Energy Stored in a Capacitor

A defibrillator does not plug the patient into the wall. It spends a few seconds sipping energy from a battery into a capacitor, then returns the whole investment through the chest in about four milliseconds: 200 joules at fifty thousand watts, from a component you can hold in one hand. A camera flash plays the same trick at pocket scale — charge slowly, dump instantly. Store slow, spend fast: that is the entire career of the capacitor as an energy device, and this module is its account book. The account book has one famous entry that trips students every year: charge a capacitor to voltage $V$ with charge $Q$ , and the energy stored is $\tfrac12 QV$ — half of the $QV$ you might expect. That half is not a convention and not rounding; it is a triangle. The first bit of charge moves across at zero voltage (free), the last bit climbs the full $V$ (expensive), and the average ride costs $V/2$ . Draw charge against voltage and the stored energy is literally the area of a triangle. You will derive that in two lines of algebra, no calculus required. The second payoff is stranger and deeper: the energy is not "on the plates" at all. It lives in the electric field filling the gap, at a density you can compute — a charged capacitor is a bottle of compressed field. For CBSE and ISC this is a named derivation; for JEE and NEET it feeds a whole family of numericals; for AP Physics 2 and A-Level it is the standard energy-storage item. And it sets up the two modules that follow: dielectrics re-price the stored field, and the E2 gym's charge-sharing problems ask where energy goes when two capacitors argue about it.

Key takeaways

Charging a capacitor costs work because each new charge climbs the voltage built by its predecessors. Total stored: $U = \tfrac12 QV = \tfrac12 CV^2 = \dfrac{Q^2}{2C}$ .
The factor $\tfrac12$ is the area of the triangle under the $Q$ – $V$ line — the average charge rode at half the final voltage.
The three forms are the same number; choose by what the problem holds fixed: $\tfrac12 CV^2$ when a battery pins $V$ , $Q^2/2C$ when a disconnected capacitor freezes $Q$ .
The energy lives in the field between the plates, with energy density $u = \tfrac12 \varepsilon_0 E^2$ (joules per cubic metre of field).
A battery charging a capacitor pays $QV$ but the capacitor banks only $\tfrac12 QV$ — the other half is lost to heat, whatever the wiring.
Key idea: U = ½CV² is the field's inventory — energy grows with the SQUARE of the voltage, which is why high-voltage capacitors are the ones that bite.

The core insight

The energy stored in a capacitor is $U = \tfrac12 QV = \tfrac12 CV^2 = Q^2/2C$ — the work of shovelling charge against a linearly growing voltage — and it physically resides in the electric field between the plates at density $\tfrac12\varepsilon_0 E^2$ .

Where this lives today

The store-slow-spend-fast trade runs modern hardware at every scale. Your phone's camera flash and the strobe of a photography studio are $\tfrac12 CV^2$ devices; so is the defibrillator on every hospital wall (100 μF at 2 kV → 200 J, delivered at 50 kW). Electric buses in several cities brake into supercapacitor banks — megajoules absorbed in seconds, returned at the next green light — because capacitors tolerate millions of such cycles where batteries would wear out. At the grid scale, capacitor banks catch millisecond power wobbles that no chemical battery can respond to in time. The field-energy picture earns its keep at the frontier. A single supercapacitor cell stores kilojoules by pushing the $u = \tfrac12\kappa\varepsilon_0 E^2$ inventory to its material limits — nanometre gaps running fields near breakdown. And the comparison every engineer carries: supercapacitors reach only a few watt-hours per kilogram against lithium-ion's ~250, because field energy is capped by how much $E$ the gap material survives — but they deliver their inventory hundreds of times faster. Battery for energy, capacitor for power: the defibrillator's division of labour, industrialised.

The physical picture

Charging a capacitor is shovelling charge uphill — onto a hill that grows as you shovel. When the plates are empty, moving the first speck of charge across costs nothing: there is no field yet to fight. By the time the capacitor holds charge $q$ , its plates sit at voltage $v = q/C$ (that is just the definition $C = Q/V$ from Capacitance & the Parallel-Plate Capacitor, rearranged), and hauling the next speck across costs that much energy per coulomb — the meaning of potential difference you locked in at Electric Potential & Potential Difference. The last speck pays the full final voltage $V$ . First speck free, last speck expensive, and because the price rises in a straight line, the average price is exactly $V/2$ .

