Energy Stored in Dielectric Systems

Take two identical capacitors and charge both to 100 V. One is empty; the other is filled with a ceramic of relative permittivity 5. The filled one holds five times the energy — the fact powering the multilayer capacitors on every circuit board you own. Now the twist: charge the empty capacitor, disconnect the battery, and slide the ceramic in. The stored energy drops — to a fifth. Same slab, same plates, opposite verdicts. "The energy of a dielectric system" is a ledger that must be kept honestly. This article is that ledger. We know from Work and Energy of a Charge Configuration and Electric Field Energy Density what assembling charge costs in vacuum. With matter present, every coulomb pushed onto a plate does two jobs at once: it builds the macroscopic field, and it stretches and aligns molecular dipoles — charging billions of microscopic springs. The remarkable result is that one compact formula, $W=\tfrac12\int\mathbf{D}\cdot\mathbf{E}\,\mathrm{d}\tau$ , books both costs in a single stroke, provided the medium is linear. The subtleties are where the physics lives, and we will own them: what this energy is (work done on free charge, springs included) and what it is not (the vacuum formula on the total field), why an isolated capacitor loses energy to a slab while a battery-backed one pays double for it, and where the missing halves go. Those energy flows become forces on dielectrics in the next article.

Key takeaways

The always-true statement is incremental: pushing free charge into a system costs $\delta W=\int\mathbf{E}\cdot\delta\mathbf{D}\,\mathrm{d}\tau$ , whatever the medium does.
For linear media the increment integrates to a state function: $W=\tfrac12\int\mathbf{D}\cdot\mathbf{E}\,\mathrm{d}\tau$ , energy density $u=\tfrac12\varepsilon E^2$ .
This already includes the polarization work: $\tfrac12\mathbf{D}\cdot\mathbf{E}=\tfrac{\varepsilon_0}{2}E^2+\tfrac12\mathbf{P}\cdot\mathbf{E}$ — field plus springs. Never add a separate "polarization energy" by hand.
Capacitor bookkeeping: at fixed charge the energy falls by $1/\varepsilon_r$ (the field does work pulling the slab in); at fixed voltage it rises by $\varepsilon_r$ — and the battery supplies exactly twice the increase.
For nonlinear and hysteretic media only the increment survives; the $E$ – $D$ loop area is heat lost per cycle.
Key idea: in a linear dielectric the stored energy is $W=\tfrac12\int\mathbf{D}\cdot\mathbf{E}\,\mathrm{d}\tau$ , polarization work included; the truly general statement is $\delta W=\int\mathbf{E}\cdot\delta\mathbf{D}\,\mathrm{d}\tau$ .

The core insight

In a linear dielectric the stored electrostatic energy is $W=\tfrac12\int\mathbf{D}\cdot\mathbf{E}\,\mathrm{d}\tau$ — the energy density is $u=\tfrac12\mathbf{D}\cdot\mathbf{E}$ — and this already includes the work spent polarizing the medium. The truly general statement is the incremental one, $\delta W=\int\mathbf{E}\cdot\delta\mathbf{D}\,\mathrm{d}\tau$ .

Where this lives today

Energy density is the figure of merit that decides which dielectric ships in a product, and Eq. (2) says the winner is $\tfrac12\varepsilon E_{\max}^2$ — permittivity times breakdown field squared. That square is why the champions are not the highest- $\varepsilon_r$ materials: polypropylene film ( $\varepsilon_r\approx2.2$ but $E_{\max}\approx650\ \mathrm{MV/m}$ ) stores about $4\ \mathrm{MJ/m^3}$ and dominates electric-vehicle inverter capacitors, roughly matching ceramics whose $\varepsilon_r$ is a thousand times larger but whose breakdown field is thirty times smaller. Both trail a lithium-ion cell's $\sim$ GJ/m³ by three orders of magnitude — capacitors win on power and cycle life, not stored energy. The incremental form earns its keep at gigahertz: in every 5G front-end, the tiny $E$ – $D$ loop a real dielectric traces per cycle dissipates $\pi\varepsilon''E_0^2$ per unit volume, quoted as the loss tangent $\tan\delta=\varepsilon''/\varepsilon'$ — vendors sell $\tan\delta=10^{-4}$ substrates as premium product, a purchasable consequence of $\oint\mathbf{E}\cdot\mathrm{d}\mathbf{D}$ . And the ledgers' virtual-work method, $F=\tfrac12V^2\,\mathrm{d}C/\mathrm{d}x$ , is the standard design equation for MEMS actuators and varactors, where nobody integrates Maxwell stresses over a fringing field.

