Capacitor Energy and Dielectrics

A defibrillator does not plug the patient into the wall. It spends five seconds sipping charge from a battery into a capacitor, then returns that investment through the chest in four milliseconds — two hundred joules delivered at fifty thousand watts, from a component you can hold in one hand. That trick — store slowly, spend instantly — is the entire reason capacitors exist as energy devices, and this article is the ledger that governs it: how much work it costs to charge a capacitor, where that energy physically sits while you wait, and why some of it can vanish without any resistor in sight. The second half of the story is the material one. Fill the gap with glass, polymer, or ceramic and the capacitance multiplies by the dielectric constant $\kappa$ — the previous two articles took that boost on faith from the dielectrics chapter. Here we make it pay: the same slab that multiplies $C$ divides the stored energy at fixed charge, multiplies it at fixed voltage, and gets physically sucked into the gap while doing either. Sorting out those three claims — and learning why "is the battery attached?" changes every answer — is the discipline this article installs. You arrive owning $C=Q/V$ and the network rules; you leave owning $U=\tfrac12CV^2$ , the field-energy picture behind it, the solved mystery of the two-capacitor paradox, and the fixed- $Q$ /fixed- $V$ bookkeeping that every energy question in circuit theory quietly assumes. That is the full energy toolkit RC transients will spend.

Key takeaways

Charging a capacitor costs work because each new charge climbs a voltage that earlier charges built: $U = \dfrac{Q^2}{2C} = \tfrac12 CV^2 = \tfrac12 QV$ — the factor $\tfrac12$ is the triangle under the $q$ – $V$ line, not a convention.
The energy lives in the field between the plates: $u = \tfrac12\varepsilon_0E^2$ integrated over the gap reproduces $\tfrac12CV^2$ exactly for parallel plates.
Connecting a charged capacitor to an uncharged one conserves charge but destroys a fixed fraction of the energy — independent of the wire's resistance. The paradox is real and solved here.
A dielectric multiplies capacitance: $C = \kappa C_0$ . At fixed $Q$ the stored energy drops to $U_0/\kappa$ ; at fixed $V$ it rises to $\kappa U_0$ — and the battery pays exactly twice the gain.
Either way, the slab is pulled into the gap: systems slide toward higher capacitance, and the fringing field does the pulling.
KEY IDEA: $U=\tfrac12CV^2$ is a field inventory, and every energy question about capacitors reduces to one discipline — decide first whether charge or voltage is held fixed, then let the ledger balance itself.

The core insight

The energy stored in a capacitor, $U = Q^2/2C = \tfrac12CV^2$ , is the work of assembling its charge and physically resides in the electric field of the gap; dielectrics re-price that energy ( $\times\kappa$ at fixed $V$ , $\div\kappa$ at fixed $Q$ ), which is why fields pull dielectrics inward.

Where this lives today

The fixed- $V$ /fixed- $Q$ ledger is daily bread in power electronics. The DC-link capacitor bank in an EV inverter is sized by $\tfrac12CV^2$ arithmetic; regenerative braking dumps into it; and "charge-recycling" logic and gate-driver designs use inductor swings — the superconducting loophole of Derivation B, engineered on purpose — to move charge between capacitances without paying the redistribution tax. At the other extreme of scale, every DRAM refresh cycle rewrites $\sim25\ \mathrm{fF}$ cells at about a volt: femtojoules per bit, times billions of bits, times dozens of refreshes per second — the two-capacitor tax collected continuously across every computer on Earth. The dielectric half of the article is the more strategic frontier. Modern transistor gates replaced silicon dioxide with hafnium-based high-κ dielectrics ( $\kappa \approx 20$ versus 3.9) precisely to keep gate capacitance up while thickening the insulator against tunnelling leakage — a straight application of $C = \kappa\varepsilon_0A/d$ at nanometre $d$ . Meanwhile supercapacitors attack the other lever, area and gap: nanometre double layers over sponge electrodes give kilofarads per cell at 2.7 V. Their energy density (~5 Wh/kg) still trails lithium-ion (~250 Wh/kg) by neat orders of magnitude — $\tfrac12CV^2$ with a capped $V$ is a hard ceiling — but their power density and cycle life are unmatched, which is why buses brake into supercapacitors and phones smooth their power rails with them.

