Capacitance Calculator
Calculate capacitance, charge, voltage, or stored energy using C = Q/V and E = ½CV².
What Is Capacitance?
A capacitor stores electrical energy by accumulating charge on two conducting plates separated by an insulating dielectric. The fundamental relationship is C = Q/V: capacitance C (farads, F) equals the charge Q (coulombs, C) stored per volt V of potential difference across the plates. A 1 F capacitor stores 1 C of charge per volt of potential — an enormous amount by practical standards.
Practical capacitors range from picofarads (pF, 10⁻¹² F) in RF circuits to millifarads (mF) in power supplies, to farads in supercapacitors. The capacitance of a parallel plate capacitor is C = ε₀εᵣA/d, where ε₀ = 8.85×10⁻¹² F/m is the permittivity of free space, εᵣ is the relative permittivity of the dielectric, A is the plate area, and d is the plate separation distance.
The energy stored in a capacitor is E = ½CV² = Q²/(2C) = ½QV. These three equivalent forms come from integrating the work done to charge the capacitor from 0 to V. A 1000 μF capacitor charged to 400 V (camera flash) stores E = ½ × 10⁻³ × 160,000 = 80 J — released in microseconds as a bright flash.
Capacitors in series: 1/C_total = 1/C₁ + 1/C₂ + ... (like resistors in parallel — reduced capacitance). Capacitors in parallel: C_total = C₁ + C₂ + ... (increased capacitance). This combination behavior is opposite to resistors in series/parallel.
Formula Reference Table
| Solve For | Formula | Notes |
|---|---|---|
| Capacitance | C = Q / V | F = C/V; 1 F = 1 coulomb/volt |
| Charge | Q = C · V | Coulombs |
| Voltage | V = Q / C | Volts |
| Stored energy | E = ½CV² | Joules |
| Energy (alt forms) | E = Q²/(2C) = ½QV | All equivalent |
| Parallel plate | C = ε₀εᵣA/d | A=area (m²), d=gap (m) |
| Series capacitors | 1/C = 1/C₁ + 1/C₂ | Smaller total C |
| Parallel capacitors | C = C₁ + C₂ | Larger total C |
3 Worked Examples
A camera flash uses 1000 μF capacitor charged to 400 V.
- C = 1000×10⁻⁶ F = 10⁻³ F
- Q = CV = 10⁻³ × 400 = 0.4 C
- E = ½CV² = 0.5 × 10⁻³ × 160,000 = 80 J
- Released in ≈1 ms → power = 80,000 W = 80 kW!
100 μF bypass capacitor at 12 V. Find stored energy.
- E = ½ × 100×10⁻⁶ × 144 = 7.2×10⁻³ J = 7.2 mJ
- Very small — adequate for short-term current supply during transients
Plates 10 cm × 10 cm, separated by 1 mm in air (εᵣ = 1).
- A = 0.01 m², d = 10⁻³ m
- C = ε₀A/d = 8.85×10⁻¹² × 0.01 / 10⁻³
- C = 8.85×10⁻¹¹ F = 88.5 pF
Real-World Applications
Common Mistakes to Avoid
C = Q/V uses C in farads, not μF. 100 μF = 100×10⁻⁶ F = 10⁻⁴ F. Using 100 directly gives Q/V 1,000,000× too large.
Energy stored = ½CV². The ½ comes from integrating (charge builds progressively from 0 to V, not all at V). Missing the ½ gives an energy value 2× too large.
Capacitors in series = reciprocals add (like resistors in parallel). In parallel = values add (like resistors in series). This is opposite to resistor rules — a common source of errors.
Charge Q (coulombs) is the total stored. Current I (amperes) is charge flow per second. They are related by I = dQ/dt. A 1 μF cap at 1 V holds Q = 1 μC; it might have zero current if the voltage is steady.
Always use volts, not mV or kV, when calculating Q = CV or E = ½CV². 1 kV = 1000 V; a 10 μF cap at 1 kV holds Q = 10⁻⁵ × 10³ = 0.01 C, not 0.00001 C.
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Interpretation: This formula links charge, voltage, current, resistance, capacitance, power or circuit time response. Assumption: Confirm DC versus AC conditions, RMS versus peak values, component topology and steady-state versus transient behavior. Ideal components may be assumed.