The physical picture

Charge a parallel-plate capacitor full of dielectric. Every increment of charge forced onto the plates fights the field of the charge already there — the vacuum cost we priced in Electric Field Energy Density. But in matter the field also grips every molecule and stretches it into a dipole (or swings a permanent one into line against thermal kicks). Stretching stores energy, exactly as compressing a spring does. A dielectric is a mattress of $10^{28}$ tiny springs per cubic metre, and charging the capacitor loads all of them.

Figure

Charging does two jobs: build the field, and load the molecular springs

Why does $\mathbf{D}$ appear in the energy formula at all? Because the work we do is booked against the free charge we control, and $\mathbf{D}$ is the field whose source is exactly that free charge — $\nabla\cdot\mathbf{D}=\rho_f$ , as established in The Electric Displacement D. The product $\mathbf{D}\cdot\mathbf{E}$ pairs "what you push" with "the field you push against." That is the deep reason the energy density takes the mixed form, rather than $E^2$ or $D^2$ alone. The picture also explains the two capacitor verdicts from the overview. At fixed voltage the field is pinned at $V/d$ ; filling the gap adds loaded springs and extra plate charge — more energy. At fixed charge the springs partially cancel the field, and the total falls. Keeping straight what is held fixed is half of this subject; the other half is the factor of $\tfrac12$ , and we now earn both.

The master increment

We charge the system the way nature does: a little free charge at a time. The configuration currently has potential $V(\mathbf{r})$ ; we ferry an extra $\delta\rho_f$ into place from infinity, paying charge times potential exactly as in Work and Energy of a Charge Configuration. Everything dielectric enters through one substitution: free charge is the source of $\mathbf{D}$ .

\delta W=\int\mathbf{E}\cdot\delta\mathbf{D}\,\mathrm{d}\tau \qquad (1)

where: $\delta\mathbf{D}$ — the change in displacement produced by the added free charge (C/m²); $\mathbf{E}$ — the macroscopic field at that moment (V/m); integral over all space. Equation (1) is the master increment: it holds for any medium — linear, nonlinear, hysteretic — because it uses only $\nabla\cdot\mathbf{D}=\rho_f$ . Every energy formula in this article descends from it.

Derivation AFrom free-charge work to the master increment

We want the work done by external agents when the free charge everywhere is nudged by $\delta\rho_f$ , and we want it expressed in fields alone, with no reference to where the charge came from. The strategy is the same integration-by-parts pivot we used in Electric Field Energy Density, but run on the increment rather than the total, and with Gauss's law in its dielectric form supplying the source identity. Watch the surface term — the argument is only as good as its vanishing.

Step 1. Ferrying charge $\delta\rho_f\,\mathrm{d}\tau$ into a location at potential $V$ costs $\delta W=\int V\,\delta\rho_f\,\mathrm{d}\tau$ , summed over all locations. Why V and not something dielectric-aware? The potential $V$ is generated by everything — free and bound charge alike. The dielectric's response is already inside $V$ ; that is what makes this bookkeeping complete.

Step 2. The added free charge changes the displacement: $\delta\rho_f=\nabla\cdot(\delta\mathbf{D})$ , which is Gauss's law in dielectrics — recalled from The Electric Displacement D, not re-derived — applied to the increment.

Step 3. Substitute and integrate by parts: $\int V\,\nabla\cdot(\delta\mathbf{D})\,\mathrm{d}\tau=\oint V\,\delta\mathbf{D}\cdot\mathrm{d}\mathbf{a}-\int\nabla V\cdot\delta\mathbf{D}\,\mathrm{d}\tau$ .