Figure

The charging triangle — each bit of charge pays the voltage built before it

Where does the paid-for energy sit while the capacitor waits? Not "on the plates" — the plates are just parking for charge. It sits in the space between them, in the electric field itself. This is not philosophy: the field between the plates is a real, stressed condition of space, and you can compute how many joules each cubic metre of it holds. Squeeze more field into the gap and you have stored more energy; let the field collapse (discharge) and the energy comes back out through the circuit. A charged capacitor is a bottle of compressed field — and thinking of it that way is what makes the next module, Dielectrics & Polarisation, feel inevitable rather than arbitrary.

The energy formula — three costumes, one number

Start from the triangle. Moving total charge $Q$ at an average voltage of $V/2$ costs

U = \tfrac12\,QV \qquad (1)

where: $Q$ is the final charge, $V$ the final voltage, and $U$ arrives in joules when $Q$ is in coulombs and $V$ in volts. Now use $Q = CV$ to dress the same result two more ways:

U = \tfrac12\,CV^2 = \frac{Q^2}{2C} \qquad (2)

where: all three expressions are numerically identical for the same capacitor in the same state — but they behave differently when something changes, and choosing the right costume is most of the skill. $\tfrac12 CV^2$ says: at fixed voltage (battery attached), more capacitance means more energy. $Q^2/2C$ says: at fixed charge (battery disconnected), more capacitance means less energy. Both are true; they describe different experiments. The quadratic dependence on $V$ is the headline: double the voltage, quadruple the energy — which is why a 2 kV defibrillator capacitor stores ten thousand times more than the same capacitance at 20 V, and why high-voltage capacitors deserve fear.

you know	use	typical situation
$Q$ and $V$	$U = \tfrac12 QV$	quick audits, mixed data
$C$ and $V$	$U = \tfrac12 CV^2$	battery attached — $V$ pinned
$Q$ and $C$	$U = Q^2/2C$	battery disconnected — $Q$ frozen

Derivation AEnergy Stored in a Capacitor from the Charging Triangle

We want the work required to charge a capacitor of capacitance $C$ from empty to final charge $Q$ and voltage $V = Q/C$ — using only algebra and one graph. The strategy: track the voltage as the charge builds, recognise that the price of each increment is the voltage at that moment, and read the total work as the area under the voltage-versus-charge line. The thing to watch is why an average is legal here: only because the price line is perfectly straight.

Step 1. At an intermediate moment the capacitor holds charge $q$ . By the definition of capacitance, its plates are at voltage $v = q/C$ — the price, in joules per coulomb, of the next increment of charge.

Step 2. Plot $v$ (vertical) against $q$ (horizontal) as charging proceeds from $q = 0$ to $q = Q$ . The graph is a straight line from the origin with slope $1/C$ , ending at the point $(Q, V)$ . Why a straight line matters: work is price × amount summed over the whole journey — the area under the price line. For a straight line that area is a triangle, and no calculus is needed to find it.

Step 3. Area of the triangle: base $Q$ , height $V$ ,

U = \tfrac12\,\text{base} \times \text{height} = \tfrac12\,QV

Equivalently: the price climbed steadily from 0 to $V$ , so the average price was $V/2$ , and $Q$ coulombs at $V/2$ joules per coulomb is the same $\tfrac12 QV$ .

