Capacitor Calculator – Q, C, V and Plate Geometry
Calculate charge, capacitance, and voltage using Q = CV, or find capacitance from parallel-plate geometry with C = ε₀εᵣA/d.
Select a calculation mode: enter any two of charge (Q), capacitance (C), and voltage (V) to solve for the third — or compute capacitance from plate area, separation, and dielectric constant.
Capacitor Calculator – Q, C, V and Plate Geometry
Calculate charge, capacitance, and voltage using Q = CV, or find capacitance from parallel-plate geometry with C = ε₀εᵣA/d.
Enter any two of capacitance (C), voltage (V), and charge (Q) to solve for the unknown third quantity using Q = C × V.
About the Capacitor Calculator
A capacitor is a passive two-terminal electrical component that stores energy in an electric field. The fundamental relationship governing a capacitor is Q = C × V, where Q is the electric charge stored in coulombs (C), C is the capacitance in farads (F), and V is the voltage across the plates in volts (V). This calculator allows you to find any one of these three quantities given the other two, covering the most common design and analysis scenarios encountered in electronics.
Capacitance is measured in farads, but practical capacitors are typically specified in microfarads (μF = 10⁻⁶ F), nanofarads (nF = 10⁻⁹ F), or picofarads (pF = 10⁻¹² F). A 1 F capacitor would be enormous at typical voltages; supercapacitors now reach hundreds of farads but are specialised components. Common through-hole electrolytic capacitors range from 1 μF to 10,000 μF; ceramic and film capacitors used for bypassing and filtering typically range from 1 pF to 100 μF.
The parallel plate capacitor model is the foundation for understanding all capacitors. Two conducting plates of area A (m²) separated by a distance d (m) filled with a dielectric material of relative permittivity εᵣ (also called the dielectric constant) form a capacitor with capacitance C = ε₀ × εᵣ × A / d, where ε₀ = 8.854187817 × 10⁻¹² F/m is the permittivity of free space. The relative permittivity εᵣ is dimensionless: it equals 1 for a vacuum, about 1.0006 for air, 2.2 for PTFE, 4.7 for FR4 PCB material, and 80 for water. High-K ceramic dielectrics (barium titanate and related compounds) can reach εᵣ > 10,000, allowing large capacitances in small packages.
In circuit analysis, capacitors serve several key roles: energy storage (in power supplies and motor drives), signal coupling (blocking DC while passing AC), decoupling (filtering supply noise at IC power pins), timing (RC time constants in oscillators and filters), and reactive power compensation in AC power systems. Understanding the Q–C–V relationship is essential for sizing capacitors for energy storage applications: the energy stored in a capacitor is E = ½ × C × V², so doubling the voltage quadruples the stored energy for the same capacitance.
Series and parallel capacitor combinations follow rules opposite to resistors: capacitors in parallel add their capacitances (Ctotal = C1 + C2 + …), while capacitors in series combine as 1/Ctotal = 1/C1 + 1/C2 + …. This is important when stacking capacitors for higher voltage ratings or combining them to achieve a target capacitance from standard values.
The dielectric strength (maximum electric field before breakdown) and temperature coefficient are also critical parameters. The voltage rating of a capacitor (its maximum working voltage, WVDC) must not be exceeded, and good practice is to operate below 80% of the rated voltage for reliability. Temperature coefficients (C0G/NP0, X7R, Y5V, etc.) describe how capacitance varies with temperature and must be considered in precision timing circuits.
