Physics · Electrostatics / field theory

Gauss’s law for the electric field

Q: What does Gauss’s law say about the electric field?

The law links total electric flux through a closed surface to the algebraic net charge inside it. Flux uses the outward normal component of the full electric field. Enclosed positive and negative charges are summed with their signs. External charges can also influence the field on the surface but contribute no net flux through the complete closed boundary. For 2 nC of enclosed charge, vacuum flux is approximately 226 N·m²/C. This single integral value does not determine every local field vector. Additional symmetry, boundary conditions or a complete solution of the field equations is required.

Q: When does Gauss’s law help me calculate an electric field?

Field calculation becomes particularly simple when sources and boundary conditions provide sufficient symmetry. For one point charge in vacuum, the field on a centred sphere is radial and has equal magnitude everywhere. Flux is therefore the signed radial component times spherical area: $\Phi_E=E_r\,4\pi r^2$. This gives the Coulomb field, with negative $E_r$ for negative charge. Inside a uniformly charged sphere, use only the charge enclosed up to the selected radius. An arbitrarily chosen cube does not have this field constancy. Gauss’s law still applies, but you cannot divide total flux by an area without justification and claim that the result is a local field value.

Q: Why must the Gaussian surface be closed?

A closed surface is the complete boundary of a volume and separates inside from outside. Only then is enclosed charge unambiguously defined and the outward flux balance complete. An open surface can certainly have electric flux, but it depends on position and orientation rather than solely on a charge behind it. Consider a uniform field passing through a charge-free cube: flux enters through one face and an equal amount leaves through the opposite face. Only summing all faces gives zero. The Gaussian surface remains an imaginary mathematical boundary, not an additional metal shell that changes the physical configuration.

Q: How do I calculate the flux of a sphere around a point charge?

Centre the sphere on the point charge and keep charge off the mathematical surface. Its radial field has equal magnitude everywhere. It is parallel to the outward normal for positive charge and opposite to that normal for negative charge. Flux is positive for a positive charge and negative for a negative charge. Field magnitude and spherical area have opposite radius dependences that cancel in their product. For 1 nC, every such radius gives approximately 112.94 N·m²/C. Doubling radius quadruples area and reduces field magnitude to one quarter. Additional external charges can destroy local symmetry without changing the total flux associated with the enclosed net charge.

Q: Does zero total flux mean the electric field is zero everywhere?

No. Zero total flux means only that incoming and outgoing normal components balance over the complete closed surface. A uniform external field passing through a charge-free cube is a simple counterexample: the field is nonzero everywhere, while opposite faces give opposite fluxes. Equally large positive and negative enclosed charges also have zero net charge but can still produce a spatially varying field. Inferring an actual zero field requires further physical assumptions. Inside a conductor in electrostatic equilibrium the field is zero, for example; that conclusion also relies on the equilibrium of mobile charges, not solely on zero net enclosed charge.

The electric flux through a closed surface depends only on the total charge it encloses. With sufficient symmetry, the law makes the electric field especially easy to calculate.

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Formula

Calculate

\oint_{\partial V} E \cdot d A = \frac{Q _{innen}}{ε _{0}}

E in N/C · area in m² · Q in C · electric flux in N·m²/C

Variables & units

Variables & units
Symbol	Meaning	Unit
$E$	Electric field on the closed surface	N/C
$dA$	Outward-directed area element	m²
$Q_{\mathrm{innen}}$	Algebraic sum of the enclosed charges	C
$ε₀$	Vacuum permittivity	C²/(N·m²)

LaTeX

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

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Imaginary Gaussian cube in a rightward uniform field. Flux enters left and leaves right; enclosed charge and net flux are zero. — A closed imaginary cube surface in a uniform external field encloses no charge. Fluxes through opposing faces cancel; the field itself is not zero.

Gauss's law calculator

Find the total electric flux through a closed surface from the enclosed charge, or the other way round, in vacuum.

Find

Enclosed net charge Q

Sum of the charges inside the surface with signs, e.g. +3 nC and −2 nC give 1 nC; outside charges do not count.