The physical picture

Charging a capacitor is shovelling charge uphill — onto a hill that grows as you shovel. The first sliver of charge $dq$ moves across at essentially zero voltage: free. The last sliver climbs the full final voltage $V$ : expensive. Averaging over the whole transfer, each of the $Q$ coulombs paid, on average, half the final voltage — so the invoice reads $\tfrac12QV$ , not $QV$ . Plot capacitor voltage against charge delivered and the picture is a straight line of slope $1/C$ ; the work is the triangle under that line. Batteries, by contrast, hold their voltage flat while charge flows, so their invoice is the full rectangle $QV$ — and the mismatch between rectangle and triangle is a gap in the accounts that this article will chase down twice.

Figure

The charging triangle: each charge increment pays the voltage built by its predecessors

Where does the paid-for energy sit? Not "on the plates" — the plates are just bookkeeping surfaces. It sits in the gap, in the field itself, at the density $u = \tfrac12\varepsilon_0E^2$ established in Electric Field Energy Density (that article owns the general derivation; we will not re-derive it, we will audit it). The capacitor is the cleanest possible testbed for the field ontology: a known uniform field filling a known volume. Multiply density by volume and, as the first derivation checks to the last digit, you recover $\tfrac12CV^2$ exactly. A charged capacitor is a bottle of compressed field. That picture makes the dielectric story physical instead of algebraic. A dielectric is polarizable matter: inserted into the field, its molecules stretch and align, building surface charge that opposes the plates' field (the mechanism is the dielectrics chapter's property — Linear Dielectrics, Susceptibility and Permittivity owns it). Less net field for the same plate charge means less voltage, which is exactly "more capacitance," and — since energy is field squared — a cheaper bottle at the same charge. Cheaper configurations are downhill: that is why the field pulls the slab in, a force you can feel on a sheet of plastic between charged plates.

Figure

The dielectric ledger: same slab, opposite energy changes, identical inward pull

The energy formulas

One quantity, three costumes. Since $Q = CV$ can eliminate any one variable, the stored energy wears whichever form matches what you know:

U \;=\; \frac{Q^2}{2C} \;=\; \tfrac12\,C V^2 \;=\; \tfrac12\,QV

where: all three are the same number for the same state — but they rank differently under change. $Q^2/2C$ says: at fixed charge, raising $C$ (inserting a slab, closing the gap) lowers the energy. $\tfrac12CV^2$ says: at fixed voltage, raising $C$ raises it. Neither statement is wrong; they answer different experiments, and choosing the wrong costume for the constraint at hand is the single most common error in this subject.

you know	use	typical scenario
$Q$ and $C$	$U = Q^2/2C$	isolated capacitor (battery disconnected)
$V$ and $C$	$U = \tfrac12CV^2$	battery attached, voltage pinned
$Q$ and $V$	$U = \tfrac12QV$	quick audits, mixed data

Derivation ACharging Work and the Field-Energy Audit

We derive the stored energy by direct accounting — integrate the cost of moving each charge increment against the instantaneous voltage — and then audit the result against the field-energy picture: does $\tfrac12\varepsilon_0E^2$ times the gap volume reproduce it? The two routes must agree if "energy in the field" is more than a metaphor, and the parallel plate is the one geometry clean enough to check by hand. Watch the factor of one-half appear from the integral, not from convention.

Step 1. At an intermediate moment the capacitor holds charge $q$ , so its plates sit at voltage $v(q) = q/C$ . Moving the next increment $dq$ from the negative plate to the positive one costs $dW = v\,dq$ — the definition of potential difference doing its job.

Step 2. Sum the increments from empty to full:

U \;=\; \int_0^{Q} \frac{q}{C}\,dq \;=\; \frac{Q^2}{2C}.

Why the half: the integrand grows linearly, so the average cost per coulomb is half the final voltage — the triangle under the $q$ – $v$ line. Substituting $Q=CV$ gives the other two costumes.