Step 4. Push the boundary to infinity. For a localized system $V\sim1/r$ and $\delta D\sim1/r^2$ while the surface grows as $r^2$ , so the surface term dies as $1/r$ — the same honesty check as in the vacuum case.

Step 5. With $\mathbf{E}=-\nabla V$ , what remains is $\delta W=\int\mathbf{E}\cdot\delta\mathbf{D}\,\mathrm{d}\tau$ . Verified: setting $\mathbf{D}=\varepsilon_0\mathbf{E}$ (vacuum) reproduces the increment of $\tfrac{\varepsilon_0}{2}\int E^2\,\mathrm{d}\tau$ , and the result matches Jackson §4.7.

Linearity collapses the increment

Equation (1) is a path statement: totals require integrating along the actual charging history, and in general depend on it. But if the medium is linear — $\mathbf{D}=\varepsilon\mathbf{E}$ , the constitutive law of Linear Dielectrics, Susceptibility and Permittivity — the differential becomes exact: the product rule gives $\delta(\mathbf{E}\cdot\mathbf{D})=\delta\mathbf{E}\cdot\mathbf{D}+\mathbf{E}\cdot\delta\mathbf{D}$ , linearity makes the two terms equal, so $\mathbf{E}\cdot\delta\mathbf{D}=\tfrac12\,\delta(\mathbf{E}\cdot\mathbf{D})$ . Integrating from zero field:

W=\frac12\int_{\text{all space}}\mathbf{D}\cdot\mathbf{E}\,\mathrm{d}\tau,\qquad u=\frac12\,\mathbf{D}\cdot\mathbf{E}=\frac12\,\varepsilon E^2=\frac{D^2}{2\varepsilon} \qquad (2)

where: $W$ — total work done on the free charges, polarization work included (J); $u$ — energy density (J/m³); $\varepsilon=\varepsilon_r\varepsilon_0$ — permittivity. The same $\tfrac12$ can be earned by the charging-parameter route: scale all fields by $\lambda$ , so $\delta\mathbf{D}=\mathbf{D}\,\mathrm{d}\lambda$ and $\mathbf{E}(\lambda)=\lambda\mathbf{E}$ , and $\int_0^1\lambda\,\mathrm{d}\lambda=\tfrac12$ — the field you fight is, on average over the charging, half the final one. Note what linearity bought: $W$ is now a state function, independent of how the charge got there. Ferroelectrics will revoke exactly this privilege — see Ferroelectrics and Spontaneous Polarization. For a capacitor the volume integral collapses to circuit variables. With $C=\varepsilon_r C_0$ from Linear Dielectrics, Susceptibility and Permittivity,

W=\frac12\,CV^2=\frac{Q^2}{2C} \qquad (3)

where: $Q$ — free charge on the plates; $V$ — voltage across them. Which form to reach for is a matter of what your process holds fixed — the whole drama of the ledgers below. A quick dispatch table for the family:

you know	use	valid when
fields in a linear medium	$u=\tfrac12\varepsilon E^2$	linear, any geometry
$D$ from free charge (a Gauss's Law in Dielectrics situation)	$u=D^2/2\varepsilon$	linear; $D$ found first
circuit variables, charge fixed	$W=Q^2/2C$	isolated capacitor
circuit variables, voltage fixed	$W=\tfrac12CV^2$	battery attached
nonlinear or hysteretic medium	$\delta W=\int\mathbf{E}\cdot\delta\mathbf{D}\,\mathrm{d}\tau$	always — integrate the actual path

Which energy is it, exactly

Here is the question that separates memorizing Eq. (2) from owning it: vacuum theory taught us that fields carry $\tfrac{\varepsilon_0}{2}E^2$ per unit volume — and the total field $\mathbf{E}$ in the dielectric is a perfectly good field. So is the energy $\tfrac{\varepsilon_0}{2}\int E^2$ , or $\tfrac12\int\mathbf{D}\cdot\mathbf{E}$ ? They differ. Both are meaningful. They answer different questions, and the difference is precisely the springs.

