Physics · Electrostatics

Electric-field energy density in vacuum

In vacuum, an electric field stores energy per unit volume proportional to the square of field strength.

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Formula

u_{E} = \frac{1}{2} ε_{0} E^{2}

uE in J/m³; ε0 in F/m; E in V/m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$uE$	Electric energy density	J/m³
$ε0$	Vacuum permittivity	F/m
$E$	Magnitude of electric field	V/m

LaTeX

u_E=\frac{1}{2}\varepsilon_0 E^2

QUANTA STUDYquanta-study.de

Quadratic normalized energy density versus normalized position. Shaded area equals one third; horizontal mean line lies at one third. — For the prescribed vacuum profile E(x)=E₀x/L, energy density grows quadratically. Its spatial mean is one third of the endpoint density, not one quarter.

Electric field energy density calculator

Find the energy density, field strength or stored energy of an electric field in vacuum, for a uniform field or one that grows linearly.

Calculation model

Find

Electric field strength |E|

Electric field strength at that point, e.g. 200 kV/m for 400 V across a 2 mm plate gap.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for uE, substitute, evaluate

1. Rearrange for uE
$u_E=\frac12\varepsilon_0|E|^2$
2. Substitute values (in SI units)
- |E| = 200 kV/m = 200 000 V/m
$\frac{8.85419\times10^{-12}\cdot200000^2}2=0.177084$
3. Result
uE = 177.1 mJ/m³

ε₀≈8.8541878188·10⁻¹² F/m: rounded measured CODATA2022 value, not an exact SI definition.

Result: Electric vacuum energy density uE

uE = 177.1 mJ/m³

With |E| = 200 kV/m, uE = 177.1 mJ/m³.

Quick tips

Twice the field strength gives four times the energy density, because u grows with E². A reversed field stores exactly as much energy.
Estimate: ½ε₀ ≈ 4.4 · 10⁻¹² F/m. At 1000 kV/m that is about 4.4 J/m³, and even just before a spark jumps in air only about 40 J/m³.

Assumptions and limits

Only electric-field energy density in vacuum, using actual total field magnitude E≥0. No magnetic contribution, energy flux or unchecked εr replacement. Regional energy is uEV only for spatially uniform density. Reversing direction leaves density unchanged.

Derivation & background

In vacuum, this is the electric component of local field energy density. Total energy is $W_E=\int_Vu_E\,d^3r$ ; a present magnetic field contributes separately. Linear reversible dielectrics include material polarization, while nonlinear media require the actual constitutive characteristic. Averaging the field before squaring does not generally give correct mean energy.

Exam blueprint

Validity range

Electric component of vacuum energy density. Integrate spatially for total energy; density times volume requires constant density. Magnetic energy is a separate contribution. Material polarization needs appropriate constitutive energy accounting.

Derivation steps

The uniform plate model gives electric energy per field volume; general fields require local integration.

1Consider large parallel vacuum plates with area A, separation d and uniform field magnitude E.
2Capacitance is $C=\varepsilon_0\frac Ad$ .
3Voltage and field magnitudes in the uniform gap obey $U=Ed$ .
4Insert both into linear stored energy $W_C=\frac12 CU^2$ .
5The result is $W_C=\frac12\varepsilon_0E^2Ad$ ; gap volume is $Ad$ .
6Dividing by that volume gives $u_E=\frac12\varepsilon_0E^2$ .
7The general vacuum expression belongs to electromagnetic energy balance; integrate local squared fields, not the square of mean field.

Rearrangements

Field strength

E=\sqrt{\frac{2u_E}{\varepsilon_0}}

Nonnegative vacuum field magnitude; density alone does not give field direction.

Task variant

A uniform vacuum field has $E=1.0\times10^5\,\mathrm{V/m}$ . Find electric energy density. How does it change if field magnitude is halved?

Approach: $u_E=\frac{\varepsilon_0E^2}{2}$ .
Rounded CODATA permittivity gives $u_E\approx0.0443\,\mathrm{J/m^3}$ .
Halving E yields $u_{E,2}=\frac14u_{E,1}\approx0.0111\,\mathrm{J/m^3}$ .

Check: density is not energy in joules; the selected volume is still required.

Common mistakes

Average the local field before squaring.

First square the actual total field locally and integrate density; opposite fields in separate volumes do not have zero total energy.

Exam context

Capacitor gaps, spatially varying fields and field superposition: distinguish density, regional energy and fixed boundary quantities.

Formula cluster

Electrostatics

Capacitor energy and geometric capacitance check the same stored energy; the electric field supplies its spatial distribution.

E = ½CU²

Energy Stored in a Capacitor

C = ε₀ εr A/d

Capacitance of a parallel-plate capacitor

E⃗ = F⃗el/q; Ehom = |U|/d

Electric Field Strength

Worked example

For $E=1.0\times10^5\,\mathrm{V/m}$ , $u_E=\frac12\varepsilon_0(10^5\,\mathrm{V/m})^2\approx0.0443\,\mathrm{J/m^3}$ . Halving field magnitude reduces density to one quarter.

Applications

Describe capacitor and electromagnetic-field energy spatially.

How do you calculate with Electric-field energy density in vacuum?