Step 4. Substitute $Q = CV$ once each way to obtain the working forms $U = \tfrac12 CV^2$ and $U = Q^2/2C$ . Why the battery's ledger differs: a battery holds its terminal voltage flat while delivering charge, so its outlay is the full rectangle QV. The capacitor banks the triangle — half. The other half is dissipated in the connecting circuit, no matter how good the wires; the pitfall box takes this up. *Verified: C = 470 μF charged to V = 12 V: Q = CV = 5.64 mC; triangle U = ½ × 5.64×10⁻³ × 12 = 3.38×10⁻² J; costume check: ½CV² = ½ × 470×10⁻⁶ × 144 = 3.38×10⁻² J ✓ — the camera-flash numbers of the first worked example.*

The energy lives in the field

For a parallel-plate capacitor the claim "energy is stored in the field" can be checked exactly, because we know everything about the gap: field $E = V/d$ , volume $A\,d$ . Take $U = \tfrac12 CV^2$ , substitute $C = \varepsilon_0 A/d$ and $V = Ed$ :

U = \tfrac12\,\frac{\varepsilon_0 A}{d}\,(Ed)^2 = \tfrac12\,\varepsilon_0 E^2 \times (A\,d) \qquad (3)

where: $A\,d$ is precisely the volume of the gap — the region the field fills. The energy is proportional to the field volume, with a fixed number of joules per cubic metre:

u = \frac{U}{\text{volume}} = \tfrac12\,\varepsilon_0 E^2 \qquad (4)

where: $u$ is the energy density of the electric field, in $\mathrm{J/m^3}$ . Read it physically: the energy tracks $E^2$ , so doubling the field quadruples the stored energy per unit volume — the same square law as equation (2), now wearing field clothing. At a typical lab field of $10^5\ \mathrm{V/m}$ , the density is $u = \tfrac12(8.85\times10^{-12})(10^5)^2 \approx 0.044\ \mathrm{J/m^3}$ — modest, which is why capacitors that store serious energy either run huge fields or huge... something. That "something" is the dielectric boost, next module. Equation (4) was derived here for the uniform plate field, but its jurisdiction is universal: any electric field, anywhere, carries this energy density — a result that will matter again when electromagnetic waves carry field energy from the Sun to your skin in chapter E8.

Figure

A charged capacitor is a bottle of compressed field

Limiting cases

case	result	meaning
$V \to 2V$ (same $C$ )	$U \to 4U$	energy is quadratic in voltage — the cheap lever and the danger
$Q \to 2Q$ (same $C$ )	$U \to 4U$	quadratic in charge too: $Q^2/2C$
battery attached, $C$ doubled	$U = \tfrac12 CV^2$ doubles	at pinned $V$ , bigger buckets bank more
battery removed, $C$ doubled	$U = Q^2/2C$ halves	at frozen $Q$ , bigger buckets relax — the dielectric module cashes this
plates pulled to $2d$ at fixed $Q$	$U$ doubles	you did mechanical work against the plates' attraction; field volume doubled at the same density
$E \to 0$	$u \to 0$	no field, no energy — the inventory is the field itself

Pitfall

Common trapThe capacitor stores all the energy the battery supplied

It stores exactly half, and no improvement in wiring can fix that. Charging a capacitor to charge $Q$ at pinned voltage $V$ , the battery's outlay is $QV$ (flat price, full rectangle); the capacitor's inventory is $\tfrac12 QV$ (rising price, triangle). The missing $\tfrac12 QV$ is dissipated in the resistance of the circuit — and if you make the resistance tiny, the loss does not shrink: the current spike grows instead, and the same half is lost faster (some of it as a radio crackle you can actually hear on an AM radio). Writing $U = QV$ for a capacitor is therefore always wrong, and "energy supplied by the battery" vs "energy stored in the capacitor" is a deliberately paired question on CBSE and JEE papers. The only way to beat the tax is to not hold the voltage fixed — charging through a gradually rising source (or via an inductor) can approach zero loss, a trick real power circuits use. At this level: battery pays $QV$ , capacitor keeps $\tfrac12 QV$ , wires get the rest.

Worked examples

Board level — the camera flash (CBSE/ISC pattern)

A 470 μF capacitor in a camera flash is charged to 12 V. (a) How much energy does it store? (b) If the flash discharges this energy in 2.0 ms, what is the average power delivered? (c) Compare that power with the 0.5 W the camera's battery can supply.

Step 1. Choose the costume: $C$ and $V$ are given, so $U = \tfrac12 CV^2$ .

