Joules to Volts Converter

Inputs

Provide energy and choose calculation pathway

Energy *

Energy Unit

Mode *

Using Charge (E = q·V)

Using Capacitance (E = ½·C·V²)

Charge (C) *

Decimal Places

Result History

Time	Energy	Unit	Mode	Parameter	Value	Decimals	Voltage (V)
No results yet. Run a conversion to populate history.

Formula Used

Using charge: E = q · V ⇒ V = E / q

E = energy in joules (J)
q = charge in coulombs (C)
V = electric potential in volts (V)

Using capacitance (capacitor): E = ½ · C · V² ⇒ V = √(2E / C)

C = capacitance in farads (F)
Assumes ideal capacitor, negligible losses, DC context.

Unit helpers: 1 Wh = 3600 J; 1 kWh = 3.6×10⁶ J; 1 kJ = 1000 J; 1 mJ = 0.001 J.

How to Use This Calculator

Enter energy and select the correct unit.
Choose a pathway: Using Charge or Using Capacitance.
Provide the required quantity: charge in coulombs or capacitance in farads.
Select decimal places for the displayed voltage.
Click Convert to compute and store the result.
Use Download CSV or Download PDF for reporting.
Clear history anytime using the provided button.

Joules to Volts Formula Using Charge (Coulombs)

Starting from the definition of potential difference as work per charge, the core relationship is E = q · V. Rearranging gives the conversion used by this tool: V = E / q.

Formula and Units

E in joules (J). Acceptable inputs include J, kJ, mJ, Wh, kWh.
q in coulombs (C). 1 C = 1 A·s; 1 Ah = 3600 C.
V in volts (V). By definition, 1 V = 1 J/C.
General integral form: E = ∫ V dq. If V varies with q, then E/q yields the average voltage over the transfer.
Constraints used here: E ≥ 0 for stored energy contexts, q > 0.

Step‑by‑Step Example

Energy: E = 3.6 J. Charge: q = 0.12 C.
Compute voltage: V = E/q = 3.6 / 0.12.
Result: 30 V. Adjust decimals using the precision selector if needed.

Practical Notes

mAh to C: convert capacity to charge first. Example: 2000 mAh = 2 Ah = 7200 C.
Energy sources: 1 Wh = 3600 J; 0.5 Wh = 1800 J, etc.
Uncertainty: for independent errors, relative uncertainty obeys (σ_V/V) ≈ √[(σ_E/E)² + (σ_q/q)²].
Non‑idealities: internal resistance or losses reduce measured voltage versus ideal E/q.

Given	Convert	Compute V
E = 0.25 Wh, q = 900 C	E = 0.25 × 3600 = 900 J	V = 900 / 900 = 1 V
E = 500 mJ, q = 0.05 C	E = 0.5 J	V = 0.5 / 0.05 = 10 V

What Are Joules and Volts? Energy per Charge

Joule (J) is the SI unit of energy. One joule is the work done when a force of one newton moves an object one meter, or equivalently one watt delivered for one second.

Volt (V) is the SI unit of electric potential (potential difference). One volt equals one joule of energy per coulomb of charge. In symbols: 1 V = 1 J / C.

This tool uses the core identity Energy per charge. If you know stored energy and the amount of charge or the capacitance that stored it, you can infer the potential difference.

Quick Facts

Energy unit: joule (J) = watt·second (W·s).
Charge unit: coulomb (C) = ampere·second (A·s).
Voltage: V = E/q (general), V = √(2E/C) (capacitor).
Dimension check: [V] = kg·m²·s⁻³·A⁻¹.
Typical scales: AA cell ≈ 1.5 V; phone pack ≈ 3.7 V nominal.

Worked Examples

Given E and q: E = 10 J, q = 2 C ⇒ V = 10/2 = 5 V.
Given E and C: E = 10 J, C = 1 F ⇒ V = √(20) ≈ 4.4721 V.
Given Wh: E = 0.5 Wh = 1800 J, q = 1800 C ⇒ V = 1 V.

Quantity	Symbol	SI Unit	Relationship
Energy	E	J (joule)	1 Wh = 3600 J; 1 kWh = 3.6×10⁶ J
Charge	q	C (coulomb)	1 C = 1 A·s
Voltage	V	V (volt)	1 V = 1 J/C

Note: Energy differs from power. Power (W) is energy per time. Voltage relates energy to charge, not to time directly.

Step-by-Step Examples: Convert Joules to Volts

Example 1 — Using Charge

Energy: E = 12 J (already in joules).
Charge: q = 0.8 C.
Apply V = E/q ⇒ V = 12 / 0.8.
Voltage: 15 V.

Example 2 — From Wh & Charge

Energy: E = 0.12 Wh.
Convert to joules: E = 0.12 × 3600 = 432 J.
Charge: q = 540 C.
Compute: V = 432 / 540 = 0.8.
Voltage: 0.8 V.

Example 3 — Using Capacitance

Energy: E = 5 J.
Capacitance: C = 4700 μF = 4.7×10⁻³ F.
Use V = √(2E/C) ⇒ V = √(10 / 0.0047).
Voltage: ≈ 46.13 V (rounded to 2 decimals).

Example 4 — kJ & Charge

Energy: E = 2 kJ = 2000 J.
Charge: q = 25 C.
Compute: V = 2000 / 25.
Voltage: 80 V.

Sanity Checks

If q increases for fixed E, V should decrease.
For capacitors, doubling E raises V by √2 if C is fixed.
Units must align: convert Wh/kWh to joules and mAh/Ah to coulombs first.