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

Working · rearrange for ΦE, substitute, evaluate

1. Rearrange for ΦE
$\Phi_E=\frac{Q}{\varepsilon_0}$
2. Substitute values (in SI units)
- Q = 1 nC = 1 · 10⁻⁹ C
$\Phi_E=\frac{1\times10^{-9}}{8.85419\times10^{-12}}=112.941$
3. Result
ΦE = 112.9 N·m²/C

Using ε₀ ≈ 8.8541878188·10⁻¹² F/m (CODATA 2022, NIST; standard uncertainty 1.4·10⁻²¹ F/m); k = 1/(4πε₀). The constant and displayed result are approximations.

Result: Total electric flux ΦE

ΦE = 112.9 N·m²/C

With Q = 1 nC, ΦE = 112.9 N·m²/C.

Quick tips

The shape and size of the surface do not matter; only the enclosed net charge counts. Charges outside contribute zero net flux.
Estimate: 1/ε₀ ≈ 1.13 · 10¹¹ N·m²/C². So each nanocoulomb inside the surface gives about 113 N·m²/C of flux.

Assumptions and limits

Gauss’s law in vacuum: oriented total flux through a closed surface from its enclosed net charge. A local electric field is not inferred from flux; that requires geometry and symmetry.

Derivation & background

Flux $\Phi_E=\oint\mathbf E\cdot d\mathbf A$ links the complete closed boundary to enclosed net charge. External sources affect the local field but not net flux. A local field value requires additional symmetry. Inside matter, the E-field form remains valid using total free and bound charge. The macroscopic D-field form instead uses free charge; for a linear isotropic material, $\mathbf D=\varepsilon\mathbf E$ .

Exam blueprint

Validity range

Electric Gauss balance holds for closed surfaces even without symmetry. The E-field uses enclosed total charge, free plus bound; a D-field balance uses free charge. Avoid charge on the mathematical surface.

Derivation steps

The Coulomb field and oriented solid angle motivate closed flux balance; symmetry is an additional aid.

1Choose the complete closed boundary of a volume and orient normals outwards.
2Flux integrates only the normal component of the complete electric field.
3For a point source, a small flux contribution is proportional to the oriented solid angle of the surface.
4An enclosed source has total solid angle 4π; an external source gives zero total contribution.
5Linear field superposition sums enclosed charges algebraically, including their signs.
6This gives $\Phi_E=\frac{Q_{\mathrm{tot,in}}}{\varepsilon_0}$ ; inside matter both free and bound charge count.
7Only additional spherical, cylindrical or planar symmetry permits extracting a local field from the integral.

Rearrangements

Find electric flux

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

The surface must be closed; dA points outward.

Field of a centred point charge

E_r(r)=\frac{q}{4\pi\varepsilon_0r^2}

A single central vacuum charge, r>0, without other sources. Negative q gives negative Er; magnitude is |Er|.

Task variant

A spherical surface of r=0.10 m encloses a single point charge q=+2.0 nC at its centre in vacuum. Find total flux, field magnitude and direction at the surface.

Approach: $\Phi_E=\frac q{\varepsilon_0}$ .
The central source also permits $E_r4\pi r^2=\Phi_E$ .
Thus $\Phi_E\approx225.882\,\mathrm{N\,m^2\,C^{-1}}$ and $E_r=\frac{q}{4\pi\varepsilon_0r^2}\approx1797.51\,\mathrm{N\,C^{-1}}$ , radially outwards.

Check: field times sphere area reproduces flux. Without the centred-source assumption, only total flux is fixed.

Common mistakes

Concluding from zero total flux that the field is zero everywhere, or using an open surface.

The law counts net flux through a closed surface. Incoming and outgoing field lines can cancel even when E is nonzero locally.

Exam context

Undergraduate electrostatics: spherical, cylindrical and planar symmetry, conductors in equilibrium and flux arguments.

Formula cluster

Electric field and flux

Gauss connects the Coulomb field with a global flux balance and simplifies symmetric field problems.

F = ke·|q₁q₂|/r²

Coulomb's Law

E⃗ = F⃗el/q; Ehom = |U|/d

Electric Field Strength

C = Q/U

Capacitance of a Capacitor

Worked example

A sphere encloses $q=2.0\,\mathrm{nC}$ : $\Phi_E=\frac q{\varepsilon_0}\approx226\,\mathrm{N\,m^2/C}$ . For a single central charge in vacuum at $r=0.10\,\mathrm m$ , $E=\frac{|q|}{4\pi\varepsilon_0r^2}\approx1.80\,\mathrm{kN/C}$ .

Applications

Electric fields of spherical, cylindrical and planar charge distributions, conductors in electrostatic equilibrium and the transition to Maxwell theory

How do you calculate with Gauss's Law for Electrostatics?