Step 2. $U = \tfrac12 (470\times10^{-6})(12)^2 = \tfrac12 (470\times10^{-6})(144) = 3.38\times10^{-2}\ \mathrm{J}$ . $\boxed{U \approx 34\ \mathrm{mJ}}$ .

Step 3. Average power: $P = U/t = 3.38\times10^{-2}/(2.0\times10^{-3}) = \boxed{16.9\ \mathrm{W}}$ .

Step 4. The comparison that makes the physics visible: 16.9 W is about 34 times what the battery can deliver directly. The capacitor is not a source of extra energy — it is a power transformer in time: it accepted energy slowly (the battery takes well over a second to fill it through the charging circuit) and released it fast. Same joules, compressed into a shorter time, hence the flash.

Step 5. Mark-scheme note: quote the answer in mJ or J with two significant figures, and state the assumption "all stored energy is delivered to the flash tube" — the examiner's scheme includes it.

Competitive level — same parts, series vs parallel (JEE-Main/NEET pattern)

Capacitors of 2 μF and 4 μF are connected across a 12 V battery, first in parallel, then in series. Find the total energy stored in each case, and the energy in each capacitor for the series case.

Step 1. Parallel: $C_{\mathrm{eq}} = 2 + 4 = 6\ \mu\mathrm{F}$ (the sum rule from Capacitors in Series & Parallel). $U_{\parallel} = \tfrac12 (6\times10^{-6})(144) = \boxed{432\ \mu\mathrm{J}}$ .

Step 2. Series: $C_{\mathrm{eq}} = \dfrac{2\times4}{2+4} = \dfrac{4}{3}\ \mu\mathrm{F}$ . $U_{\mathrm{series}} = \tfrac12 \left(\tfrac43\times10^{-6}\right)(144) = \boxed{96\ \mu\mathrm{J}}$ .

Step 3. The ratio: $U_{\parallel}/U_{\mathrm{series}} = 6/(4/3) = 4.5$ . Same two components, same battery — 4.5 times the energy, purely from wiring. At fixed $V$ , energy follows $C_{\mathrm{eq}}$ , and parallel maximises $C_{\mathrm{eq}}$ .

Step 4. Energy split in the series case. Series members share charge: $Q = C_{\mathrm{eq}}V = (4/3)(12) = 16\ \mu\mathrm{C}$ on each. Use the frozen- $Q$ costume per capacitor: $U_2 = Q^2/2C = (16\times10^{-6})^2/(2\times2\times10^{-6}) = 64\ \mu\mathrm{J}$ and $U_4 = (16\times10^{-6})^2/(2\times4\times10^{-6}) = 32\ \mu\mathrm{J}$ . Check: $64 + 32 = 96\ \mu\mathrm{J}$ ✓.

Step 5. Read the moral off the numbers: in series, the smaller capacitor stores the larger energy (it also carries the larger voltage: $V_2 = 16/2 = 8$ V vs $V_4 = 16/4 = 4$ V). In a mismatched series stack the runt is the most stressed member on every axis — voltage, energy, failure risk. Timing note: 3 minutes. The only decisions are (i) $C_{\mathrm{eq}}$ per configuration, (ii) costume per sub-question — $\tfrac12 CV^2$ while the battery pins $V$ , then $Q^2/2C$ inside the series chain where $Q$ is the shared quantity. If you find yourself hunting for each capacitor's individual voltage first, you are doing it the slow way.

Exam traps

Writing $U = QV$ . The factor $\tfrac12$ is the physics — the triangle against the rectangle. Boards set "energy stored" vs "energy supplied by the battery" side by side precisely to catch the missing half.

Wrong costume for the constraint. Battery attached → $V$ fixed → use $\tfrac12 CV^2$ ; battery disconnected → $Q$ frozen → use $Q^2/2C$ . Applying $\tfrac12 CV^2$ with the old voltage after the geometry changed is the classic follow-up-question killer. Decide the constraint before touching the algebra.

