0.0072
J
7.2
mJ
0.0012
C
1,200
uC
0.000333
mAh
0.000002
Wh
0.0072
J
7.2
mJ
0.0012
C
1,200
uC
0.000333
mAh
0.000002
Wh
The Capacitor Energy Calculator computes the energy stored in a capacitor, the charge on its plates, and verifies the result using all three equivalent energy formulas. When a capacitor is charged to voltage V, it stores energy in the electric field between its plates. The primary formula is $$E = \frac{1}{2}CV^2$$, but equivalent expressions include $$E = \frac{Q^2}{2C}$$ and $$E = \frac{1}{2}QV$$, where Q = CV is the stored charge. The factor of 1/2 arises because the voltage across the capacitor builds up linearly from 0 to V during charging — the average voltage is V/2. Capacitor energy storage is critical in camera flash units, defibrillators, power supplies, pulsed lasers, electric vehicles, and grid-scale energy storage. This calculator provides both joule and millijoule outputs along with charge in coulombs and microcoulombs for practical convenience.
The calculator uses these fundamental capacitor equations:
$$E = \frac{1}{2}CV^2$$
where:
The charge stored on the plates is:
$$Q = CV$$
The energy can also be computed from the charge:
$$E = \frac{Q^2}{2C} = \frac{1}{2}QV$$
All three energy formulas give identical results — they are algebraically equivalent via Q = CV. The energy is physically stored in the electric field between the plates. For a parallel plate capacitor with uniform field E = V/d, the energy density (energy per unit volume) is:
$$u = \frac{1}{2}\varepsilon_0 E^2$$
Integrating this over the volume between the plates (A × d) recovers E = ½CV². The quadratic dependence on voltage means doubling the voltage quadruples the stored energy, which has important safety implications for high-voltage capacitors.
The energy output tells you how many joules are stored in the capacitor's electric field at the specified voltage. The millijoule conversion is convenient for smaller capacitors used in electronics. The stored charge shows the total charge separated between the plates — equal and opposite charges reside on each plate. The energy from Q²/(2C) output serves as a verification, confirming consistency between the two calculation methods. Note that energy scales with the square of voltage: a capacitor charged to 100 V stores 100 times more energy than the same capacitor at 10 V. This is why high-voltage capacitors can be extremely dangerous even with small capacitance values.
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A typical camera flash capacitor (330 μF at 300 V) stores about 14.85 J of energy and holds 99 mC of charge. This energy is released in a few milliseconds to produce the bright flash, delivering several kilowatts of instantaneous power.
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A 100 nF ceramic capacitor at 5 V stores only 1.25 μJ — a tiny amount of energy, typical for decoupling capacitors in digital circuits. The stored charge is 0.5 μC. Despite the small energy, these capacitors are essential for filtering high-frequency noise in electronic circuits.
The factor of ½ arises because the capacitor charges gradually. Initially, with no charge on the plates, the voltage is zero and transferring charge requires no work. As charge accumulates, the voltage rises, requiring progressively more work to add each additional increment of charge. Mathematically, E = ∫₀ᑫ (q/C)dq = Q²/(2C) = ½CV². The average voltage during charging is V/2, giving E = (average voltage) × Q = ½QV.
A charged capacitor can deliver its stored energy almost instantaneously, producing lethal currents. Capacitors storing more than about 1 J at voltages above 50 V are considered potentially dangerous. Camera flash capacitors (≈15 J at 300 V) can cause painful shocks. Industrial capacitor banks storing kilojoules can be lethal. Always discharge capacitors safely through a resistor before handling them. Even after disconnection, capacitors can retain charge for hours or days.
Conventional capacitors have much lower energy density (0.01–0.1 Wh/kg) compared to lithium-ion batteries (150–250 Wh/kg). However, capacitors excel in power density — they can charge and discharge thousands of times faster than batteries. Supercapacitors bridge the gap with 5–10 Wh/kg but retain high power density, making them ideal for regenerative braking and burst-power applications.
Energy scales with the square of voltage (E = ½CV²), so doubling the voltage quadruples the stored energy. This is why high-voltage capacitors store disproportionately more energy. However, each capacitor has a maximum voltage rating determined by its dielectric breakdown strength. Exceeding this rating causes permanent damage or catastrophic failure. Always select capacitors with voltage ratings well above the maximum expected circuit voltage.
Energy (joules) is the total amount stored; power (watts) is the rate of energy delivery. A capacitor storing 10 J that discharges in 1 ms delivers 10,000 W (10 kW) of instantaneous power. The same energy released over 1 second is only 10 W. Capacitors excel at high-power, short-duration applications precisely because they can release energy extremely rapidly with very low internal resistance.
Medical defibrillators charge a capacitor (typically 32–200 μF) to 1000–5000 V, storing 50–360 J of energy. When triggered, this energy is delivered to the patient's heart in about 4–20 ms through electrode paddles, producing a current pulse that depolarizes the heart muscle and allows normal rhythm to resume. The capacitor's ability to store energy slowly and release it as a brief, high-power pulse is perfectly suited for this life-saving application.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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