Energy Stored in a Capacitor

Insert capacitance and charging voltage into E = ½·C·U². Example: a flash capacitor with C = 1,000 µF = 10⁻³ F charged to U = 300 V stores E = ½ × 10⁻³ × 90,000 = 45 J, enough for a powerful flash of light. The voltage enters squared and is the most effective lever: double the voltage, four times the energy. If the charge is given instead of the voltage, use the equivalent forms E = ½·Q·U or E = Q²/(2C). Mind the unit prefixes: convert microfarads to farads first (µ = 10⁻⁶), otherwise the result is off by orders of magnitude.

From the charging process: the voltage on the capacitor is not fully there from the start but grows with the charge from 0 to U. The first portion of charge is moved almost without opposing voltage, the last against the full voltage. On average only U/2 acts, so the charging work is W = Q·(U/2) = ½QU = ½CU². Graphically this is the triangular area under the Q-U line. The contrast makes it clear: pushing the charge Q through a fixed voltage U (say from a battery) takes energy Q·U, without the factor ½. Leaving it out wrongly doubles the capacitor energy, the standard exam mistake.

Because the energy grows quadratically with the voltage: E = ½CU². To store a lot of energy in a compact component it pays more to raise the voltage than to enlarge the capacitance; ten times the voltage brings a hundred times the energy. A defibrillator typically charges C = 150 µF to about 2,000 V: E = ½ × 1.5×10⁻⁴ × 4×10⁶ = 300 J, delivered within a few milliseconds. This shows the capacitor second strength: it can release its energy extremely fast, briefly reaching enormous power (here ~100 kW), which no battery manages. That is why capacitors serve wherever short, strong bursts of energy are needed.

In the electric field between the plates, not "in the charges" themselves. The modern view assigns the field an energy density: w = ½·ε₀·E² per unit volume (E being the field strength here). Integrated over the field volume of a parallel-plate capacitor this gives exactly ½CU²; both descriptions are equivalent. This field picture is more than formalism: it explains why pulling the plates apart requires energy input (the field volume grows) and extends all the way to electromagnetic waves, which transport field energy through empty space. In final exams this appears as a concept question: the carrier of the energy is the field, the capacitor merely its container.

Charging a capacitor from a fixed voltage source U through a resistor, the source delivers in total W = Q·U = C·U². Yet only E = ½CU² arrives in the capacitor; the other half inevitably becomes heat in the resistor, no matter how large R is. The reason: at the start almost the entire source voltage sits across the resistor, only at the end across the capacitor; integrated over the whole process the two share the energy exactly half and half. A smaller resistance only shortens the charging time (time constant τ = R·C) but does not change the 50% balance. In practice the loss is avoided with switched charging circuits or inductors, a popular advanced question.