Step 3. Now the audit. For plates of area $A$ and gap $d$ : field $E = V/d$ fills the volume $Ad$ (fringing ignored, as usual at the percent level). The field-energy inventory, using the density owned by Electric Field Energy Density:

U_{\text{field}} \;=\; \tfrac12\,\varepsilon_0 E^2 \times (Ad) \;=\; \tfrac12\,\varepsilon_0 \frac{V^2}{d^2}\,Ad \;=\; \tfrac12\,\underbrace{\frac{\varepsilon_0 A}{d}}_{C}\,V^2 .

Step 4. The two ledgers agree identically — no approximation beyond fringing, no adjustable constant. The work you did against the growing voltage is stored, joule for joule, as compressed field in the gap. Why this matters beyond reassurance: the field form generalizes where the circuit form cannot. Between the plates of a charging capacitor there is no charge in transit, yet energy is arriving — flowing in through the sides of the gap, as the Poynting-vector article shows in its famous capacitor example. The triangle integral is the accountant's view; the field inventory is the physicist's. *Verified: $A = 0.1\ \mathrm{m^2}$ , $d = 1\ \mathrm{mm}$ , $V = 100\ \mathrm{V}$ : $C = 0.885\ \mathrm{nF}$ , $\tfrac12CV^2 = 4.43\ \mu\mathrm{J}$ ; field route: $E = 10^5\ \mathrm{V/m}$ , $u = \tfrac12\varepsilon_0E^2 = 4.43\times10^{-2}\ \mathrm{J/m^3}$ , volume $10^{-4}\ \mathrm{m^3}$ , product $4.43\ \mu\mathrm{J}$ — equal to all displayed digits.*

Go deeper · Derivation B · 2 minThe Two-Capacitor Paradox, Solved

Derivation BThe Two-Capacitor Paradox, Solved

The previous article ended on a cliffhanger: capacitor $C_1$ charged to $V_0$ is connected across uncharged $C_2$ ; charge is conserved, yet energy goes missing — with no resistor specified. We now solve the paradox in general: compute the loss, prove its fraction is fixed by the capacitances alone, and explain where the energy physically goes and why the wire's resistance cannot appear in the answer. The strategy is to let charge conservation pin both endpoint states — energy accounting then has no freedom left.

Step 1. Endpoints from conservation. Initially $Q_0 = C_1V_0$ , $U_i = Q_0^2/2C_1$ . Finally the pair is in parallel at a common voltage, sharing the same total charge:

V_f \;=\; \frac{Q_0}{C_1+C_2}, \qquad U_f \;=\; \tfrac12 (C_1+C_2)V_f^2 \;=\; \frac{Q_0^2}{2(C_1+C_2)}.

Step 2. The deficit:

\Delta U \;=\; U_i - U_f \;=\; \frac{Q_0^2}{2}\left(\frac1{C_1} - \frac1{C_1+C_2}\right) \;=\; \frac{C_2}{C_1+C_2}\;U_i

where: the fraction lost, $C_2/(C_1+C_2)$ , contains no resistance, no time, no mechanism — only the two capacitances. Equal capacitors lose exactly half; a large empty $C_2$ swallows nearly everything.

Step 3. Why $R$ cannot matter. Both endpoint states are fixed by charge conservation before any wire is chosen; the deficit is a difference of state functions. Resistance decides how fast and in what form the energy leaves — big $R$ means a slow trickle of gentle heat, small $R$ means a violent spike of heat plus a burst of electromagnetic radiation from the fast current transient — but the total leaving is already decided. (The explicit dissipation integral $\int i^2R\,dt$ , evaluated along the actual exponential transient, reproduces $\Delta U$ exactly and $R$ -independently — RC Circuits performs that integral as its energy audit.) Why can't a superconducting wire ( $R = 0$ exactly) save the energy? Then nothing dissipates — and nothing damps: the charge overshoots and oscillates back and forth forever through the wire's residual inductance. The "final" parallel state is never reached, so the theorem's premise fails rather than its arithmetic. Add back any loss channel, however small, and the oscillation decays to $V_f$ , having shed exactly $\Delta U$ . The LC-oscillations article picks up precisely this loophole — and real charge-recycling circuits exploit it, swinging charge across through an inductor and catching it at peak.

