How this instrument works
That factor of one half is not a fudge, it is an average. Charge does not arrive all at once at final voltage: a first coulomb slides on while plates sit near zero and costs almost nothing, while every coulomb after it must be shoved against whatever voltage its predecessors already built. Add up q/C dq from empty to full and ½CV² falls out. What you hold is exactly half of what moving that same charge at final voltage would have cost — and it lives in an electric field inside the dielectric, not on metal.
Joseph Henry noticed in 1842 that a Leyden jar discharge magnetised steel needles in either direction, hinting that stored energy sloshed rather than simply drained away. William Thomson put numbers to that behaviour in 1853, deriving when a discharge rings and when it merely decays. Since voltage is squared, it became designers' favourite lever. A 1000 µF electrolytic at 100 V holds 5 J — the pop and recycle whine of a camera flash. Same part at 200 V holds 20 J: quadruple energy for double volts. Defibrillators deliver on order of 150 to 360 J, and pulsed-power banks at fusion laboratories are rated in megajoules.
Two caveats outrank arithmetic here. First, ½CV² counts field energy, not spendable energy: a converter that quits below 0.8 V strands everything under that line, and draining from V only down to V/2 releases three quarters of your total, since energy tracks a square. Second, C must genuinely stay constant. Big film and electrolytic parts show dielectric absorption — short one, walk away, return to find a live terminal as slow polarisation hands charge back. Service manuals earn their bleeder-resistor warnings honestly.
- Type your part's value into Capacitance. Unit menu runs from farads down to picofarads, so a 470 µF electrolytic needs no arithmetic from you.
- Put your working figure into Voltage across it — volts, millivolts or kilovolts. Use what that part actually sits at, never its printed rating.
- Read Stored energy in joules, or switch to watt-hours when comparing against a battery datasheet.
- Rerun at your circuit's cut-off voltage and subtract. That difference, not the first number, is what you can genuinely draw back out.
Worked example — dawn charge on a 2 F supercapacitor
A solar sensor node among vineyard rows survives each night on its 2 F supercapacitor. At first light the panel has nudged that part up to just 1 volt, and firmware wants to know whether such a trickle is worth waking for. E = ½ × 2 × 1² = 1 joule, exactly — nothing rounded anywhere, because 2 and 1 are kind numbers. Against the node's 3 mW average draw, one joule buys a little over five minutes of runtime, provided your boost converter still starts down there.
Let sunshine finish that job and a square law pays out. At its full 2.7 V rating the same part holds ½ × 2 × 2.7² = 7.29 joules — seven times a dawn figure for under three times as much voltage. Put differently: climbing from 0 to 1 V banked 1 joule, while a single further volt, 1 V to 2 V, banked three more. On any capacitor, a last volt is always worth most.
Questions
Why is there a factor of one half?
Because voltage climbs as you fill it. Work done pushing charge dq onto a capacitor already at q/C is (q/C)dq; integrate from empty to Q and you land on Q²/2C, which is ½CV². Final charge cost full voltage, first charge cost nothing, average sits halfway. That also explains a famous result: charge a capacitor through any resistor from a fixed supply and your resistor dissipates exactly ½CV² as heat — precisely as much as ends up stored — regardless of how large or small that resistance happens to be.
How much energy can I actually get back out?
Less than ½CV², and how much less depends on your cut-off voltage. Energy scales with V², so draining from full V down to half V releases 75% of your total; down to a quarter releases about 94%. Whatever sits below a threshold your regulator can still start from is stranded. Real parts also surrender a slice to equivalent series resistance during discharge, which shows up as warmth rather than useful work.
Why do capacitors hold so much less than batteries?
Because capacitors separate charge across physical gaps while batteries store it in chemical bonds, and bonds pack far denser. Good supercapacitors manage roughly 5 to 10 Wh/kg; lithium-ion cells reach 150 to 250 Wh/kg, twenty times better. Capacitors win on power rather than energy: they surrender their joules in milliseconds and tolerate millions of cycles, where batteries would overheat or wear out. That trade decides which belongs in the camera flash and which belongs in the camera.
Should I enter working voltage or rated voltage?
Working voltage — what your part actually sits at in circuit. A 450 V marking on an electrolytic that spends its life at 320 V is safety margin, not a storage figure, and feeding it here inflates your answer by a factor of two. Where voltage swings, as with rectifier ripple, enter peak for worst-case stored energy and trough for what survives a discharge, then subtract one from the other.
Is a charged capacitor dangerous?
It can be, and a joule figure is your honest way to judge. Tens of joules through a chest lands in defibrillator territory, which is why microwave-oven and photoflash parts kill people who assume unplugged means safe. Large electrolytics also recover voltage on their own minutes after a brief short, as dielectric absorption releases charge still relaxing inside. Discharge deliberately through a resistor, measure, then short across those terminals and leave that short in place while you work.
How does this compare with energy stored in an inductor?
They mirror each other. A capacitor holds ½CV² in an electric field and resists sudden voltage changes; an inductor holds ½LI² in a magnetic field and resists sudden current changes. Wire them together and energy trades back and forth at 1/√(LC) radians per second — that LC oscillation Thomson analysed in 1853. Within a lossless resonant tank, total energy stays put at every instant while each half of it breathes.