Here is how to work through a typical Electric-field energy density in vacuum (uE = 1/2 ε₀E²) task step by step:

1
Task
A uniform vacuum field has $E=1.0\times10^5\,\mathrm{V/m}$ . Find electric energy density. How does it change if field magnitude is halved?
Solution path
1. Approach: $u_E=\frac{\varepsilon_0E^2}{2}$ .
2. Rounded CODATA permittivity gives $u_E\approx0.0443\,\mathrm{J/m^3}$ .
3. Halving E yields $u_{E,2}=\frac14u_{E,1}\approx0.0111\,\mathrm{J/m^3}$ .
Check: density is not energy in joules; the selected volume is still required.

Understand and apply the equation

Energy density assigns electric vacuum-field energy per volume to each location. It is proportional to squared field magnitude and unchanged by reversing direction. Volume integration then gives energy. A uniform plate field and a spatially increasing model field distinguish local density, its average and total energy.

What does electric energy density in vacuum describe?

Energy is not confined to moving objects. An electric field also contains energy, for example between charged capacitor plates. Energy density tells how much occupies a small volume. Total energy therefore also needs the volume and spatial field distribution. We first examine a uniform vacuum field, then a varying field and a finite region around a charge.[1]

u_E(\mathbf r)=\frac12\varepsilon_0|\mathbf E(\mathbf r)|^2

$u_E$ is the electric component of vacuum energy density; a simultaneous magnetic field has its own contribution.[1][2]

The vector is dotted with itself: $|\mathbf E|^2=E_x^2+E_y^2+E_z^2$ . Negative components thus do not make energy density negative. Its unit is J/m³ rather than joules. Doubling field magnitude under the same model multiplies local energy density by four. Doubling the observed volume at unchanged density instead doubles energy.[1]

The basic relation gives electric-field energy in vacuum. In a material, simply replacing the vacuum constant requires justification: polarization and what is held fixed belong to the model. The later extension explains this distinction.[1][6]

Obtain density from ideal plate energy

\begin{aligned}C&=\varepsilon_0\frac Ad\\U&=Ed\\W_C&=\frac12 CU^2=\frac12\varepsilon_0E^2Ad\end{aligned}

For the ideal uniform vacuum gap, $Ad$ is the field volume.[1]

Insert geometric capacitance and uniform voltage relation into linear capacitor energy. Remaining $\frac{W_C}{Ad}$ is $\frac12\varepsilon_0 E^2$ , motivating the local vacuum expression through the plate model. Its general application to electric vacuum fields belongs to electromagnetic energy balance; one component model alone does not prove every spatial or time-dependent field. Energy in the capacitor here checks the same vacuum region rather than providing an additional energy contribution.[1][2]

W_E(V)=\int_V\frac12\varepsilon_0|\mathbf E|^2\,d^3r

Only constant field magnitude permits density to be taken outside the integral; otherwise evaluate it locally.[1]

Do not confuse volume density with radiation flux

\begin{aligned}[\varepsilon_0E^2]&=\frac{\mathrm F}{\mathrm m}\frac{\mathrm{V^2}}{\mathrm{m^2}}=\frac{\mathrm J}{\mathrm{m^3}}\\E&=\sqrt{\frac{2u_E}{\varepsilon_0}}\end{aligned}

The rearrangement yields only nonnegative field magnitude, not direction.

For uniform density, regional energy is $u_E$ times volume in cubic metres, not surface area. Density alone determines no transported power. Electromagnetic energy flux also requires field geometry and magnetic field or the Poynting vector. At $E=10^5\,\mathrm{V/m}$ , the result is approximately 0.0443 J/m³. We use rounded measured CODATA-2022 vacuum permittivity rather than calling it an exact SI definition. The Poynting vector and electromagnetic energy flux describe that transport rather than simply stored volume density.[4][2]

Uniform field and spatial variation

Energy in a vacuum plate gap

Given: Large parallel plates: $A=0.00600\,\mathrm{m^2}$ , $d=2.00\,\mathrm{mm}$ , $U=400\,\mathrm V$ . Ideal uniform vacuum gap, fringe fields neglected.
Find: Find E, $u_E$ , gap volume and stored energy.

Approach: A uniform vacuum field has the same electric energy density throughout the gap, permitting multiplication by the actual gap volume.
$\begin{aligned}E&=\frac Ud\\u_E&=\frac12\varepsilon_0E^2\\W_E&=u_EAd\end{aligned}$
Find field and volume from geometry.
$\begin{aligned}E&=\frac{400\,\mathrm V}{0.00200\,\mathrm m}=2.00\times10^5\,\mathrm{V/m}\\V_{\mathrm{gap}}&=Ad=1.20\times10^{-5}\,\mathrm{m^3}\end{aligned}$
Multiply density by volume.
$\begin{aligned}u_E&\approx0.1770837564\,\mathrm{J/m^3}\\W_E&=u_EV_{\mathrm{gap}}\approx2.125005077\,\mathrm{\mu J}\end{aligned}$

Result: Local density ≈ 0.177084 J/m³ and gap energy ≈ 2.12501 μJ. Model surface area is not confused with volume.

Check: Independently: $C=\varepsilon_0\frac Ad$ ≈ 26.5626 pF. $\frac12CU^2$ gives 2.12501 μJ. Both calculations describe the same uniform gap energy, not separate terms to be added.

