capacitors are rated in microfarads (10⁻⁶ F), nanofarads (10⁻⁹ F), or picofarads (10⁻¹² F). Supercapacitors are the main parts measured in whole farads.