Go deeper · Derivation B · 1 minThe which-energy decomposition and the spring ledger

Derivation BThe which-energy decomposition and the spring ledger

We show that the linear-dielectric energy splits identically into pure field energy plus the microscopic work of polarization, so that nothing is double-counted and nothing is missing. The strategy is bare algebra on the definition of $\mathbf{D}$ , followed by a microscopic audit of the leftover term against the energy of one stretched molecule. The payoff is the doctrine: $\tfrac12\int\mathbf{D}\cdot\mathbf{E}$ is the work done on free charges; $\tfrac{\varepsilon_0}{2}\int E^2$ is the energy residing in the field; the gap is stored in the matter.

Step 1. Split the displacement by its definition $\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}$ :

\tfrac12\,\mathbf{D}\cdot\mathbf{E}=\tfrac{\varepsilon_0}{2}E^2+\tfrac12\,\mathbf{P}\cdot\mathbf{E}

Step 2. Audit the second term microscopically. A single molecule of polarizability $\alpha$ stretched to $\mathbf{p}=\alpha\mathbf{E}$ stored work $\int_0^E \alpha E'\,\mathrm{d}E'=\tfrac12\alpha E^2$ — the spring cost, with the $\tfrac12$ coming from linear ramp-up exactly as in a Hooke spring.

Step 3. Sum over the number density $N$ : the stored spring energy per volume is $N\cdot\tfrac12\alpha E^2=\tfrac12(N\alpha E)E=\tfrac12 PE$ — precisely the leftover term of Step 1. Why is this not the dipole's interaction energy, which we know is negative? Because $-\mathbf{p}\cdot\mathbf{E}$ prices a permanent dipole placed in an external field. An induced dipole must first be built: interaction $-\alpha E^2$ plus spring $+\tfrac12\alpha E^2$ nets $-\tfrac12\alpha E^2$ . The spring half-share is real, stays in the molecule, and is what the $\tfrac12\mathbf{P}\cdot\mathbf{E}$ term collects.

Step 4. Conclusion: $\tfrac12\int\mathbf{D}\cdot\mathbf{E}\,\mathrm{d}\tau=\tfrac{\varepsilon_0}{2}\int E^2\,\mathrm{d}\tau+\tfrac12\int\mathbf{P}\cdot\mathbf{E}\,\mathrm{d}\tau$ : field energy plus polarization work, one formula, no double counting. Verified: for a linear medium both terms are positive, $\tfrac{\varepsilon_0}{2}E^2+\tfrac12(\varepsilon_0\chi_e E)E=\tfrac12\varepsilon E^2$ ✓, consistent with Zangwill §6.7 and the induced-dipole energy of Griffiths §4.1.

The doctrine, stated bluntly: "the energy of the system" is ill-posed until you say which work you are counting. "How much work did the supply do on the free charges, dielectric responding?" — that is $\tfrac12\int\mathbf{D}\cdot\mathbf{E}$ , what engineers mean by stored energy, springs and all. "How much energy sits in the field itself?" — that is $\tfrac{\varepsilon_0}{2}\int E^2$ on the total field, the shortfall living in stretched molecules. Neither is wrong; adding them together is. One warning flows downstream: the formula prices work done through free charge. A frozen-polarization electret — no free charge anywhere, as in Uniformly Polarized Sphere — has $\tfrac12\int\mathbf{D}\cdot\mathbf{E}$ with no interpretation as its cost of manufacture; indeed inside that sphere $\mathbf{D}\cdot\mathbf{E}$ is negative ( $\mathbf{D}=\tfrac23\mathbf{P}$ along $\mathbf{P}$ , $\mathbf{E}=-\mathbf{P}/3\varepsilon_0$ against it). The energetics of frozen-in $\mathbf{P}$ belong to the material's chemistry, outside electrostatics' jurisdiction.