Here is how to work through a typical Gauss's Law for Electrostatics (∮ E·dA = Q_innen / ε₀) task step by step:

1
Task
A spherical surface of r=0.10 m encloses a single point charge q=+2.0 nC at its centre in vacuum. Find total flux, field magnitude and direction at the surface.
Solution path
1. Approach: $\Phi_E=\frac q{\varepsilon_0}$ .
2. The central source also permits $E_r4\pi r^2=\Phi_E$ .
3. Thus $\Phi_E\approx225.882\,\mathrm{N\,m^2\,C^{-1}}$ and $E_r=\frac{q}{4\pi\varepsilon_0r^2}\approx1797.51\,\mathrm{N\,C^{-1}}$ , radially outwards.
Check: field times sphere area reproduces flux. Without the centred-source assumption, only total flux is fixed.

Understand and apply the equation

The net electric flux leaving a closed volume depends on its enclosed net charge. Gauss’s law balances the whole boundary. It determines field at an individual location only when the source arrangement also provides suitable symmetry.

Imagine a boundary around the charges

Imagine a closed boundary around some charges. At some places the field points outwards; elsewhere it may point inwards or along the surface. Flux counts outward normal components positively and inward components negatively. Every surface patch must be included. The balance depends on enclosed positive minus negative charge; one patch alone is insufficient.[1]

Prior knowledge: electric field is a vector. Its dot product with the outward surface normal selects the normal component.
A Gaussian surface is selected for calculation. It adds no conductor and does not change the original field.
Separate total flux from local field. The law answers the first question without symmetry.

[1]

What does Gauss’s law state, and what are the units of flux?

Total electric flux through a closed surface equals enclosed net charge divided by $\varepsilon_0$ . Use the full field and outward normal. Electric flux is measured in $\mathrm{N\,m^2/C}$ ; it is not magnetic flux.[1]

\Phi_E=\oint_{\partial V}\mathbf E\cdot\hat{\mathbf n}\,dA=\frac{Q_{\mathrm{tot,in}}}{\varepsilon_0}

Normal points outward; enclosed positive and negative charges are summed algebraically.[1]

A Gaussian surface is a mathematical surface, not an added conducting shell. The integrand uses the full field on it, including contributions from external charges. Those sources affect local field values but give no net flux through the complete closed surface. Zero enclosed net charge therefore means zero total flux, not necessarily zero field.[1]

How do you calculate electric flux through a surface?

d\Phi_E=E\cos\theta\,dA

θ is the angle between local field and outward normal, not between field and the surface.[1]

Outward contribution is positive, inward contribution negative and tangential contribution zero. Only a constant normal projection across a surface part permits its integral to become projection times area. An open measurement rectangle may have field flux, but that flux is not automatically $\frac{Q_{\mathrm{in}}}{\varepsilon_0}$ because an open surface has no enclosed interior.[1]

Explore further: solid angle and local form

\begin{aligned}d\Phi_E&=\frac{q}{4\pi\varepsilon_0}\,d\Omega\\\oint d\Omega&=4\pi\quad(\mathbf r_q\in V)\end{aligned}

For a point source, signed solid-angle projection connects inverse-square field to surface-area scaling.[1]

Coulomb’s law supplies the point-charge field used here. For a central point charge on a sphere the check is direct: $E_r=\frac{q}{4\pi\varepsilon_0r^2}$ , area $4\pi r^2$ and unit normal projection give $\frac q{\varepsilon_0}$ . More generally, oriented solid-angle balance over a closed surface is 4π for an enclosed point source and 0 for an external one. Vector superposition sums all contributions. This establishes the law within the Coulomb model rather than proving a universal natural fact from the flux definition alone. Avoid charges exactly on the chosen surface in this simple source-counting argument.[1]

\nabla\cdot\mathbf E=\frac{\rho_{\mathrm{tot}}}{\varepsilon_0}

The local form follows from the divergence theorem for suitably smooth fields; ideal point sources require a distributional interpretation.[1]

How do you choose a Gaussian surface to calculate the field?