A linear vacuum field over a selected region

Given: A prescribed quasistatic model has $E(x)=E_0\frac xL$ along +x for $0\le x\le L$ . $E_0=30000\,\mathrm{V/m}$ , $L=0.0300\,\mathrm m$ and constant cross-section $A=0.0120\,\mathrm{m^2}$ . Required source distribution is prescribed; calculate only energy in this vacuum region.
Find: Integrate energy density and compare with incorrectly squaring mean field.

Approach: Square the prescribed linear field locally first. Constant cross-section then reduces the volume integral to a position integral.
$\begin{aligned}u_E(x)&=\frac12\varepsilon_0E_0^2\frac{x^2}{L^2}\\W_E&=A\int_0^Lu_E(x)\,\mathrm dx\end{aligned}$
Local density grows quadratically with x.
$u_E(x)=\frac12\varepsilon_0E_0^2\frac{x^2}{L^2}$
Here the volume integral becomes A times the x integral.
$\begin{aligned}W_E&=A\int_0^L u_E(x)\,dx\\&=\frac{\varepsilon_0E_0^2AL}{6}\\&\approx0.4781261422\,\mathrm{\mu J}\end{aligned}$
Correct mean of E² differs from square of mean E.
$\begin{aligned}\langle E^2\rangle&=\frac{E_0^2}{3}\\\langle E\rangle^2&=\frac{E_0^2}{4}\\W_{\mathrm{mean}}&=\frac34W_E\end{aligned}$

Result: Regional energy is ≈ 0.478126 μJ. Substituting mean field $15000\,\mathrm{V/m}$ would underestimate it by 25%.

Check: Maximum density 0.0039843845 J/m³ times volume 0.000360 m³ divided by 3 gives the same result. The linear field is not charge free: $\frac{\partial E_x}{\partial x}=\frac{E_0}{L}$ , so Gauss requires $\rho=\varepsilon_0\frac{E_0}{L}$ . This is not claimed to be an arbitrary source-free vacuum Laplace solution.

Why must you square the field before spatial averaging?

Linear field, quadratically growing energy density

QUANTA STUDYquanta-study.de

Original curve for the second example, with local density divided by its endpoint value. Mean is one third rather than one quarter: squaring mean field would underestimate energy.[1][5]

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For a spatial distribution, first form $E_x^2+E_y^2+E_z^2$ and then integrate. Equal regions with opposite field directions can have mean field vector zero while both contain energy. Squaring a superposed field also produces cross terms. Simply adding individual source-field energies therefore need not equal total-field energy; interaction contributions may matter.[2][3]

|\mathbf E_1+\mathbf E_2|^2=|\mathbf E_1|^2+|\mathbf E_2|^2+2\mathbf E_1\cdot\mathbf E_2

The cross term explains why field energy is not merely a sum of positive individual self-energies.

Explore further: Field energy in reversible dielectrics

For a linear reversible isotropic dielectric, suitable electrostatic energy density can be written $\frac12\mathbf E\cdot\mathbf D$ with $\mathbf D=\varepsilon\mathbf E$ . It then includes material polarization in that model rather than unchanged pure vacuum-field energy. Nonlinear reversible response requires actual construction accounting $\int\mathbf E\cdot d\mathbf D$ ; with hysteresis or loss, stored energy is not all supplied work.[3][6]

Total electromagnetic vacuum energy also includes magnetic term $\frac{B^2}{2\mu_0}$ . In a plane vacuum wave electric and magnetic contributions are equal, whereas a static electric capacitor does not automatically have that magnetic contribution. Distinguish total field energy from electric-field energy when comparing values.[2][6]

Five density and volume exercises

Calculate · Foundation

A vacuum field has magnitude $E=50000\,\mathrm{V/m}$ . Find $u_E$ .

Show hint

Square field magnitude and multiply by $\frac{\varepsilon_0}2$ .

Learning goal: Calculate vacuum energy density from field magnitude.

Show solution and calculation

Approach: $u_E=\frac{\varepsilon_0E^2}{2}$ .
$u_E$ ≈ 0.0110677 J/m³.
It is one quarter of the upper 10⁵ V/m example because field is halved.

\begin{aligned}u_E&=\frac{\varepsilon_0}{2}(50000\,\mathrm{V/m})^2\\&\approx0.0110677\,\mathrm{J/m^3}\end{aligned}

[1]

Rearrange · Application

Uniform electric energy density 0.0200 J/m³ fills the selected vacuum volume 8.00×10⁻⁴ m³. Find energy and field magnitude.

Show hint

Energy is density times volume; E requires a separate square root.

Learning goal: Infer energy and field magnitude from density and volume.

Show solution and calculation

Approach: $W_E=u_EV$ .
$W_E=1.60\cdot10^{-5}\,\mathrm J=16.0\,\mathrm{\mu J}$ .
From $E=\sqrt{\frac{2\cdot0.0200\,\mathrm{J/m^3}}{\varepsilon_0}}$ , magnitude is approximately 67213.4 V/m.
Field direction remains unknown.

\begin{aligned}W_E&=(0.0200)(8.00\times10^{-4})\,\mathrm J=16.0\,\mathrm{\mu J}\\E&=\sqrt{\frac{2u_E}{\varepsilon_0}}\approx6.72134\times10^4\,\mathrm{V/m}\end{aligned}

[1]

Interpret · Application

Two equal vacuum subvolumes have fields $+E_0$ and $-E_0$ with $E_0=20000\,\mathrm{V/m}$ . What is the volume-averaged energy density?