The capacitor ledgers

Now the bookkeeping that decides real design questions. Insert a dielectric into a capacitor — slab sliding in, or liquid rising. Two protocols, two ledgers.

Fixed charge (battery disconnected). $Q$ cannot change; $C$ grows to $\varepsilon_r C_0$ , so $W=Q^2/2C$ falls to $W_0/\varepsilon_r$ . The difference is paid out as mechanical work done by the field on the slab, which is pulled inward (that pull is Forces on Dielectrics, next; here we audit the books).

Fixed voltage (battery attached). $W=\tfrac12CV^2$ rises to $\varepsilon_r W_0$ . The battery is on the hook — and here sits the classic trap, worth its own derivation, because the battery pays more than the stored increase.

Go deeper · Derivation C · 1 minThe battery ledger at fixed voltage pays exactly double

Derivation CThe battery ledger at fixed voltage pays exactly double

We prove a completely general little theorem: in any quasistatic process at constant voltage in which the capacitance changes by $\mathrm{d}C$ , the battery delivers exactly twice the increase in stored energy — the other half leaves as mechanical work. The strategy is to differentiate the two energy accounts separately and compare; the theorem's charm is that $\varepsilon_r$ , geometry, and even the cause of $\mathrm{d}C$ drop out.

Step 1. At fixed $V$ , a capacitance change $\mathrm{d}C$ pulls extra charge $\mathrm{d}Q=V\,\mathrm{d}C$ from the battery.

Step 2. The battery moves that charge through its full EMF: it delivers $\mathrm{d}W_{\text{bat}}=V\,\mathrm{d}Q=V^2\,\mathrm{d}C$ .

Step 3. The stored energy changes by $\mathrm{d}W_{\text{stored}}=\mathrm{d}\big(\tfrac12CV^2\big)=\tfrac12V^2\,\mathrm{d}C$ — exactly half of Step 2.

Step 4. The unaccounted half, $\tfrac12V^2\,\mathrm{d}C$ , is mechanical work done on whatever changed $C$ — the inward pull on the entering slab. If nothing moves quasistatically (slab slammed home, charge surging through wire resistance), that half is dissipated instead: Joule heat. Why doesn't the fixed-charge case have a hidden partner? With the battery gone there is no external source: the field's loss is the slab's gain, one-to-one — two entries, not three. Verified: integrating Step 2 minus Step 3 from $C_0$ to $\varepsilon_rC_0$ gives mechanical output $\tfrac12V^2C_0(\varepsilon_r-1)$ , which matches the worked numbers below and Griffiths §4.4.4's discussion.

Figure

The two ledgers, with the numbers of Worked Example 1

Limiting cases

case	result	meaning
vacuum	$W=\tfrac{\varepsilon_0}{2}\int E^2\,\mathrm{d}\tau$	no springs; Eq. (2) collapses to the vacuum energy of Electric Field Energy Density
linear isotropic medium	$u=\tfrac12\varepsilon E^2=D^2/2\varepsilon$	at fixed $E$ a high- $\varepsilon$ medium stores more; at fixed $D$ , less
linear anisotropic crystal	$u=\tfrac12E_i\varepsilon_{ij}E_j$	a positive-definite quadratic form; the tensor $\varepsilon_{ij}$ is the closing topic of this chapter
$\varepsilon_r\to1$	all dielectric corrections vanish	the springs stiffen away; both ledgers reduce to vacuum bookkeeping
nonlinear or hysteretic	only $\delta W=\int\mathbf{E}\cdot\delta\mathbf{D}\,\mathrm{d}\tau$ survives	$W$ is path-dependent; loop area $\oint\mathbf{E}\cdot\mathrm{d}\mathbf{D}$ is heat per cycle

Worked examples

Example 1 — one capacitor, two protocols (confidence builder). A parallel-plate capacitor has $C_0=100\ \mathrm{pF}$ and is charged to $V_0=100\ \mathrm{V}$ , so $Q_0=C_0V_0=10\ \mathrm{nC}$ and $W_0=\tfrac12C_0V_0^2=0.500\ \mathrm{\mu J}$ . A ceramic slab with $\varepsilon_r=5$ fills the gap. Find the energy ledger for both protocols.