A convenient surface does not create symmetry. First inspect the charge distribution and other sources. Only when these are, for example, spherically symmetric is radial field constant on a concentric sphere. Otherwise total flux is still known, but varying field cannot be pulled outside the integral as a common value.[2]

Spherical symmetry about a centre gives a radial field of equal magnitude on each concentric sphere. The electric field of an infinite line charge has cylindrical symmetry: it gives radial side flux, while end-cap flux is zero. The field is tangential to those faces; its magnitude need not be zero. Planar symmetry of an ideal infinite sheet gives a normal field on both sides. Merely drawing a spherical Gaussian surface does not make an arbitrary charge distribution spherically symmetric.[2]

\begin{aligned}E_r(r)\,4\pi r^2&=\frac{Q_{\mathrm{in}}(r)}{\varepsilon_0}\\E_r(r)&=\frac{Q_{\mathrm{in}}(r)}{4\pi\varepsilon_0r^2}\end{aligned}

$E_r$ here is a signed radial component. Enclosed charge may depend on radius.[2]

Interior field of a uniformly charged insulating sphere

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For the negative charge in the second example field points inwards. Scale distance by sphere radius R and field by its positive surface magnitude. Charge is uniformly distributed through the volume; an equilibrium conducting sphere has a different interior field.[2]

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Sphere and cube: distinguish charge, field and flux

Zero enclosed net charge does not necessarily mean zero field: the third example has a uniform field whose flux contributions cancel. The charged sphere in the second example is insulating with fixed volume charge; a conductor in electrostatic equilibrium has a different interior field model.[1][2]

Flux through a cube without field symmetry

Given: A cube of side 0.120 m encloses fixed point charges +3.00 nC and −0.500 nC at different interior locations. No charge lies on a face; external sources may be present. Vacuum model.
Find: Find total flux and state whether this determines E on every face.

Approach: Total flux requires only algebraic enclosed charge; cube geometry without symmetry does not supply local field values.
$\begin{aligned}Q_{\mathrm{in}}&=\sum_iq_i\\\Phi_E&=\frac{Q_{\mathrm{in}}}{\varepsilon_0}\end{aligned}$
Sum interior sources with signs.
$Q_{\mathrm{in}}=(3.00-0.500)\,\mathrm{nC}=2.50\times10^{-9}\,\mathrm C$
Total flux is independent of cube shape and side length.
$\Phi_E=\frac{Q_{\mathrm{in}}}{\varepsilon_0}\approx282.35227\,\mathrm{N\,m^2/C}$

Result: $\Phi_E$ ≈282.352 N·m²/C. Without further symmetry this determines neither local field nor equal flux through the six faces.

Check: A larger closed surface enclosing the same charges has the same total flux. External sources do not change the sum even though they can change its distribution among faces.

Uniformly charged insulating model sphere

Given: A nonconducting sphere has fixed uniform volume charge density, radius $R=0.0800\,\mathrm m$ and total charge $Q=-6.00\,\mathrm{nC}$ . Vacuum environment and no other sources. Find fields at $r=0.0400\,\mathrm m$ and $r=0.160\,\mathrm m$ .
Find: Find enclosed charge and signed radial fields.

Approach: Uniform fixed volume charge and spherical symmetry give enclosed charge first, then the signed radial field component.
$\begin{aligned}Q_{\mathrm{in}}(r<R)&=Q\left(\frac rR\right)^3\\E_r(r)&=\frac{Q_{\mathrm{in}}(r)}{4\pi\varepsilon_0r^2}\end{aligned}$
Inside, enclosed charge grows with sphere volume.
$\begin{aligned}Q_{\mathrm{in}}(r)&=Q\left(\frac rR\right)^3\\Q_{\mathrm{in}}(0.0400\,\mathrm m)&=-0.750\,\mathrm{nC}\end{aligned}$
The interior Gauss solution is proportional to radius.
$\begin{aligned}E_r(r<R)&=\frac{Qr}{4\pi\varepsilon_0R^3}\\E_r(0.0400\,\mathrm m)&\approx-4212.915\,\mathrm{N/C}\end{aligned}$
Outside, all charge is enclosed.
$\begin{aligned}E_r(r\ge R)&=\frac Q{4\pi\varepsilon_0r^2}\\E_r(0.160\,\mathrm m)&\approx-2106.457\,\mathrm{N/C}\end{aligned}$

Result: Both fields point inward. At half the radius only one eighth of charge is enclosed; outside the inverse-square relation holds.

Check: At $r=R$ , both forms give the same value of about −8425.8298 N/C. Inside, $E_r\,4\pi r^2=\frac{-0.750\,\mathrm{nC}}{\varepsilon_0}$ . A conducting equilibrium sphere is a different model with zero interior field.