Show hint

Both local squares stay positive.

Learning goal: Distinguish mean squared field from square of mean field.

Show solution and calculation

Approach: $\langle u_E\rangle=\frac{\varepsilon_0}{2}\langle E^2\rangle$ .
Mean field vector is zero, but mean energy density is $\frac12\varepsilon_0 E_0^2$ ≈ 0.00177084 J/m³.
Averaging field before squaring would incorrectly give zero.

\begin{aligned}\langle u_E\rangle&=\frac12\left[\frac{\varepsilon_0E_0^2}{2}+\frac{\varepsilon_0(-E_0)^2}{2}\right]\\&=\frac{\varepsilon_0}{2}(20000\,\mathrm{V/m})^2\\&\approx0.00177084\,\mathrm{J/m^3}\end{aligned}

[1]

Interpret · Application

Two uniform fields of magnitude $E_0$ overlap in the same vacuum volume. Compare energy density for parallel co-directed, perpendicular and antiparallel fields with $u_0=\frac12\varepsilon_0E_0^2$ . These are not separate subvolumes.

Show hint

The squared sum vector includes a cross term.

Learning goal: Add overlapping fields vectorially before squaring.

Show solution and calculation

Approach: $u_E=\frac{\varepsilon_0}{2}|\mathbf E_1+\mathbf E_2|^2$ .
Co-directed fields give magnitude $2E_0$ and density $4u_0$ .
Perpendicular fields give squared magnitude $2E_0^2$ and density $2u_0$ .
Antiparallel fields cancel, so total electric-field energy density vanishes.

Check: the cross term is positive, zero or negative. Simply adding separate densities works for the perpendicular case, not general spatial overlap.

\begin{aligned}u_E&=\frac12\varepsilon_0|\mathbf E_1+\mathbf E_2|^2\\&=\frac12\varepsilon_0\left(E_1^2+E_2^2+2\mathbf E_1\cdot\mathbf E_2\right)\end{aligned}

[1]

Error analysis · Challenge

An ideal isolated parallel-plate capacitor keeps the same free plate charge. A linear dielectric with $\varepsilon=4\varepsilon_0$ completely replaces vacuum, geometry unchanged, fringing neglected. A solution claims: “ $u=\frac12\varepsilon E^2$ , so density quadruples.” Check it.

Show hint

Fixed free charge fixes D, not E.

Learning goal: Account for boundary conditions when permittivity changes.

Show solution and calculation

Approach: $D=\frac{Q_{\mathrm{free}}}{A}=\varepsilon E$ .
$D=\varepsilon E$ stays constant for fixed free plate charge and area.
Thus E falls to one quarter and $u=\frac{D^2}{2\varepsilon}$ also falls to one quarter.

Check: capacitance quadruples, so fixed-Q energy $\frac{Q^2}{2C}$ falls to one quarter. A factor-four increase would fit fixed E, such as constant voltage with unchanged geometry. Field energy is distinct from the complete mechanical and source balance.

\begin{aligned}E_2&=\frac{D}{4\varepsilon_0}=\frac{E_1}{4}\\\frac{u_2}{u_1}&=\frac{4\varepsilon_0(E_1/4)^2}{\varepsilon_0E_1^2}=\frac14\end{aligned}

[3][1]

Ideal point sources expose a self-energy limit

For a point Coulomb field, $u_E$ grows as $\frac1{r^4}$ . Radial volume integration contributes $r^2\,dr$ , leaving divergent self-energy as r approaches zero. The local vacuum expression alone therefore gives no finite classical self-energy for an ideal point electron. Extended sources, a prescribed excluded radius or a different physical description change the problem. The finite region in the second example is explicitly a selected portion rather than automatically total energy of all sources.[3][1]

Example: energy in a spherical shell

Finite energy between two radii

Given: A conducting sphere of radius $a=0.0200\,\mathrm m$ carries $Q=2.00\,\mathrm{nC}$ . Other sources vanish. Find only vacuum field energy from a to $b=0.0400\,\mathrm m$ , not all energy outside the sphere.
Find: Find shell energy and its share of total exterior field energy.

Approach: spherical symmetry gives the radial field; a volume shell has size $4\pi r^2\,\mathrm dr$ .
$\begin{aligned}E(r)&=\frac{Q}{4\pi\varepsilon_0r^2}\\W_{a,b}&=\int_a^b\frac{\varepsilon_0E^2(r)}2\,4\pi r^2\,\mathrm dr\\&=\frac{Q^2}{8\pi\varepsilon_0}\left(\frac1a-\frac1b\right)\end{aligned}$
Substitution: a finite inner radius avoids the point-source divergence.
$\begin{aligned}W_{a,b}&=\frac{(2.00\times10^{-9})^2}{8\pi\varepsilon_0}\left(\frac1{0.0200}-\frac1{0.0400}\right)\,\mathrm J\\&\approx0.449378\,\mathrm{\mu J}\\W_{a,\infty}&=\frac{Q^2}{8\pi\varepsilon_0a}\approx0.898755\,\mathrm{\mu J}\end{aligned}$

Result: The shell contains about 0.449 μJ, half the total exterior field energy.