Step 1. Battery disconnected first. The capacitance becomes $C=\varepsilon_rC_0=500\ \mathrm{pF}$ ; the charge is stuck at $10\ \mathrm{nC}$ . Then $W=Q_0^2/2C=(10^{-8})^2/(2\times5\times10^{-10})=0.100\ \mathrm{\mu J}$ — one fifth of $W_0$ , as the $1/\varepsilon_r$ rule demands. The voltage reads $Q_0/C=20\ \mathrm{V}$ .

Step 2. Audit the missing $0.400\ \mathrm{\mu J}$ : it was delivered as mechanical work by the inward pull on the slab as it entered. Nothing is lost; it changed accounts.

Step 3. Battery attached instead. Now $W=\tfrac12CV_0^2=\tfrac12(500\ \mathrm{pF})(100\ \mathrm{V})^2=2.500\ \mathrm{\mu J}$ — five times $W_0$ , a stored gain of $+2.000\ \mathrm{\mu J}$ .

Step 4. The battery moved $\Delta Q=(C-C_0)V_0=40\ \mathrm{nC}$ through $100\ \mathrm{V}$ : it spent $\Delta Q\,V_0=4.000\ \mathrm{\mu J}$ . Stored: $2.000$ . Mechanical work on the slab: the other $2.000\ \mathrm{\mu J}$ — the double-payment theorem, live. Fixed Q: $0.5\to0.1\ \mathrm{\mu J}$ ; fixed V: battery pays $4.0$ , field keeps $2.0$ , slab gets $2.0\ \mathrm{\mu J}$ .

Example 2 — where the energy sits around the dielectric sphere (exam level). Recall the polystyrene sphere of Dielectric Sphere in a Uniform Field: $\varepsilon_r=2.6$ , $R=2\ \mathrm{cm}$ , in $E_0=1.0\times10^{5}\ \mathrm{V/m}$ , with interior field $E_{\text{in}}=3E_0/(\varepsilon_r+2)=6.52\times10^{4}\ \mathrm{V/m}$ and dipole moment $p=3.1\times10^{-11}\ \mathrm{C\cdot m}$ . Localize the energy.

Step 1. Interior energy density: $u_{\text{in}}=\tfrac12\varepsilon_r\varepsilon_0E_{\text{in}}^2=\tfrac12(2.6)(8.854\times10^{-12})(6.52\times10^{4})^2=49.0\ \mathrm{mJ/m^3}$ , against $u_0=\tfrac12\varepsilon_0E_0^2=44.3\ \mathrm{mJ/m^3}$ before the sphere arrived. Ratio: $u_{\text{in}}/u_0=9\varepsilon_r/(\varepsilon_r+2)^2=1.106$ . Why more, when the field went down? The field dropped by $0.652$ , its square by $0.425$ — but $u=\tfrac12\varepsilon E^2$ multiplies back one power of $\varepsilon_r=2.6$ . The competition is close, and non-monotonic.

Step 2. Map the competition: $f(\varepsilon_r)=9\varepsilon_r/(\varepsilon_r+2)^2$ equals $1$ at $\varepsilon_r=1$ and at $\varepsilon_r=4$ , peaks at $\varepsilon_r=2$ ( $f=1.125$ ), and falls far below $1$ for strong dielectrics — water at $\varepsilon_r=80$ manages only $f=0.107$ . A weak dielectric slightly concentrates energy; a strong one expels it (its interior field is nearly dead, $E_{\text{in}}=0.037E_0$ for water).

Step 3. Totals: volume $33.5\ \mathrm{cm^3}$ , so $W_{\text{in}}=1.64\ \mathrm{\mu J}$ against $1.48\ \mathrm{\mu J}$ pre-insertion. Outside, the dipole field adds its own store, $\int_{r>R}\tfrac{\varepsilon_0}{2}E_{\text{dip}}^2\,\mathrm{d}\tau=p^2/(12\pi\varepsilon_0R^3)=0.359\ \mathrm{\mu J}$ .

