Capacitance Calculator
This calculator finds the capacitance of a parallel plate capacitor from three physical measurements: the relative permittivity of the dielectric material between the plates, the area of each plate, and the distance separating them. It also calculates the charge stored on the capacitor and the energy held in its electric field if you supply the voltage across it. A capacitor is one of the three passive components at the heart of electronic circuits, alongside resistors and inductors. It stores energy in an electric field between two conductive plates separated by an insulating material called a dielectric. The capacitance tells you how much charge the capacitor can hold per volt of applied voltage. The formula is C = epsilon_r x epsilon_0 x A / d, where epsilon_0 is the permittivity of free space at 8.854 x 10 to the minus 12 farads per metre, epsilon_r is the relative permittivity of the dielectric (1 for air, 2.1 for PTFE, 4.5 for typical FR4 circuit board material), A is the plate area in square metres, and d is the plate separation in metres. Because real capacitors have very small dimensions, the result is typically in picofarads or nanofarads. Once you have the capacitance, Q = CV gives the stored charge for a given voltage, and E = 0.5 x C x V squared gives the energy in joules. These relationships are used in filter design, energy storage, timing circuits and sensor design. Enter the dielectric constant, plate area, separation, and optionally the voltage, and all outputs update instantly.
epsilon_0 = 8.854 x 10^-12 F/m. Results assume a uniform field between infinite parallel plates; real capacitors differ slightly from this ideal model.
How it works
Capacitance is calculated from C = epsilon_r x epsilon_0 x A / d, where epsilon_0 = 8.854187817 x 10 to the minus 12 F/m. The area and separation must both be in SI units (square metres and metres). Once C is known, stored charge is Q = C x V coulombs and stored energy is E = 0.5 x C x V squared joules. Results are displayed in picofarads for typical small capacitors, with a nanofarad conversion alongside. If the voltage field is zero, Q and E both show zero.
Worked example
A PCB trace pair acts as a parallel plate capacitor. The dielectric constant of FR4 board material is 4.5, the overlapping plate area is 0.01 m squared and the separation (board thickness) is 0.001 m. Capacitance is C = 4.5 x 8.854 x 10^-12 x 0.01 / 0.001 = 398.4 pF (0.3984 nF). With 12 V applied, the stored charge is Q = 398.4 x 10^-12 x 12 = 4.781 nC and the stored energy is E = 0.5 x 398.4 x 10^-12 x 144 = 28.69 nJ. These match the default values above.
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