Check: Check: sphere capacitance $C=4\pi\varepsilon_0a$ gives $\frac{Q^2}{2C}=W_{a,\infty}$ , not just shell energy. This small shell holds half the energy because density increases strongly towards the inner radius.

What to take away

Square the actual total field and integrate density over the selected volume. Squaring a mean field does not replace averaging its square. Vacuum density alone is neither a material-energy balance nor transported power. Also specify whether electric or total electromagnetic energy is required.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Electric-field energy density in vacuum. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrische-feldenergiedichte

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Frequently asked questions about Electric-field energy density in vacuum

What does Electric-field energy density in vacuum describe physically?+

The vacuum relation describes electric field energy per local volume. It is proportional to squared field magnitude; field direction does not introduce a sign into density. A 100000 V/m field has approximately 0.04427 J/m³ of electric energy density. This is neither energy alone nor power. A volume integral is required to obtain energy. With a spatially constant field, density multiplied by volume suffices; otherwise integrate local density. A simultaneous magnetic field has an additional energy contribution. Electric density must therefore not be interpreted as the total electromagnetic energy density without checking the physical field configuration.

How can I check a calculation involving Electric-field energy density in vacuum?+

First find field magnitude in the selected vacuum volume and square it. Multiply by half the vacuum permittivity. In the plate example, 400 V across 2 mm gives 200000 V/m and approximately 0.177084 J/m³ density. The gap volume from 0.00600 m² area and 2 mm separation is 1.20×10⁻⁵ m³. This yields about 2.125 µJ of electric energy. Independently check with geometric capacitance and capacitor energy. Both routes must agree under identical assumptions. With a varying field, density at the boundary multiplied by total volume would instead fail to give the correct integral.

What assumptions and limits apply to Electric-field energy density in vacuum?+

The stated local density applies to the electric component of a vacuum field. For a linear reversible isotropic dielectric, material polarization enters an adjusted energy balance; arbitrarily replacing the vacuum constant is insufficient for every real material. Nonlinear or lossy media require further characteristics. For ideal point charges, classical self-energy integrated down to the origin diverges, so the formula does not produce a finite self-energy for a point electron. A finite selected volume can nevertheless contain finite energy. Clearly distinguish that partial energy from the total energy of the complete source configuration when defining the calculation.

How do I check units and scale for Electric-field energy density in vacuum?+

Vacuum permittivity has units of farads per metre, and electric field has volts per metre. Multiplying permittivity by field squared gives joules per cubic metre. Doubling field magnitude quadruples electric density; halving it leaves one quarter. At 50000 V/m, the result is approximately 0.01107 J/m³, one quarter of the value at 100000 V/m. A negative field component does not make density negative. Total energy additionally requires a volume factor in cubic metres. Density and total energy have different units and must therefore not be compared directly as if they were the same physical quantity.

When should I combine Electric-field energy density in vacuum with another relation?+

To obtain total energy, combine local density with a volume integral and the actual field distribution. For the second example’s linear profile, mean squared field is one third of maximum field squared. Squaring the mean field instead gives only one quarter and underestimates energy by 25 percent. Order matters: square first, then average spatially. With plates, capacitance and voltage provide an additional check. A single travelling plane vacuum wave also has magnetic energy; its electric and magnetic densities are equal. A standing superposition requires evaluating its total fields separately. A static electrical arrangement does not automatically possess that equally large additional magnetic contribution to its energy.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §8.3: Energy Stored in a Capacitor

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Faraday’s law of induction

\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

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Symbol

Meaning

Unit

uE

Electric energy density

J/m³

ε0

Vacuum permittivity

F/m

E

Magnitude of electric field

V/m

Derivation & background

Exam blueprint

Validity range

Derivation steps

The uniform plate model gives electric energy per field volume; general fields require local integration.

1Consider large parallel vacuum plates with area A, separation d and uniform field magnitude E.
2Capacitance is $C=\varepsilon_0\frac Ad$ .
3Voltage and field magnitudes in the uniform gap obey $U=Ed$ .
4Insert both into linear stored energy $W_C=\frac12 CU^2$ .
5The result is $W_C=\frac12\varepsilon_0E^2Ad$ ; gap volume is $Ad$ .
6Dividing by that volume gives $u_E=\frac12\varepsilon_0E^2$ .
7The general vacuum expression belongs to electromagnetic energy balance; integrate local squared fields, not the square of mean field.

How do you calculate with Electric-field energy density in vacuum?

Here is how to work through a typical Electric-field energy density in vacuum (uE = 1/2 ε₀E²) task step by step:

Task

A uniform vacuum field has $E=1.0\times10^5\,\mathrm{V/m}$ . Find electric energy density. How does it change if field magnitude is halved?

Solution path

Approach: $u_E=\frac{\varepsilon_0E^2}{2}$ .
Rounded CODATA permittivity gives $u_E\approx0.0443\,\mathrm{J/m^3}$ .
Halving E yields $u_{E,2}=\frac14u_{E,1}\approx0.0111\,\mathrm{J/m^3}$ .

Check: density is not energy in joules; the selected volume is still required.