Flux with no enclosed charge

Given: A cube of side 4.00 cm lies in a uniform field $E=250\,\mathrm{N\,C^{-1}}$ along +x. It contains no charge; its faces are perpendicular to the coordinate axes.
Find: Find flux through the right and left faces and through the whole boundary.

Approach: each flat face has $\Phi=EA\cos\theta$ . Outward normals of the two x-faces point oppositely.
$\begin{aligned}\Phi_{+x}&=+EA\\\Phi_{-x}&=-EA\\\Phi_{\mathrm{other}}&=0\\\Phi_{\mathrm{closed}}&=\Phi_{+x}+\Phi_{-x}=0\end{aligned}$
Substitute: convert side length to metres before squaring.
$\begin{aligned}A&=(0.0400\,\mathrm m)^2=0.00160\,\mathrm{m^2}\\\Phi_{+x}&=250(0.00160)\,\mathrm{N\,m^2\,C^{-1}}=+0.400\,\mathrm{N\,m^2\,C^{-1}}\\\Phi_{-x}&=-0.400\,\mathrm{N\,m^2\,C^{-1}}\end{aligned}$

Result: Net flux is zero even though field magnitude is 250 N/C everywhere. The other four faces contribute zero because the field is tangent to them.

Check: Gauss independently gives $\frac{Q_{\mathrm{in}}}{\varepsilon_0}=0$ . Equal incoming and outgoing flux confirms the balance, not a vanishing field.

Which charge belongs in the flux laws for E and D?

For actual electric field E, Gauss’s law contains total enclosed charge, both free and bound. Do not simply replace $\varepsilon_0$ by $\varepsilon_r\varepsilon_0$ while still using the same total charge. In macroscopic material modelling, displacement field D satisfies the flux balance for free charge. Only with a suitable linear isotropic constitutive law does $\mathbf D=\varepsilon\mathbf E$ hold.[4][6]

Inside conducting material in electrostatic equilibrium, $\mathbf E=\mathbf0$ . A Gaussian surface there therefore encloses zero net charge. This does not mean every region associated with a conductor is charge free: excess charge can reside on surfaces and cavities may contain sources. The uniformly charged insulating example must not be confused with mobile charges in a conductor.[2]

Five flux and symmetry exercises

Calculate · Foundation

A closed surface encloses +4.00 nC and −7.00 nC. Find total flux.

Show hint

Add charges algebraically, not by magnitude.

Learning goal: Convert enclosed net charge to total flux.

Show solution and calculation

Model and approach: Add enclosed charges algebraically, not their magnitudes.
$\Phi_E=\frac{Q_++Q_-}{\varepsilon_0}$ $Q_{\mathrm{in}}$ =−3.00 nC.
$\Phi_E$ ≈−338.823 N·m²/C.
Negative sign means net inward flux.

Check: $\varepsilon_0\Phi_E=-3.00\,\mathrm{nC}$ .

\Phi_E=\frac{-3.00\times10^{-9}\,\mathrm C}{\varepsilon_0}\approx-338.823\,\mathrm{N\,m^2/C}

[1]

Interpret · Application

A central positive charge +1.00 nC lies inside concentric spheres of radii 0.0500 m and 0.100 m. Compare fields and total fluxes.

Show hint

Field scales as $\frac1{r^2}$ ; total flux counts charge.

Learning goal: Separate radius-dependent field from radius-independent total flux.

Show solution and calculation

Model and approach: The positive point charge is at the common centre, so field is radial.
$\begin{aligned}E(r)&=\frac{q}{4\pi\varepsilon_0r^2}\\\Phi_E&=\frac q{\varepsilon_0}\end{aligned}$ The larger sphere has one quarter of the field magnitude.
Both fluxes are ≈112.941 N·m²/C.
Field and flux are distinct quantities.

Check: $E_1\approx3595.02\,\mathrm{N\,C^{-1}},\quad E_2\approx898.755\,\mathrm{N\,C^{-1}}$ .

\begin{aligned}\frac{E_2}{E_1}&=\frac14\\\Phi_2&=\Phi_1\end{aligned}

[2]

Interpret · Application

A uniform external field points along +x through an empty cube. Is E zero because $Q_{\mathrm{in}}$ =0?

Show hint

Compare entry and exit faces.