Step 4. The battery version of the same accounting. Charge an empty capacitor directly from a battery at fixed $V$ : the battery delivers charge $Q$ at constant voltage, spending $QV$ ; the capacitor banks $\tfrac12QV$ . Exactly half is gone — same mechanism, same $R$ -independence, the rectangle-minus-triangle from the physical picture. Slow "adiabatic" charging (raising the source voltage gradually so charge never falls through a finite voltage drop) is the engineering escape hatch, and the two-capacitor limit of that trick is the inductor swing above. Verified: $C_1 = 4\ \mu\mathrm{F}$ at $V_0 = 12\ \mathrm{V}$ onto $C_2 = 8\ \mu\mathrm{F}$ (the previous article's numbers): $U_i = 288\ \mu\mathrm{J}$ , $V_f = 4\ \mathrm{V}$ , $U_f = 96\ \mu\mathrm{J}$ ; deficit $192\ \mu\mathrm{J} = \tfrac{8}{12}\times288$ — the formula's $C_2/(C_1+C_2) = 2/3$ on the nose.

Go deeper · Derivation C · 2 minThe Dielectric Ledger: Fixed Q versus Fixed V

Derivation CThe Dielectric Ledger: Fixed Q versus Fixed V

Insert a dielectric slab of constant $\kappa$ so it fills the gap of a parallel-plate capacitor, and account for every joule — twice. Once with the battery disconnected (charge $Q$ frozen), once with it attached (voltage $V$ pinned). The capacitance boost $C = \kappa C_0$ is imported from the dielectrics chapter (Linear Dielectrics owns its microscopic derivation); our job is strictly the energy bookkeeping, and the punchline is that the two protocols give opposite-signed energy changes yet the same inward force on the slab. Keep asking: who paid?

Step 1. Fixed $Q$ (battery disconnected first). With $C \to \kappa C_0$ and $Q$ frozen, use the charge costume:

U_Q \;=\; \frac{Q^2}{2\,\kappa C_0} \;=\; \frac{U_0}{\kappa}.

The stored energy fell. Microscopically: the slab's polarization charge cancels part of the field ( $E = E_0/\kappa$ ), and energy is field squared. The books balance only if the missing $U_0(1-1/\kappa)$ went somewhere — it went into mechanical work done by the field on the slab as it was drawn in (plus kinetic energy, then heat, if you let it slam). The pull is real: you must hold the slab back to insert it slowly.

Step 2. Fixed $V$ (battery attached). Now use the voltage costume: $U_V = \tfrac12\kappa C_0V^2 = \kappa U_0$ — the stored energy rose by $\Delta U_{\text{store}} = (\kappa-1)U_0$ . But the capacitor also drew fresh charge $\Delta Q = (\kappa-1)C_0V$ from the battery, which paid the full rectangle:

W_{\text{batt}} \;=\; V\,\Delta Q \;=\; (\kappa-1)\,C_0V^2 \;=\; 2\,(\kappa-1)\,U_0.

Step 3. Balance the fixed- $V$ books. The battery spent $2(\kappa-1)U_0$ ; storage banked $(\kappa-1)U_0$ ; the remaining $(\kappa-1)U_0$ is again the field's mechanical work on the incoming slab. Half to the vault, half to the pull — the same 50/50 split as battery charging, and for the same reason: charge fell through the pinned voltage.

\underbrace{2(\kappa-1)U_0}_{\text{battery}} \;=\; \underbrace{(\kappa-1)U_0}_{\text{stored gain}} \;+\; \underbrace{(\kappa-1)U_0}_{\text{work on slab}}

where: every term is positive for $\kappa > 1$ : the battery finances both the fatter field and the tug on the slab.