Understand and apply the equation

What does electric energy density in vacuum describe?

u_E(\mathbf r)=\frac12\varepsilon_0|\mathbf E(\mathbf r)|^2

$u_E$ is the electric component of vacuum energy density; a simultaneous magnetic field has its own contribution.[1][2]

Obtain density from ideal plate energy

\begin{aligned}C&=\varepsilon_0\frac Ad\\U&=Ed\\W_C&=\frac12 CU^2=\frac12\varepsilon_0E^2Ad\end{aligned}

For the ideal uniform vacuum gap, $Ad$ is the field volume.[1]

W_E(V)=\int_V\frac12\varepsilon_0|\mathbf E|^2\,d^3r

Only constant field magnitude permits density to be taken outside the integral; otherwise evaluate it locally.[1]

Do not confuse volume density with radiation flux

\begin{aligned}[\varepsilon_0E^2]&=\frac{\mathrm F}{\mathrm m}\frac{\mathrm{V^2}}{\mathrm{m^2}}=\frac{\mathrm J}{\mathrm{m^3}}\\E&=\sqrt{\frac{2u_E}{\varepsilon_0}}\end{aligned}

The rearrangement yields only nonnegative field magnitude, not direction.

Uniform field and spatial variation

Energy in a vacuum plate gap

Given: Large parallel plates: $A=0.00600\,\mathrm{m^2}$ , $d=2.00\,\mathrm{mm}$ , $U=400\,\mathrm V$ . Ideal uniform vacuum gap, fringe fields neglected.
Find: Find E, $u_E$ , gap volume and stored energy.

Approach: A uniform vacuum field has the same electric energy density throughout the gap, permitting multiplication by the actual gap volume.
$\begin{aligned}E&=\frac Ud\\u_E&=\frac12\varepsilon_0E^2\\W_E&=u_EAd\end{aligned}$
Find field and volume from geometry.
$\begin{aligned}E&=\frac{400\,\mathrm V}{0.00200\,\mathrm m}=2.00\times10^5\,\mathrm{V/m}\\V_{\mathrm{gap}}&=Ad=1.20\times10^{-5}\,\mathrm{m^3}\end{aligned}$
Multiply density by volume.
$\begin{aligned}u_E&\approx0.1770837564\,\mathrm{J/m^3}\\W_E&=u_EV_{\mathrm{gap}}\approx2.125005077\,\mathrm{\mu J}\end{aligned}$

Result: Local density ≈ 0.177084 J/m³ and gap energy ≈ 2.12501 μJ. Model surface area is not confused with volume.

Check: Independently: $C=\varepsilon_0\frac Ad$ ≈ 26.5626 pF. $\frac12CU^2$ gives 2.12501 μJ. Both calculations describe the same uniform gap energy, not separate terms to be added.

A linear vacuum field over a selected region

Given: A prescribed quasistatic model has $E(x)=E_0\frac xL$ along +x for $0\le x\le L$ . $E_0=30000\,\mathrm{V/m}$ , $L=0.0300\,\mathrm m$ and constant cross-section $A=0.0120\,\mathrm{m^2}$ . Required source distribution is prescribed; calculate only energy in this vacuum region.
Find: Integrate energy density and compare with incorrectly squaring mean field.

Approach: Square the prescribed linear field locally first. Constant cross-section then reduces the volume integral to a position integral.
$\begin{aligned}u_E(x)&=\frac12\varepsilon_0E_0^2\frac{x^2}{L^2}\\W_E&=A\int_0^Lu_E(x)\,\mathrm dx\end{aligned}$
Local density grows quadratically with x.
$u_E(x)=\frac12\varepsilon_0E_0^2\frac{x^2}{L^2}$
Here the volume integral becomes A times the x integral.
$\begin{aligned}W_E&=A\int_0^L u_E(x)\,dx\\&=\frac{\varepsilon_0E_0^2AL}{6}\\&\approx0.4781261422\,\mathrm{\mu J}\end{aligned}$
Correct mean of E² differs from square of mean E.
$\begin{aligned}\langle E^2\rangle&=\frac{E_0^2}{3}\\\langle E\rangle^2&=\frac{E_0^2}{4}\\W_{\mathrm{mean}}&=\frac34W_E\end{aligned}$

Result: Regional energy is ≈ 0.478126 μJ. Substituting mean field $15000\,\mathrm{V/m}$ would underestimate it by 25%.

Why must you square the field before spatial averaging?

Linear field, quadratically growing energy density

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Original curve for the second example, with local density divided by its endpoint value. Mean is one third rather than one quarter: squaring mean field would underestimate energy.[1][5]

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|\mathbf E_1+\mathbf E_2|^2=|\mathbf E_1|^2+|\mathbf E_2|^2+2\mathbf E_1\cdot\mathbf E_2

The cross term explains why field energy is not merely a sum of positive individual self-energies.

Explore further: Field energy in reversible dielectrics

Five density and volume exercises

Calculate · Foundation

A vacuum field has magnitude $E=50000\,\mathrm{V/m}$ . Find $u_E$ .

Show hint

Square field magnitude and multiply by $\frac{\varepsilon_0}2$ .

Learning goal: Calculate vacuum energy density from field magnitude.