Example Data Table

#	Energy	Unit	Mode	Parameter	Value	Voltage (V)	Explanation
1	10	J	Using Charge	Charge (C)	2	5	V = E/q = 10/2 = 5 V
2	10	J	Using Capacitance	Capacitance (F)	1	4.4721	V = √(2E/C) = √(20) ≈ 4.4721 V
3	0.5	Wh	Using Charge	Charge (C)	1800	1	E=0.5×3600=1800 J, then V=E/q=1800/1800=1 V

Common Use Cases: Batteries, Capacitors, Circuits

Batteries

Approximate relation: E ≈ ∫ V dq. If voltage is nearly flat, use E ≈ V_nom · q.
Capacity conversion: 1 Ah = 3600 C. So 2000 mAh = 2 Ah = 7200 C.
Voltage estimate from energy and charge: V ≈ E/q.
Real packs have discharge curves and internal resistance; results are idealized.

Example: E = 10 Wh ⇒ 36,000 J; capacity = 2 Ah ⇒ 7200 C. Then V ≈ 36,000/7200 = 5 V.

Capacitors

Stored energy: E = ½ · C · V². Solve for voltage: V = √(2E/C).
Useful for pulse power, camera flashes, ESD simulators, and lab supplies.
Always respect component voltage rating with margin.

Example: E = 0.5 J, C = 2200 μF = 2.2×10⁻³ F ⇒ V = √(1 / 0.0022) ≈ 21.33 V.

Circuits

For a steady current I over time t: charge moved q = I · t. Then V = E / (I · t).
For a resistor R: energy in time t is E = V² t / R ⇒ V = √(E R / t).
Choose the pathway that matches what you measure: current/time or capacitance.

Example (current/time): E = 12 J, I = 60 mA, t = 10 s ⇒ q = 0.6 C ⇒ V = 12 / 0.6 = 20 V.

Example (resistive load): E = 18 J dissipated by R = 10 Ω in t = 2 s ⇒ V = √(18×10/2) = √90 ≈ 9.49 V.

Safety & Practical Notes

Verify ratings: capacitor voltage, battery C‑rate, resistor power, wiring.
Account for efficiency and losses; ideal formulas give an upper bound.
Convert units first (Wh → J, mAh/Ah → C) to avoid inconsistent inputs.

Component	Typical Quantity	Notes
AA alkaline cell	~1.5 V; 2–3 Wh	Voltage droops under load; capacity depends on discharge rate.
Lithium‑ion pack (phone)	~3.7 V nominal; 10–15 Wh	Voltage spans ~3.0–4.35 V across state of charge and model.
Electrolytic capacitor	100 μF–4700 μF; 6.3–50 V	Observe polarity and ripple current limits; derate voltage.
Supercapacitor	1–50 F; 2.7–3.0 V per cell	Series balancing needed for higher voltages; high ESR variants exist.

Conversion Table and Quick Reference

Energy Units → Joules

Unit	Symbol	Multiply by → J	Notes
Joule	J	1	Base SI unit
Kilojoule	kJ	1,000	1 kJ = 10³ J
Millijoule	mJ	0.001	1 mJ = 10⁻³ J
Watt-hour	Wh	3,600	1 Wh = 3600 J
Kilowatt-hour	kWh	3,600,000	1 kWh = 3.6×10⁶ J

Charge Units → Coulombs

Unit	Symbol	Multiply by → C	Notes
Coulomb	C	1	Base SI unit
Milliamp-hour	mAh	3.6	1 mAh = 3.6 C
Amp-hour	Ah	3,600	1 Ah = 3600 C
Millicoulomb	mC	0.001	1 mC = 10⁻³ C
Microcoulomb	μC	1e−6	1 μC = 10⁻⁶ C

Voltage from Inputs — Quick Formulas

Inputs Known	Formula for V	When to Use
Energy E, Charge q	`V = E / q`	General “energy per charge” situations; batteries, charge transfer
Energy E, Capacitance C	`V = √(2E / C)`	Stored energy in capacitors, pulse systems, lab setups
Energy E, Current I, Time t	`V = E / (I · t)`	When you measure current over time; q = I·t
Energy E, Resistance R, Time t	`V = √(E R / t)`	Resistive loads where energy over interval is known

Quick Tips

Always convert Wh/kWh → J and mAh/Ah → C before applying formulas.
If q increases at fixed E, V must decrease; sanity‑check results.
For capacitors, doubling E scales V by √2 when C is constant.
Choose decimals to match measurement uncertainty; avoid false precision.
Dimension check: V has units J/C = kg·m²·s⁻³·A⁻¹.

FAQs

No direct factor exists. You need charge (q) for general systems, or capacitance (C) for capacitors, via V = E/q or V = √(2E/C) respectively.

Use it for capacitor energy problems, pulsed power storage, or lab setups where energy is stored electrostatically and C is known or measured.

Yes. Select Wh or kWh and the tool converts to joules internally using 1 Wh = 3600 J and 1 kWh = 3.6×10⁶ J.

Choose decimals consistent with your measurement uncertainty. For rough estimates, 2–3 decimals are typical; for precise labs, choose higher if your inputs justify it.

No. The formulas assume ideal behavior. Real systems have losses and internal resistance; measured voltages can be lower than ideal values.

Yes, rearrange: q = E/V for general systems, and C = 2E/V² for capacitors. This tool focuses on finding voltage.

Zero or negative charge/capacitance is invalid. Ensure energy is non‑negative and required parameters are strictly positive to avoid division by zero or square‑root issues.