Compute energy in joules directly from voltage and charge (E = V × Q) or from capacitance and voltage (E = ½ C V²). Real‑time results with full formula traceability.
The joule (J) is the SI unit of energy, and the volt (V) measures electric potential. The relationship derives directly from the definition: one volt equals one joule per coulomb (1 V = 1 J/C). Hence, moving a charge Q through a potential difference V transfers energy E = V × Q. In capacitive systems, energy is stored in the electric field: E = ½ C V², where C is capacitance. Both formulas are cornerstones of electromagnetism and electrical engineering.
Fundamental energy relations:
E (J) = V (V) × Q (C) | E (J) = ½ × C (F) × V² (V²)
Derived from work done against electric field: dW = V·dq
The volt is defined as the potential difference that causes one ampere of current to do one watt of work (1 V = 1 W/A). Since power P = V·I and energy = P·t, integrating over time: E = ∫ V·I dt = V·∫ I dt = V·Q (for constant V). For a capacitor, I = C·dV/dt, leading to stored energy E = ∫ V·C·dV = ½ C V². These formulas are experimentally verified and used in every electrical engineering discipline. The work of James Prescott Joule (1818–1889) and Alessandro Volta (1745–1827) established the quantitative link between electrical and thermal energy.
Modern applications include renewable energy systems (solar + supercapacitors), electric vehicle battery packs (joules correspond to range), and high‑voltage transmission line studies. This calculator applies double‑precision arithmetic, ensuring accuracy beyond typical engineering needs (±1e‑12 relative error).
| System / Component | Voltage | Charge / Capacitance | Energy (Joules) | Practical context |
|---|---|---|---|---|
| Lightning bolt | ~100 MV | ~15 C | 1.5 × 10⁹ J | ~ 416 kWh, enough for 14 houses/day |
| Camera flash capacitor | 330 V | 120 µF | 6.53 J | Single high‑intensity flash |
| Smartphone battery (Li‑ion) | 3.7 V | 10 Ah ≈ 36000 C | 133,200 J | ~ 37 Wh, 1 day moderate use |
| Electron accelerated by 1 V | 1 V | 1.602 × 10⁻¹⁹ C | 1.602 × 10⁻¹⁹ J | 1 electronvolt (eV) = 1.602e-19 J |
| Power line capacitor bank | 15 kV | 50 µF | 5,625 J | Power factor correction |
A medical defibrillator stores energy in a capacitor bank charged to a specific voltage. Typical biphasic defibrillators deliver 150–360 J. Using E = ½ C V², if a capacitor of 60 µF is charged to 3000 V, the stored energy = 0.5 × 60×10⁻⁶ × (3000)² = 270 J. This energy is then discharged through the patient's chest to restore cardiac rhythm. Our calculator allows biomedical engineers to verify energy thresholds rapidly, ensuring device safety and clinical efficacy.