Show solution and calculation

Approach: $u_E=\frac{\varepsilon_0E^2}{2}$ .
$u_E$ ≈ 0.0110677 J/m³.
It is one quarter of the upper 10⁵ V/m example because field is halved.

\begin{aligned}u_E&=\frac{\varepsilon_0}{2}(50000\,\mathrm{V/m})^2\\&\approx0.0110677\,\mathrm{J/m^3}\end{aligned}

[1]

Rearrange · Application

Uniform electric energy density 0.0200 J/m³ fills the selected vacuum volume 8.00×10⁻⁴ m³. Find energy and field magnitude.

Show hint

Energy is density times volume; E requires a separate square root.

Learning goal: Infer energy and field magnitude from density and volume.

Show solution and calculation

Approach: $W_E=u_EV$ .
$W_E=1.60\cdot10^{-5}\,\mathrm J=16.0\,\mathrm{\mu J}$ .
From $E=\sqrt{\frac{2\cdot0.0200\,\mathrm{J/m^3}}{\varepsilon_0}}$ , magnitude is approximately 67213.4 V/m.
Field direction remains unknown.

\begin{aligned}W_E&=(0.0200)(8.00\times10^{-4})\,\mathrm J=16.0\,\mathrm{\mu J}\\E&=\sqrt{\frac{2u_E}{\varepsilon_0}}\approx6.72134\times10^4\,\mathrm{V/m}\end{aligned}

[1]

Interpret · Application

Two equal vacuum subvolumes have fields $+E_0$ and $-E_0$ with $E_0=20000\,\mathrm{V/m}$ . What is the volume-averaged energy density?

Show hint

Both local squares stay positive.

Learning goal: Distinguish mean squared field from square of mean field.

Show solution and calculation

Approach: $\langle u_E\rangle=\frac{\varepsilon_0}{2}\langle E^2\rangle$ .
Mean field vector is zero, but mean energy density is $\frac12\varepsilon_0 E_0^2$ ≈ 0.00177084 J/m³.
Averaging field before squaring would incorrectly give zero.

\begin{aligned}\langle u_E\rangle&=\frac12\left[\frac{\varepsilon_0E_0^2}{2}+\frac{\varepsilon_0(-E_0)^2}{2}\right]\\&=\frac{\varepsilon_0}{2}(20000\,\mathrm{V/m})^2\\&\approx0.00177084\,\mathrm{J/m^3}\end{aligned}

[1]

Interpret · Application

Two uniform fields of magnitude $E_0$ overlap in the same vacuum volume. Compare energy density for parallel co-directed, perpendicular and antiparallel fields with $u_0=\frac12\varepsilon_0E_0^2$ . These are not separate subvolumes.

Show hint

The squared sum vector includes a cross term.

Learning goal: Add overlapping fields vectorially before squaring.

Show solution and calculation

Approach: $u_E=\frac{\varepsilon_0}{2}|\mathbf E_1+\mathbf E_2|^2$ .
Co-directed fields give magnitude $2E_0$ and density $4u_0$ .
Perpendicular fields give squared magnitude $2E_0^2$ and density $2u_0$ .
Antiparallel fields cancel, so total electric-field energy density vanishes.

Check: the cross term is positive, zero or negative. Simply adding separate densities works for the perpendicular case, not general spatial overlap.

\begin{aligned}u_E&=\frac12\varepsilon_0|\mathbf E_1+\mathbf E_2|^2\\&=\frac12\varepsilon_0\left(E_1^2+E_2^2+2\mathbf E_1\cdot\mathbf E_2\right)\end{aligned}

[1]

Error analysis · Challenge

An ideal isolated parallel-plate capacitor keeps the same free plate charge. A linear dielectric with $\varepsilon=4\varepsilon_0$ completely replaces vacuum, geometry unchanged, fringing neglected. A solution claims: “ $u=\frac12\varepsilon E^2$ , so density quadruples.” Check it.

Show hint

Fixed free charge fixes D, not E.

Learning goal: Account for boundary conditions when permittivity changes.

Show solution and calculation

Approach: $D=\frac{Q_{\mathrm{free}}}{A}=\varepsilon E$ .
$D=\varepsilon E$ stays constant for fixed free plate charge and area.
Thus E falls to one quarter and $u=\frac{D^2}{2\varepsilon}$ also falls to one quarter.

\begin{aligned}E_2&=\frac{D}{4\varepsilon_0}=\frac{E_1}{4}\\\frac{u_2}{u_1}&=\frac{4\varepsilon_0(E_1/4)^2}{\varepsilon_0E_1^2}=\frac14\end{aligned}

[3][1]

Ideal point sources expose a self-energy limit

Example: energy in a spherical shell

Finite energy between two radii

Given: A conducting sphere of radius $a=0.0200\,\mathrm m$ carries $Q=2.00\,\mathrm{nC}$ . Other sources vanish. Find only vacuum field energy from a to $b=0.0400\,\mathrm m$ , not all energy outside the sphere.
Find: Find shell energy and its share of total exterior field energy.