Step 4. Yet the configuration energy went down: for fixed distant sources, inserting a linear body changes the total by $\Delta W=-\tfrac12pE_0=-\tfrac12(3.1\times10^{-11})(10^5)=-1.55\ \mathrm{\mu J}$ . The local gains of Steps 1–3 are outweighed by the negative interaction with the source field — the induced-dipole ledger $-\alpha E_0^2+\tfrac12\alpha E_0^2=-\tfrac12\alpha E_0^2$ in macroscopic clothes. The system lowers its energy by hosting the sphere — exactly why dielectrics are sucked toward strong field, quantified in Forces on Dielectrics.

Figure

Interior energy density versus permittivity: the 9ε over squared competition

Origins · 1 minjames prescott joule and how this was discovered

james prescott joule

1818–1889 · Manchester brewer-physicist who priced every energy conversion

Joule never wrote a line about dielectrics — yet every ledger in this article is audited by his rules. A brewer's son who measured temperature to a fiftieth of a degree, he established in the 1840s that heat, mechanical work, and electrical work are one currency: his $i^2R$ law priced pushing charge through resistance, and his paddle-wheel experiments fixed the work–heat exchange rate. That is precisely the bookkeeping we used on the battery's missing half: energy that neither stays in the field nor moves the slab must appear as Joule heat in the wires. When the fixed-voltage ledger refused to balance until we found the mechanical-work account, we were doing pure Joule: find the other account; there is always another account. The SI unit of every $W$ in this article was named for him in the year he died.

History

1840

— Joule announces the

i^2R

heating law: electrical work has a measurable price in heat

1843

— Mechanical equivalent of heat measured; energy becomes a single convertible currency

1847

— Helmholtz's Über die Erhaltung der Kraft makes conservation of energy a universal audit rule

1873

— Maxwell's Treatise locates electrostatic energy in the medium, with density following D and E

1881

— Second law meets dielectrics: hysteresis in E–D loops recognized as dissipated work per cycle

1889

— The international electrical congress names the unit of energy the joule

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §4.4.3 · Jackson, Classical Electrodynamics, 3rd ed., §4.7 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 10 · Zangwill, Modern Electrodynamics, §6.7

Provenance. Energy formula and force examples from Griffiths Introduction to Electrodynamics 4th ed. §4.4.3 and Example 4.10/Problem 4.28; incremental derivation and hysteresis from Jackson Classical Electrodynamics 3rd ed. §4.7; decomposition and loss tangent from Zangwill Modern Electrodynamics §6.7; vacuum limit cross-checks with Purcell & Morin Ch. 10. A verifier should recompute: (1) the path integral $\int_0^1\lambda\,\mathrm{d}\lambda=\tfrac12$ giving the energy prefactor; (2) the slab force $F=\varepsilon_0(\varepsilon_r-1)wV^2/2d$ from $\mathrm{d}C/\mathrm{d}x$ ; (3) the decomposition $\tfrac12\mathbf{D}\cdot\mathbf{E}=\tfrac{\varepsilon_0}{2}E^2+\tfrac12\mathbf{P}\cdot\mathbf{E}$ and the induced-dipole work $\tfrac12\alpha E^2$ .

Builds on: The Electric Displacement D · Gauss's Law in Dielectrics · Linear Dielectrics, Susceptibility and Permittivity · Polarization Density P Leads to: Boundary Conditions for D and E · Torque and Force on a Dipole in a Field · Forces on Dielectrics

Why does the energy formula pair D with E instead of using E squared or D squared?

Because the two fields play different roles in the bookkeeping. The work is done on the free charge you control, and free charge sources $\mathbf{D}$ ( $\nabla\cdot\mathbf{D}=\rho_f$ ); but the force doing the pushing is the total field $\mathbf{E}$ . The increment $\mathbf{E}\cdot\delta\mathbf{D}$ pairs "what you push" with "what pushes back." In a linear medium you may of course rewrite $\tfrac12\mathbf{D}\cdot\mathbf{E}$ as $\tfrac12\varepsilon E^2$ or $D^2/2\varepsilon$ — but the mixed form is the one that survives when linearity fails.