Step 4. The force, from either ledger. Both protocols say the system's available energy decreases as the slab advances, so the generalized force points inward; run the fixed- $Q$ ledger with the slab inserted a variable distance $x$ and $F = -\,dU/dx$ gives the standard inward pull, uniform in $x$ across the flat interior. The mechanism — the non-uniform fringing field at the gap's mouth acting on the slab's induced dipoles, the only place a horizontal force can arise — together with the honest resolution of "energy method versus field method," belongs to Forces on Dielectrics, which owns that derivation. Our ledger only certifies the sign and the magnitude. Why physicists trust the ledger even without the mechanism: energy methods are exactly Helmholtz's audit discipline — if the books balance in both protocols and both give the same force, the microscopic story must comply. When the fixed-V books seem to show energy appearing from nowhere, the forgotten line item is always the battery. *Verified: $C_0 = 200\ \mathrm{pF}$ , $V_0 = 1000\ \mathrm{V}$ , $\kappa = 4.7$ (glass): $U_0 = 100\ \mu\mathrm{J}$ . Fixed $Q$ : $U = 21.3\ \mu\mathrm{J}$ (drop $78.7\ \mu\mathrm{J}$ ). Fixed $V$ : $U = 470\ \mu\mathrm{J}$ , battery pays $740\ \mu\mathrm{J} = 2\times370\ \mu\mathrm{J}$ , stored gain $370\ \mu\mathrm{J}$ , slab work $370\ \mu\mathrm{J}$ — both ledgers close.*

Limiting cases

case	result	meaning
$\kappa \to 1$	all dielectric effects vanish	vacuum is the $\kappa=1$ member of the family
$\kappa \to \infty$ (conductor slab)	fixed $Q$ : $U \to 0$ ; slab fills gap	a conductor is the ultimate dielectric: field expelled, energy fully cashed out as pull
$C_2 \to 0$ (redistribution)	$\Delta U \to 0$	connecting to nothing loses nothing — the loss fraction $C_2/(C_1+C_2)$ interpolates smoothly
$C_2 \to \infty$ (redistribution)	$\Delta U \to U_i$	dumping into a huge empty bank forfeits essentially everything — "grounding through a wire" as the extreme case
slow adiabatic charging	loss $\to 0$	the half-loss is a fixed- $V$ artifact: ramp the source and the rectangle hugs the triangle
gap doubled at fixed $Q$	$U$ doubles	you did work pulling the plates apart against their attraction; field volume doubled at fixed density

Pitfall

Common trapThe battery's work all ends up stored in the capacitor

Never at fixed voltage. Charging an empty capacitor from a battery banks exactly half of what the battery spends — $\tfrac12QV$ stored out of $QV$ paid — and the other half is dissipated or radiated regardless of the circuit's resistance. Making the wires better does not improve the ratio; it only makes the loss faster and more violent. The same 50/50 rule reappears whenever charge falls through a pinned voltage: attaching a dielectric-boosted capacitor to its battery, paralleling a charged capacitor onto an empty one (there the lost fraction is $C_2/(C_1+C_2)$ ). If you catch yourself writing $U = QV$ for a capacitor — or crediting the full battery invoice to storage — the ledger is already broken. Only a changing source voltage that tracks the capacitor (adiabatic charging, or an inductor doing the tracking automatically) can beat the tax.

Worked examples

Example 1 — the defibrillator (confidence builder). A defibrillator stores its shot in a $C = 100\ \mu\mathrm{F}$ capacitor charged to $V = 2.0\ \mathrm{kV}$ . Find the stored energy, the charge, and the average power if the useful discharge lasts $4.0\ \mathrm{ms}$ .

Step 1. Energy: $U = \tfrac12CV^2 = \tfrac12(100\times10^{-6})(2000)^2 = \boxed{200\ \mathrm{J}}$ — the "200 J" setting on the front panel is literally this formula.

Step 2. Charge: $Q = CV = (100\times10^{-6})(2000) = 0.20\ \mathrm{C}$ . Modest charge; the voltage does the heavy lifting, because energy grows as $V^2$ .

Step 3. Average power: $P = U/t = 200/0.004 = \boxed{50\ \mathrm{kW}}$ — about forty household kettles, sustainable for four milliseconds precisely because the energy was banked in advance. The battery that refills the capacitor works at a leisurely $\sim40\ \mathrm{W}$ for five seconds: same energy, three orders of magnitude apart in power. Capacitors are power converters as much as energy stores.