Learning goal: Explain zero total flux despite nonzero field.

Show solution and calculation

Model and approach: Use outward normals, opposite on the two x-faces.
$\Phi_E=\sum_{\mathrm{faces}}\mathbf E\cdot\hat{\mathbf n}\,A$ No.
The two x faces contribute −EA and +EA; the other four have tangential field and zero flux.
Total flux is zero despite nonzero field.

\Phi_E=-EA+EA=0

[1]

Interpret · Application

A spherical surface encloses one point charge q. The charge moves from the sphere centre to another interior point without crossing the surface. Compare total flux before and after. May field still be treated as uniform and normal over the sphere?

Show hint

Gauss counts enclosed charge; pulling E outside the integral needs additional symmetry.

Learning goal: Recognize flux invariance without a uniform surface field.

Show solution and calculation

Approach: enclosed charge remains q, so total flux remains $\frac q{\varepsilon_0}$ .
Off centre, distances and angles vary across the surface; using $E_r\,4\pi R^2$ with one magnitude is unjustified.

Check: the charge has not crossed the surface, so the flux balance still applies. Individual surface regions can change contributions while the total stays fixed.

\Phi_E^{\mathrm{before}}=\Phi_E^{\mathrm{after}}=\frac q{\varepsilon_0},\qquad\Phi_E=\oint\mathbf E\cdot\hat{\mathbf n}\,dA

[1][2]

Error analysis · Challenge

A closed mathematical surface encloses +2.00 nC free and −0.50 nC net bound charge. A solution uses only $\frac{2.00\,\mathrm{nC}}{\varepsilon_0}$ for ∮E·dA. Correct using ε₀=8.8541878×10⁻¹² F/m. The requested quantity is E flux, not D flux.

Show hint

The electric Gauss law with ε₀ uses total enclosed charge.

Learning goal: Distinguish total charge in E flux from free charge.

Show solution and calculation

Approach: $Q_{\mathrm{tot}}=Q_{\mathrm{free}}+Q_{\mathrm{bound}}=1.50\,\mathrm{nC}$ .
E flux is about 169.411 N·m²/C, not 225.882 N·m²/C.

Check: $\Phi_E\varepsilon_0$ returns 1.50 nC. The displacement-field D balance instead counts free charge; its quantity and units cannot be equated to E flux. These net charges determine total flux but no local surface field without symmetry.

\begin{aligned}Q_{\mathrm{tot}}&=Q_{\mathrm{free}}+Q_{\mathrm{bound}}=1.50\,\mathrm{nC}\\\Phi_E&=\frac{Q_{\mathrm{tot}}}{\varepsilon_0}\\&=\frac{1.50\cdot10^{-9}\,\mathrm C}{8.8541878\cdot10^{-12}\,\mathrm{F/m}}\\&\approx169.411\,\mathrm{N\,m^2/C}\end{aligned}

[1][4]

Distinguish electric and magnetic Gauss laws

Electric Gauss balance remains a Maxwell law for asymmetric or time-varying sources; static symmetries do not automatically solve a time-dependent problem. An enclosed charge of 2.0 nC gives $\Phi_E=\frac{2.0\,\mathrm{nC}}{\varepsilon_0}\approx226\,\mathrm{N\,m^2/C}$ . For the sole central charge, field magnitude at 0.10 m is about 1.80 kN/C. Finite asymmetric systems additionally require source geometry, boundary conditions or a field solution; net flux must not be distributed uniformly over an arbitrary surface. Maxwell’s equations in integral form supplement electric charge accounting with the magnetic law: net $\mathbf B$ flux through a closed surface is zero. This likewise does not mean that the field vanishes everywhere.[1][2][7]

What to take away

Closed flux counts enclosed net charge; external sources can still change local field.
Zero net flux does not mean zero field. Incoming and outgoing contributions may cancel.
Calculating local field requires additional symmetry of the source arrangement. The E-field balance counts free plus bound charge.

[1][2][4]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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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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Change history

October 1, 2026Clarified total flux, units and symmetry; distinguished zero end-cap flux from zero field and made existing examples and links to line charge, Coulomb and Maxwell accessible.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

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Quanta. Gauss's Law for Electrostatics. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/gauss-gesetz-elektrisches-feld

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Frequently asked questions about Gauss's Law for Electrostatics

What does Gauss’s law say about the electric field?+

The law links total electric flux through a closed surface to the algebraic net charge inside it. Flux uses the outward normal component of the full electric field. Enclosed positive and negative charges are summed with their signs. External charges can also influence the field on the surface but contribute no net flux through the complete closed boundary. For 2 nC of enclosed charge, vacuum flux is approximately 226 N·m²/C. This single integral value does not determine every local field vector. Additional symmetry, boundary conditions or a complete solution of the field equations is required.

