Capacitance & Capacitor Energy Calculator
Capacitance, charge, voltage and stored energy — plus the parallel-plate geometry formula.
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Parallel-plate geometry (optional)
Formulas
E = ½ · C · V² = Q²/(2C) = ½ · Q · V
Parallel plate: C = ε₀ · εr · A / d (ε₀ = 8.854×10⁻¹² F/m)
Physics behind capacitors
A capacitor stores electrical energy in the electric field between two conductors separated by an insulator (dielectric). Its capacitance is a purely geometric and material property — bigger plates and thinner gaps give more capacitance; inserting a dielectric with εr > 1 increases it further. Charge and voltage are proportional (Q = C·V), and the energy scales with voltage squared.
Worked example
C = 100 μF, V = 12 V
Q = C·V = 100×10⁻⁶ × 12 = 1.2×10⁻³ C = 1.2 mC E = ½·C·V² = ½·100×10⁻⁶·144 = 7.2×10⁻³ J = 7.2 mJ
Related tools
FAQs
How to calculate energy stored in a capacitor?
The energy a capacitor holds is given by E = ½ CV², where C is capacitance in farads and V is voltage across the plates in volts. The result comes out in joules. Take a 100 µF capacitor charged to 12 V: E = ½ × 100×10^-6 × 144 = 7.2 mJ. The square on V is the important part — doubling the voltage quadruples the energy stored. That's also why charged high-voltage capacitors deserve respect in the lab: even small ones can pack a real punch when discharged across a screwdriver.
How do I use capacitor energy formula calculator?
Feed in the capacitance and the voltage, and the tool returns stored energy using E = ½ CV². Keep capacitance in farads (so 47 µF becomes 47×10^-6 F) and voltage in volts. Output is in joules, though the calculator may show millijoules or microjoules for small values. A 1000 µF capacitor at 50 V stores ½ × 0.001 × 2500 = 1.25 J, enough to feel a sharp shock. Use it when you're designing flash circuits, defibrillators in problems, or working backwards from energy storage requirements.
How to find capacitance from energy and voltage?
Start from E = ½ CV² and rearrange for C: C = 2E / V². Energy goes in as joules, voltage as volts, and you get capacitance in farads. Suppose a circuit needs to store 5 J at 100 V — then C = 2 × 5 / 10000 = 0.001 F, or 1000 µF. This kind of reverse-solving comes up often when you're sizing a capacitor for camera flash energy, audio amplifier rails, or a backup supply. Always square the voltage first to avoid silly arithmetic slips.
How much energy is stored in a 1 farad capacitor?
There's no single answer because a 1 F capacitor's stored energy depends entirely on the voltage you put across it. Using E = ½ CV²: at 1 V it holds 0.5 J, at 5 V it holds 12.5 J, and at 12 V it holds a hefty 72 J. Real-world supercapacitors of 1 F at car-audio voltages can store enough energy to weld a screwdriver tip. So whenever someone says 'a 1 farad cap', I always ask: charged to what voltage? That's where the energy lives.
Why is capacitor energy one half CV squared?
When you charge a capacitor from 0 to V, the voltage rises gradually as charge builds up. Each tiny bit of charge dq is added against the current voltage, so the work isn't qV but the integral ∫V dq. Since V = q/C, this becomes ∫(q/C)dq from 0 to Q, which evaluates to Q²/2C, equal to ½ CV². The factor of half comes from averaging the voltage over the charging process — it starts at 0 and ends at V, so the mean is V/2. The other half of the energy from the source ends up as heat in the wires.
How to calculate charge stored in a capacitor?
Charge on a capacitor obeys the very tidy relation Q = CV, where C is capacitance in farads and V is voltage in volts. The charge comes out in coulombs. A 220 µF capacitor at 9 V holds Q = 220×10^-6 × 9 ≈ 1.98×10³ C, or about 2 millicoulombs. This is the building-block formula for everything else — energy, current flow during charging, and time constants in RC circuits all start here. Just remember coulombs are big units, so capacitor charges usually appear in micro or milli range.
How do I use capacitors in parallel equivalent capacitance formula?
Capacitors in parallel simply add up: C_total = C + C + C + … This is the opposite of resistors in parallel, which trips up students every term. The reason is geometric — putting capacitors in parallel is like increasing the plate area, and bigger plates store more charge at the same voltage. So three 100 µF caps in parallel give 300 µF total. In series, they behave like resistors in parallel and the reciprocals add. Easy way to remember: in parallel, capacitance grows; in series, it shrinks.