Example 2 — the slab ledger, audited (exam level). A $C_0 = 200\ \mathrm{pF}$ air capacitor is charged to $V_0 = 1.0\ \mathrm{kV}$ . A glass slab ( $\kappa = 4.7$ ) is slowly inserted, fully filling the gap. Account for all energy (a) with the battery disconnected, (b) with it attached.

Step 1. Common starting point: $Q_0 = C_0V_0 = 0.20\ \mu\mathrm{C}$ , $U_0 = \tfrac12C_0V_0^2 = 100\ \mu\mathrm{J}$ .

Step 2. (a) Fixed $Q$ : capacitance rises to $\kappa C_0 = 940\ \mathrm{pF}$ ; voltage collapses to $V_0/\kappa = 213\ \mathrm{V}$ ; energy falls to $U_0/\kappa = \boxed{21.3\ \mu\mathrm{J}}$ . The missing $78.7\ \mu\mathrm{J}$ was delivered by the field to whatever restrained the slab during insertion — remove the restraint and it becomes kinetic energy, then heat, as the slab oscillates and settles.

Step 3. (b) Fixed $V$ : energy rises to $\kappa U_0 = \boxed{470\ \mu\mathrm{J}}$ (gain $370\ \mu\mathrm{J}$ ). The battery supplied $\Delta Q\,V_0 = (\kappa-1)C_0V_0^2 = 740\ \mu\mathrm{J}$ : half banked, half again spent pulling the slab in. Both protocols pulled the slab inward; they merely disagree about who financed the tug.

Step 4. Sanity: the fractions are universal. Fixed $Q$ always keeps $1/\kappa$ ; fixed $V$ always stores half the battery's outlay. Only the absolute joules needed the numbers.

History

Origins · 1 minMichael Faraday and how this was discovered

Michael Faraday

1791–1867 · London, Royal Institution

Faraday enters this article for the 1837 experiments that put the material inside the capacitor on physics' books. With twin concentric-sphere capacitors — one air-gapped, one filled — he compared charges at matched voltages and found the filled sphere held more by a factor that depended only on the filling: shellac 2, glass around 1.8, sulfur higher still. He named the effect specific inductive capacity — our $\kappa$ — and read it, characteristically, as evidence that "induction" propagates through the particles of the dielectric rather than acting at a distance: the polarization picture, seventy years early and in prose. He had, without knowing it, replicated Cavendish's buried measurements of the 1770s; unlike Cavendish, he published, and the dielectric constant entered science. The SI unit of capacitance honours exactly this circle of work — the farad, for the man who showed that what stands between the plates is as much a part of the capacitor as the plates themselves.

History

1745

— Leyden jar: glass, water, and a nail store the first "electric fire" — a dielectric capacitor before either word existed

1837

— Faraday's concentric spheres establish specific inductive capacity: the dielectric constant enters physics

1853

— William Thomson's analysis of the Leyden jar discharge finds the stored energy oscillating through an inductance — the loophole in the two-capacitor tax

1881

— The farad is adopted internationally as the unit of capacity, fixing Faraday's name to the subject

1957

— Becker's General Electric patent on the electrolytic double-layer capacitor opens the road to farad-scale "supercapacitors"

Going further

References. Halliday, Resnick & Walker, Fundamentals of Physics, ch. 25 (energy, dielectrics); Griffiths, Introduction to Electrodynamics, §2.4.4 and §4.4.3–4.4.4 (energy in fields and in dielectric systems); Purcell & Morin, Electricity and Magnetism, §3.7–3.9 and §10.5 (energy accounting, dielectric slabs); Zangwill, Modern Electrodynamics, ch. 6 for the careful treatment of forces from energy methods.

Provenance. Authored fresh for the SuperSTEM E&M track's circuits chapter (recon-§d gap heal); sources: v3 circuit pilots (rewritten to UG register) + legacy wiki circuit articles (narrative) — no AllQFT base exists for this topic. This article completes the "Capacitance and Stored Energy" reference that earlier electrostatics articles cite; the field-energy density, dielectric microscopics, and slab-force mechanism remain owned by their home articles (Electric Field Energy Density, Linear Dielectrics, Forces on Dielectrics), which this ledger recalls rather than re-derives.