Worked Examples
Three representative capacitor problems illustrating different uses of the Q = CV relationship and the parallel-plate formula.
| Given Values | Calculated Result | Notes |
|---|---|---|
| C = 100 μF = 1×10⁻⁴ F, V = 12 V | Q = C × V = 1.2 × 10⁻³ C = 1.2 mC | Charge stored in a 100 μF electrolytic capacitor charged to 12 V — typical for a small DC supply filter. |
| Q = 50 μC = 5×10⁻⁵ C, V = 5 V | C = Q / V = 1×10⁻⁵ F = 10 μF | Finding the required capacitance to store 50 μC at 5 V — useful when specifying an energy storage cap. |
| Plate: A = 0.01 m², d = 0.1 mm = 0.0001 m, εᵣ = 4.7 (FR4) | C = ε₀ × 4.7 × 0.01 / 0.0001 ≈ 4.16 nF | Capacitance of a 10 cm × 10 cm PCB trace over a ground plane 0.1 mm away with FR4 dielectric. |
How to Use the Capacitor Calculator
- Choose your calculation mode. Select 'Charge / Voltage (Q = CV)' to work with electrical quantities, or 'Parallel Plate Geometry' to compute capacitance from physical dimensions.
- In Q = CV mode, fill in any two of the three fields (capacitance, voltage, charge) and leave the third blank. The calculator solves for the missing quantity. The solved value is highlighted in blue.
- In Parallel Plate mode, enter the plate area in square metres, the separation in metres, and the dielectric constant (εᵣ). Use εᵣ = 1 for air/vacuum, 4.7 for FR4, 2.2 for PTFE, or look up the value for your specific material.
- Click Calculate to see the result. Click Reset to clear all fields and start a new calculation.
- Note that all capacitance values are in farads (F). Convert as needed: 1 μF = 1×10⁻⁶ F, 1 nF = 1×10⁻⁹ F, 1 pF = 1×10⁻¹² F.
Frequently Asked Questions
What is capacitance and how is it measured?
Capacitance is the ability of a component to store electric charge per unit voltage, defined as C = Q/V. It is measured in farads (F), where 1 farad = 1 coulomb per volt. In practice, most capacitors are rated in microfarads (μF), nanofarads (nF), or picofarads (pF) because a 1 F capacitor would be extremely large at ordinary voltages.
What is the dielectric constant (εᵣ)?
The relative permittivity εᵣ (dielectric constant) describes how much a material increases capacitance compared to a vacuum (εᵣ = 1). Air ≈ 1.0006, polypropylene ≈ 2.2, FR4 PCB ≈ 4.7, barium titanate ceramics ≈ 1,000–10,000. Higher εᵣ means more capacitance for the same geometry, which is why ceramic capacitors can pack large capacitance into tiny packages.
How does charge relate to current and voltage?
Charge Q = C × V, and current is the rate of change of charge: I = dQ/dt = C × dV/dt. This means the current through a capacitor depends on how quickly the voltage changes, not the voltage itself. At constant (DC) voltage, dV/dt = 0 and no steady current flows. At rapidly changing voltages (high frequency AC), the current is large, which is why capacitors pass AC but block DC.
What is the electric field between capacitor plates?
For a parallel plate capacitor, the electric field E = V/d (volts per metre), where V is the applied voltage and d is the plate separation. The charge density on the plates is σ = ε₀ × εᵣ × E = ε₀ × εᵣ × V / d. The maximum electric field before breakdown depends on the dielectric material: air breaks down at about 3 MV/m, while many solid dielectrics withstand 10–100 MV/m.
How do I combine capacitors in series and parallel?
Capacitors in parallel add: Ctotal = C1 + C2 + C3 + … (voltages are equal, charges add). Capacitors in series combine reciprocally: 1/Ctotal = 1/C1 + 1/C2 + … (charges are equal, voltages add). Parallel combination is used to increase capacitance or share ripple current; series combination is used to achieve a higher voltage rating or reduce capacitance to a non-standard value.
What is the energy stored in a capacitor?
The energy stored is E = ½ × C × V² (in joules). This is derived by integrating the work done to move charge from one plate to the other. Doubling the voltage quadruples the stored energy. For example, a 1000 μF capacitor charged to 400 V stores E = ½ × 0.001 × 400² = 80 J — enough to deliver a painful or dangerous electric shock, which is why large capacitors in power supplies must be discharged before servicing.