If the isolated capacitor's energy drops when I insert the slab, could I extract that energy?

Yes — that is not a paradox but a machine. The 0.4 μJ in Worked Example 1 leaves as mechanical work on the slab: the fringing field pulls it in, and a slab on a rail could raise a (very small) weight. Run backwards, you must do 0.4 μJ dragging the slab out, recharging the field to 0.5 μJ. Electrostatic energy harvesters run exactly this cycle.

Is half the battery's energy always lost when charging through a dielectric change?

The battery always pays double the stored increase at fixed voltage — but where the other half goes depends on the process. Quasistatic insertion: it becomes mechanical work on the slab, fully recoverable. Sudden connection or slamming the slab: the surge dissipates it in wire resistance and slab oscillations — Joule's account collects. The theorem fixes the split of payment; thermodynamics decides whether the second half is work or heat.

Can I use W equals one half integral D dot E for an electret or a uniformly polarized sphere?

No meaningful way. That formula tallies work done through free charge, and an electret has none — its polarization is frozen in by manufacture, not induced by charging. Inside the Uniformly Polarized Sphere, $\mathbf{E}$ opposes $\mathbf{P}$ while $\mathbf{D}$ follows it, so $\tfrac12\mathbf{D}\cdot\mathbf{E}$ is negative there — a loud hint it is not anyone's cost of assembly. The electret's creation energy belongs to the material's internal chemistry.

Why does a ferroelectric break the compact energy formula?

The collapse from $\int\mathbf{E}\cdot\delta\mathbf{D}$ to $\tfrac12\mathbf{D}\cdot\mathbf{E}$ needed $\mathbf{D}$ single-valued and linear in $\mathbf{E}$ , so the charging integral wouldn't care about the path. A ferroelectric's $D(E)$ is a hysteresis loop: multivalued, history-dependent. The increment must be integrated along the actual path, and each cycle leaves the loop's area $\oint\mathbf{E}\cdot\mathrm{d}\mathbf{D}>0$ behind as heat. Details in Ferroelectrics and Spontaneous Polarization.

Does inserting a dielectric increase or decrease the field energy in the region it occupies?

Delightfully, it depends. The occupied region's density changes by the factor $9\varepsilon_r/(\varepsilon_r+2)^2$ for a sphere: up for $1<\varepsilon_r<4$ (peak $+12.5\%$ at $\varepsilon_r=2$ ), back to unity at $\varepsilon_r=4$ , and sharply down beyond — water retains only 11%. Weak dielectrics dilute the field a little while multiplying by $\varepsilon_r$ ; strong ones nearly expel the field ( $E_{\text{in}}=3E_0/(\varepsilon_r+2)\to0$ ), and no permittivity factor can rescue a dead field. The total configuration energy nonetheless always falls ( $\Delta W=-\tfrac12pE_0$ ), which is why dielectrics are universally pulled inward.

At 2 am: energy is stored in the field, or in the springs, or in the battery — how do I not mix accounts on an exam?

Use the three-question ritual. One: what is held fixed — Q (isolated) or V (battery)? That picks $Q^2/2C$ or $\tfrac12CV^2$ . Two: whose energy is asked — supply-side ( $\tfrac12\int\mathbf{D}\cdot\mathbf{E}$ , the default) or field-only ( $\tfrac{\varepsilon_0}{2}\int E^2$ , rare and flagged)? Three: does the ledger close — initial plus battery input equals final plus mechanical work plus heat? A missing term is almost always the battery's double payment or the slab's work.

Energy Stored in Dielectric Systems

Where this lives today

The physical picture

The master increment

Linearity collapses the increment

Which energy is it, exactly

The capacitor ledgers

Limiting cases

Worked examples

Going further

Before this

Where this leads