Prerequisites. Capacitance · Capacitors in Series and Parallel · Electric Field Energy Density.

Leads to. RC Circuits · Energy Stored in Dielectric Systems · Forces on Dielectrics.

Why is the stored energy ½QV and not QV — didn't every charge cross voltage V?

No: only the last sliver of charge crossed the full $V$ . The first crossed at essentially zero voltage, and the average ride was $V/2$ because the voltage grows linearly with the charge already delivered. Geometrically the work is the triangle under the $q$ – $v$ line. A battery, whose voltage does not grow as it delivers charge, really does pay $QV$ — which is exactly why capacitor-charging at fixed voltage wastes half.

Is the energy really "in the field," or is that just a way of speaking?

For electrostatics you can keep the books either way — charges-and-potentials or field-density — and every total comes out identical, as the derivation's audit showed. The field ontology becomes obligatory once things move: electromagnetic waves carry energy through vacuum with no charges anywhere near, and the energy arriving in a charging capacitor demonstrably flows in through the sides of the gap (the Poynting analysis). Circuit theory lets you ignore this; physics does not.

In the two-capacitor problem, can I identify exactly where the lost energy went?

Into heat in whatever resistance exists, plus electromagnetic radiation from the current transient — in proportions that depend on the details, summing to a total that does not. With ordinary wire, almost all becomes heat along the wire during the RC transient. With deliberately tiny resistance, the current spike is enormous and brief, and a real fraction leaves as a radio-frequency burst (connect capacitors near an AM radio and you can hear the energy leaving). The invariance of the total, mechanism-blind, is the point.

If inserting a dielectric at fixed Q lowers the energy, why doesn't the capacitor just "want" dielectrics in all circumstances?

It does — that is exactly what the inward force means, in both protocols. What changes between fixed $Q$ and fixed $V$ is only the financing: at fixed $Q$ the field pays for the pull out of its own stored energy (which therefore drops), while at fixed $V$ the battery pays for both the pull and an increase in storage. "System slides toward higher capacitance" is the reliable slogan; "stored energy always decreases" is not.

Does the ½ε₀E² audit still work when a dielectric fills the gap?

Yes, with the upgraded density $u = \tfrac12\varepsilon_0\kappa E^2$ (equivalently $\tfrac12\,\mathbf D\cdot\mathbf E$ ) — the form derived in Energy Stored in Dielectric Systems, which carefully separates field energy from the energy of stretching the molecules. Run the same audit: $E = V/d$ , density $\tfrac12\kappa\varepsilon_0E^2$ , volume $Ad$ , product $\tfrac12(\kappa C_0)V^2$ . It closes exactly, which is a compact way to see why $C$ must scale as $\kappa$ .

Why do capacitors store so much less energy than batteries of the same weight?

Because a capacitor stores energy in field configuration, capped by dielectric breakdown ( $E_{\max}$ sets $u_{\max} = \tfrac12\kappa\varepsilon_0E_{\max}^2$ , of order $10^6\ \mathrm{J/m^3}$ for the best materials), while a battery stores it in chemical bonds at densities thousands of times higher. The capacitor's compensations are power and immortality: no chemistry means nearly unlimited charge–discharge cycles and the ability to dump its whole inventory in milliseconds — the defibrillator division of labour, battery for energy, capacitor for power.

What does "the smallest capacitor takes the most voltage" from the last article mean for energy?

Combine the rules: in series each capacitor holds the same $Q$ , so $U_i = Q^2/2C_i$ — the smallest capacitance stores the most energy as well as the most voltage. In a mismatched stack the runt is simultaneously the most stressed and the most loaded, which is why it fails first and why balancing matters doubly. In parallel the roles flip: $U_i = \tfrac12C_iV^2$ , and the biggest member dominates the inventory.

Capacitor Energy and Dielectrics

Where this lives today

The physical picture

The energy formulas

Limiting cases

Pitfall

Worked examples

History

Going further

Before this

Where this leads