Prefix arithmetic. $\tfrac12 CV^2$ with $C$ in μF and $V$ in volts gives μJ directly. Mixing μF with kV: $(10^{-6})(10^3)^2 = 1$ , so μF × kV² lands in joules — the defibrillator check: $\tfrac12(100\ \mu\mathrm{F})(2\ \mathrm{kV})^2 = \tfrac12 \times 100\times10^{-6} \times 4\times10^6 = 200\ \mathrm{J}$ . Practise the prefix chains; exponent slips here cost more marks than physics errors.

Squaring the wrong thing. $\tfrac12 CV^2$ : only $V$ is squared. Under pressure, students square $C$ or forget to square at all. Anchor: doubling $V$ must quadruple $U$ — if your formula does not do that, it is mis-copied.

Series energy ranking. "Which capacitor stores more energy in series?" — the smaller one ( $U = Q^2/2C$ at shared $Q$ ). In parallel it flips: the larger one ( $U = \tfrac12 CV^2$ at shared $V$ ). Memorising one case and applying it to both is the planted MCQ error.

Constants per board. CBSE/ISC/NEET: $\varepsilon_0 = 8.85\times10^{-12}$ supplied; energy-density questions expect (4) quoted. JEE-Main: clean numbers, usually μJ or mJ answers. AP-2 and A-Level: formula sheets carry $U_C = \tfrac12 CV^2$ and (for AP) $u = \tfrac12\varepsilon_0 E^2$ — know which sheet you get.

Origins · 1 minHermann von Helmholtz and how this was discovered

Hermann von Helmholtz

1821–1894 · Berlin

The auditor-general of energy. In 1847, aged twenty-six, army surgeon Helmholtz read a paper to the Berlin Physical Society titled "On the Conservation of Force" — the first fully general statement that energy is a single currency convertible between mechanical work, heat, chemistry, and electricity, but never created or destroyed. He worked out the electrical entries of the ledger explicitly: the energy of charged conductors ( $\tfrac12 QV$ — the very triangle of this module), what a battery's chemistry pays, what heat the circuit collects. The style of accounting you used in the pitfall box — "the battery paid $QV$ ; the capacitor shows $\tfrac12 QV$ ; find the missing half before signing off" — is Helmholtz's discipline, applied by every physicist and engineer since. He went on to measure the speed of nerve signals, invent the ophthalmoscope, and train a generation that included Hertz, who used capacitor discharges to make the first radio waves — stored field energy, spent as electromagnetic radiation.

History

1746

— The Leyden jar delivers the first stored-energy electric shocks; nobody yet has a word for the energy involved

1847

— Helmholtz's conservation-of-energy memoir puts charged conductors on the universal energy ledger with the ½QV entry

1853

— William Thomson (Lord Kelvin) analyses the Leyden jar discharge and finds the stored energy oscillating back and forth — capacitor energy has dynamics

1905

— Einstein's light-quantum paper begins with the energy density of the electromagnetic field — the ½ε₀E² picture taken utterly seriously

1957

— The double-layer supercapacitor patent starts the race to farad-scale energy storage

Going further

Texts. NCERT Physics Part 1 (Class 12), Ch. 2 §2.15 — the board derivation via the charging work · H.C. Verma, Concepts of Physics 2, Ch. 31 · Halliday, Resnick & Walker, Ch. 25 §25-4 · For the full ledger — two-capacitor paradox solved, dielectric energy accounting, field-vs-circuit views: Capacitor Energy and Dielectrics (UG deep dive).

Provenance. All numbers recomputed: flash $\tfrac12(470\ \mu\mathrm{F})(12)^2 = 33.8$ mJ, $P = 16.9$ W over 2 ms; defibrillator $\tfrac12(100\ \mu\mathrm{F})(2000)^2 = 200$ J, 50 kW over 4 ms, $Q = 0.20$ C; parallel 6 μF → 432 μJ vs series 4/3 μF → 96 μJ (ratio 4.5), series split 64/32 μJ at 16 μC; $u = 0.044\ \mathrm{J/m^3}$ at $10^5$ V/m; plate audit $\tfrac12 CV^2 = \tfrac12\varepsilon_0E^2(Ad) = 4.43\ \mu\mathrm{J}$ for $A = 0.1\ \mathrm{m^2}$ , $d = 1$ mm, $V = 100$ V.