Approach: spherical symmetry gives the radial field; a volume shell has size $4\pi r^2\,\mathrm dr$ .
$\begin{aligned}E(r)&=\frac{Q}{4\pi\varepsilon_0r^2}\\W_{a,b}&=\int_a^b\frac{\varepsilon_0E^2(r)}2\,4\pi r^2\,\mathrm dr\\&=\frac{Q^2}{8\pi\varepsilon_0}\left(\frac1a-\frac1b\right)\end{aligned}$
Substitution: a finite inner radius avoids the point-source divergence.
$\begin{aligned}W_{a,b}&=\frac{(2.00\times10^{-9})^2}{8\pi\varepsilon_0}\left(\frac1{0.0200}-\frac1{0.0400}\right)\,\mathrm J\\&\approx0.449378\,\mathrm{\mu J}\\W_{a,\infty}&=\frac{Q^2}{8\pi\varepsilon_0a}\approx0.898755\,\mathrm{\mu J}\end{aligned}$

Result: The shell contains about 0.449 μJ, half the total exterior field energy.

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
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Electric-field energy density in vacuum

Variables & units

Electric field energy density calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrostatics

Worked example

Applications

How do you calculate with Electric-field energy density in vacuum?

What does electric energy density in vacuum describe?

Obtain density from ideal plate energy

Do not confuse volume density with radiation flux

Uniform field and spatial variation

Energy in a vacuum plate gap

A linear vacuum field over a selected region

Why must you square the field before spatial averaging?

Linear field, quadratically growing energy density

Explore further: Field energy in reversible dielectrics

Five density and volume exercises

A vacuum field has magnitude E=50000 V/mE=50000\,\mathrm{V/m}E=50000V/m. Find uEu_EuE​.

Uniform electric energy density 0.0200 J/m3 fills the selected vacuum volume 8.00×10−4 m3. Find energy and field magnitude.

Two equal vacuum subvolumes have fields +E0+E_0+E0​ and −E0-E_0−E0​ with E0=20000 V/mE_0=20000\,\mathrm{V/m}E0​=20000V/m. What is the volume-averaged energy density?

Two uniform fields of magnitude E0E_0E0​ overlap in the same vacuum volume. Compare energy density for parallel co-directed, perpendicular and antiparallel fields with u0=12ε0E02u_0=\frac12\varepsilon_0E_0^2u0​=21​ε0​E02​. These are not separate subvolumes.

Ideal point sources expose a self-energy limit

Example: energy in a spherical shell

Finite energy between two radii

What to take away

Sources for this explanation

Frequently asked questions about Electric-field energy density in vacuum

Subject references and further reading

More formulas: Electricity and magnetism

Retain Electric-field energy density in vacuum for exams

Electric-field energy density in vacuum

Variables & units

Electric field energy density calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrostatics

Worked example

Applications

How do you calculate with Electric-field energy density in vacuum?

What does electric energy density in vacuum describe?

Obtain density from ideal plate energy

Do not confuse volume density with radiation flux

Uniform field and spatial variation

Energy in a vacuum plate gap

A linear vacuum field over a selected region

Why must you square the field before spatial averaging?

Linear field, quadratically growing energy density

Explore further: Field energy in reversible dielectrics

Five density and volume exercises

A vacuum field has magnitude E=50000 V/mE=50000\,\mathrm{V/m}E=50000V/m. Find uEu_EuE​.

Uniform electric energy density 0.0200 J/m3 fills the selected vacuum volume 8.00×10−4 m3. Find energy and field magnitude.

Two equal vacuum subvolumes have fields +E0+E_0+E0​ and −E0-E_0−E0​ with E0=20000 V/mE_0=20000\,\mathrm{V/m}E0​=20000V/m. What is the volume-averaged energy density?

Two uniform fields of magnitude E0E_0E0​ overlap in the same vacuum volume. Compare energy density for parallel co-directed, perpendicular and antiparallel fields with u0=12ε0E02u_0=\frac12\varepsilon_0E_0^2u0​=21​ε0​E02​. These are not separate subvolumes.

Ideal point sources expose a self-energy limit

Example: energy in a spherical shell

Finite energy between two radii

What to take away

Sources for this explanation

Frequently asked questions about Electric-field energy density in vacuum

Subject references and further reading

More formulas: Electricity and magnetism

Retain Electric-field energy density in vacuum for exams

A vacuum field has magnitude $E=50000\,\mathrm{V/m}$ . Find $u_E$ .

Uniform electric energy density 0.0200 J/m³ fills the selected vacuum volume 8.00×10⁻⁴ m³. Find energy and field magnitude.

Two equal vacuum subvolumes have fields $+E_0$ and $-E_0$ with $E_0=20000\,\mathrm{V/m}$ . What is the volume-averaged energy density?

Two uniform fields of magnitude $E_0$ overlap in the same vacuum volume. Compare energy density for parallel co-directed, perpendicular and antiparallel fields with $u_0=\frac12\varepsilon_0E_0^2$ . These are not separate subvolumes.

A vacuum field has magnitude $E=50000\,\mathrm{V/m}$ . Find $u_E$ .

Uniform electric energy density 0.0200 J/m³ fills the selected vacuum volume 8.00×10⁻⁴ m³. Find energy and field magnitude.

Two equal vacuum subvolumes have fields $+E_0$ and $-E_0$ with $E_0=20000\,\mathrm{V/m}$ . What is the volume-averaged energy density?

Two uniform fields of magnitude $E_0$ overlap in the same vacuum volume. Compare energy density for parallel co-directed, perpendicular and antiparallel fields with $u_0=\frac12\varepsilon_0E_0^2$ . These are not separate subvolumes.