Convert energy in joules to voltage in volts using the relationship V = J/C (Voltage = Energy / Charge). Enter joules and coulombs for instant voltage results with full explanations.
✓ Verified: SI Units — NIST Reference on Constants, Units, and Uncertainty — April 2026
J
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Energy in joules (J)
C
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Electric charge in coulombs (C)
Convert energy in joules to voltage in volts using the relationship V = J/C (Voltage = Energy / Charge). Enter joules and coulombs for instant voltage results with full explanations.
Voltage (Volts)
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Sources & Methodology
✓ Formulas verified against authoritative sources listed below.
IEEE reference for electrical energy, voltage, and charge unit definitions
Methodology: Voltage (V) = Energy (J) / Charge (C). This is the SI definition of the volt: 1 volt = 1 joule per coulomb. Equivalently: energy = charge x voltage (J = C x V). Electron-volt conversion: 1 eV = 1.602176634 x 10^-19 joules (exact, 2019 SI redefinition).
⏱ Last reviewed: April 2026
How to Convert Joules to Volts
The relationship between joules (energy), volts (electric potential), and coulombs (charge) is one of the fundamental equations of electrostatics. The volt is defined as one joule of energy per coulomb of charge: 1 V = 1 J/C. This means voltage represents the energy transferred per unit of charge, making this conversion essential for understanding batteries, capacitors, and electrical circuits.
The Definition of the Volt
One volt is defined as the potential difference that causes one joule of energy to be transferred per coulomb of charge: V = J/C. This is the SI base definition. Equivalently: one joule of work is done when one coulomb of charge moves through a potential difference of one volt. All three quantities are related: J = C x V = Q x V.
Worked Example: Battery Energy
A 12V car battery delivers energy to move charge. If 3 coulombs of charge pass through the circuit: Energy = Q x V = 3C x 12V = 36 joules. Conversely, if a capacitor stores 36 joules and has 3 coulombs of charge, its voltage = 36/3 = 12V. This bidirectional relationship is used in capacitor energy calculations.
Joules, Volts, and Electron-Volts in Physics
In atomic and particle physics, the electron-volt (eV) is the common energy unit: 1 eV = 1.602 x 10⁻¹⁹ joules. This is the energy gained by one electron (charge 1.602 x 10⁻¹⁹ coulombs) moving through a 1-volt potential difference. Particle accelerators are rated in MeV, GeV, and TeV — millions, billions, and trillions of electron-volts.
Capacitor Energy: J and V Relationship
A capacitor stores energy as E = (1/2) x C x V² = Q²/(2C) where C is capacitance in farads, V is voltage, and Q is charge. For a 100μF capacitor charged to 50V: E = 0.5 x 100x10⁻⁶ x 2500 = 0.125 joules. Rearranging: V = sqrt(2E/C) lets you find voltage from stored energy and capacitance.
V = J / C | J = C x V | C = J / V | 1 Volt = 1 Joule per Coulomb (SI definition)
V = voltage in volts. J = energy in joules. C = charge in coulombs. This is exact by SI definition. Electron-volt: 1 eV = 1.602176634 x 10^-19 J (exact since 2019 SI redefinition). For capacitors: E = Q*V/2 = C*V^2/2 where C is capacitance.
Joules to Volts Conversion Examples
Energy (J)
Charge (C)
Voltage (V)
Context
1 J
1 C
1.000 V
SI definition reference
12 J
1 C
12.000 V
Car battery reference
36 J
3 C
12.000 V
Car battery with 3C charge
0.001 J
0.001 C
1.000 V
Millijoule / millicoulomb
1.602e-19 J
1.602e-19 C
1.000 V
Electron through 1V (1 eV)
100 J
50 C
2.000 V
Capacitor discharge example
500 J
100 C
5.000 V
Electronics power supply
⚡ Physics Note: The joules-to-volts conversion requires knowing the charge in coulombs, which is often the part you do not directly measure. In practical circuits, you more commonly measure voltage and current (amperes). Since current = charge / time (A = C/s), you can derive charge as C = A x t (coulombs = amps x seconds). For a 2A circuit running 5 seconds: charge = 10 coulombs.
Frequently Asked Questions
Voltage (V) = Energy (J) / Charge (C). This requires knowing the charge in coulombs. If you know joules and coulombs: V = J/C. Example: 10 joules / 2 coulombs = 5 volts.
The fundamental relationship is: J = C x V (Energy = Charge x Voltage). Rearranged: V = J/C (Voltage = Energy/Charge) and C = J/V (Charge = Energy/Voltage). The volt is defined as 1 joule per coulomb by SI.
You cannot convert joules to volts without knowing the charge in coulombs. For 1 joule with 1 coulomb: V = 1/1 = 1 volt. For 1 joule with 2 coulombs: V = 1/2 = 0.5 volts. The voltage depends on both joules and coulombs.
Joules depend on charge: J = C x V. At 12V with 1 coulomb: 12 joules. With 5 coulombs: 60 joules. The energy in joules = voltage x charge in coulombs.
An electron-volt (eV) is the energy gained by one electron (charge 1.602176634 x 10^-19 coulombs) moving through a potential difference of 1 volt: 1 eV = 1.602176634 x 10^-19 joules. Used in atomic and particle physics.
From E = C*V^2/2 (where C is capacitance in farads): V = sqrt(2E/C). Example: 0.5 joules in a 100 microfarad capacitor: V = sqrt(2 x 0.5 / 0.0001) = sqrt(10000) = 100 volts.
Joules measure energy (how much work is done or energy transferred). Volts measure electric potential (energy per unit charge). They are related by charge: energy = voltage x charge. A battery rated at 12V can deliver different amounts of energy depending on how much charge flows.
Power (watts) = Voltage x Current. Energy (joules) = Power x Time = Voltage x Current x Time. Since Current x Time = Charge: Energy = Voltage x Charge. This confirms V = J/C. A 5V circuit through which 4 coulombs of charge flows delivers 20 joules.
Yes. Negative joules means energy is being absorbed rather than released (like charging a battery). Voltage can also be negative, indicating direction of potential difference. V = J/C handles signed values correctly: if both J and C are negative, V is positive.
1 coulomb = charge of approximately 6.242 x 10^18 electrons (1 / (1.602 x 10^-19)). In practical terms: a 1-ampere current flows when 1 coulomb passes a point per second. A typical AA battery stores about 3,000 coulombs of charge (approximately 2.78 ampere-hours x 3600 seconds).