Prerequisites. Capacitance & the Parallel-Plate Capacitor · Capacitors in Series & Parallel. Leads to. Dielectrics & Polarisation · Charge Sharing and the Lost Energy (coming in the E2 gym) · Capacitor Energy and Dielectrics (UG deep dive).

Why is the stored energy ½QV and not QV? Every coulomb crossed a potential difference of V, didn't it?

No — only the last one did. The potential difference was built by the charging itself: the first coulomb crossed when the capacitor was empty and the voltage was zero, the last crossed the full $V$ , and each intermediate coulomb paid the voltage that existed at its moment. Since the price rose in a straight line (that is exactly what $C = Q/V$ constant means), the average fare was $V/2$ , and $Q$ coulombs at average fare $V/2$ costs $\tfrac12 QV$ . A battery is different: its voltage sits flat at $V$ while charge flows through it, so its energy transactions genuinely are $QV$ . Same journey, different pricing schemes — that mismatch is the missing-half story.

Where physically is the energy? On the plates, in the charges, or in the field?

In the field, and the gap volume is the receipt. Equation (3) showed the stored energy equals $\tfrac12\varepsilon_0 E^2$ times exactly the volume the field occupies — not the plate volume, not the wire volume. At this level you could call that a bookkeeping choice, but physics settles the question decisively later: electromagnetic waves carry energy across empty space with no charges anywhere nearby (sunlight is field energy in transit), so the field must be a genuine energy container. The capacitor is simply the cleanest place to first count that energy, because its field is uniform and its volume is known.

If I pull the plates of a charged, disconnected capacitor apart, the energy increases. Where does it come from?

From your hands. Opposite plates attract, so separating them requires real mechanical work, and that work is banked as extra field energy: at frozen $Q$ the field $E = \sigma/\varepsilon_0$ stays the same but now fills twice the volume, so $U = Q^2/2C$ doubles when $d$ doubles ( $C$ halves). Run the same move with the battery attached and the story inverts: $V$ is pinned, charge drains back to the battery, the field thins, and the stored energy drops — the battery, not your hands, closes the books. One geometry change, two opposite energy verdicts, decided entirely by what is held fixed: this pairing is a JEE-Advanced favourite.

Is the energy density formula only for capacitors?

It was derived here inside a capacitor because the uniform field makes the audit exact, but $u = \tfrac12\varepsilon_0 E^2$ is a property of the electric field itself, valid point by point in any electrostatic field — around a point charge, near a charged sphere, anywhere. Wherever the field is stronger, space holds more energy per cubic metre, always as $E^2$ . With a dielectric present it upgrades to $\tfrac12\kappa\varepsilon_0 E^2$ (next module), and its magnetic sibling $B^2/2\mu_0$ arrives in chapter E6 — together they price the energy of light.

A capacitor and a battery can both light an LED. What is actually different about them as energy stores?

Mechanism and consequences. A battery stores energy in chemical bonds — enormous density (a AAA cell holds ~5000 J) but slow conversion, limited charge-discharge cycles, and a voltage that chemistry pins nearly constant. A capacitor stores energy as field — tiny density by comparison (a 1 F supercapacitor at 2.7 V holds ~3.6 J), but it can absorb or release that inventory in milliseconds, survives millions of cycles, and its voltage falls as it drains ( $V = Q/C$ — the "sagging" that circuit designers must manage). That is why the two are partners, not rivals: battery for endurance, capacitor for bursts.

What happens to the energy when a charged capacitor is connected to an uncharged one?

Some of it vanishes — genuinely. Charge is conserved when the two share out, but recompute $\sum Q^2/2C$ afterwards and the total is less than before; the deficit went to heat and a little radio-frequency radiation during the rush of redistribution, and (strangest of all) the lost fraction does not depend on the resistance of the connecting wire. That paradox — where the energy goes, why the wire's quality cannot save it, and the exam patterns built on it — is precisely the business of Charge Sharing and the Lost Energy, coming in the E2 gym. For now, bank the warning: charge conservation and energy conservation are separate audits, and capacitor problems must pass both.