When does Gauss’s law help me calculate an electric field?+

Field calculation becomes particularly simple when sources and boundary conditions provide sufficient symmetry. For one point charge in vacuum, the field on a centred sphere is radial and has equal magnitude everywhere. Flux is therefore the signed radial component times spherical area: $\Phi_E=E_r\,4\pi r^2$ . This gives the Coulomb field, with negative $E_r$ for negative charge. Inside a uniformly charged sphere, use only the charge enclosed up to the selected radius. An arbitrarily chosen cube does not have this field constancy. Gauss’s law still applies, but you cannot divide total flux by an area without justification and claim that the result is a local field value.

Why must the Gaussian surface be closed?+

A closed surface is the complete boundary of a volume and separates inside from outside. Only then is enclosed charge unambiguously defined and the outward flux balance complete. An open surface can certainly have electric flux, but it depends on position and orientation rather than solely on a charge behind it. Consider a uniform field passing through a charge-free cube: flux enters through one face and an equal amount leaves through the opposite face. Only summing all faces gives zero. The Gaussian surface remains an imaginary mathematical boundary, not an additional metal shell that changes the physical configuration.

How do I calculate the flux of a sphere around a point charge?+

Centre the sphere on the point charge and keep charge off the mathematical surface. Its radial field has equal magnitude everywhere. It is parallel to the outward normal for positive charge and opposite to that normal for negative charge. Flux is positive for a positive charge and negative for a negative charge. Field magnitude and spherical area have opposite radius dependences that cancel in their product. For 1 nC, every such radius gives approximately 112.94 N·m²/C. Doubling radius quadruples area and reduces field magnitude to one quarter. Additional external charges can destroy local symmetry without changing the total flux associated with the enclosed net charge.

Does zero total flux mean the electric field is zero everywhere?+

No. Zero total flux means only that incoming and outgoing normal components balance over the complete closed surface. A uniform external field passing through a charge-free cube is a simple counterexample: the field is nonzero everywhere, while opposite faces give opposite fluxes. Equally large positive and negative enclosed charges also have zero net charge but can still produce a spatially varying field. Inferring an actual zero field requires further physical assumptions. Inside a conductor in electrostatic equilibrium the field is zero, for example; that conclusion also relies on the equilibrium of mobile charges, not solely on zero net enclosed charge.

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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]Ling, S. J., Sanny, J., & Moebs, W. University Physics Volume 2, §6.2: Explaining Gauss’s Law. OpenStax.

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U = R \cdot I

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P = U \cdot I

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Retain Gauss's Law for Electrostatics for exams

Create a curated FSRS exam set for ∮ E·dA = Q_innen / ε₀: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Gauss’s law for the electric field

The electric flux through a closed surface depends only on the total charge it encloses. With sufficient symmetry, the law makes the electric field especially easy to calculate.

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Symbol

Meaning

Unit

E

Electric field on the closed surface

N/C

dA

Outward-directed area element

m²

Q_{\mathrm{innen}}

Algebraic sum of the enclosed charges

ε₀

Vacuum permittivity

C²/(N·m²)

Derivation & background

Exam blueprint

Validity range

Derivation steps

The Coulomb field and oriented solid angle motivate closed flux balance; symmetry is an additional aid.

1Choose the complete closed boundary of a volume and orient normals outwards.
2Flux integrates only the normal component of the complete electric field.
3For a point source, a small flux contribution is proportional to the oriented solid angle of the surface.
4An enclosed source has total solid angle 4π; an external source gives zero total contribution.
5Linear field superposition sums enclosed charges algebraically, including their signs.
6This gives $\Phi_E=\frac{Q_{\mathrm{tot,in}}}{\varepsilon_0}$ ; inside matter both free and bound charge count.
7Only additional spherical, cylindrical or planar symmetry permits extracting a local field from the integral.

How do you calculate with Gauss's Law for Electrostatics?

Here is how to work through a typical Gauss's Law for Electrostatics (∮ E·dA = Q_innen / ε₀) task step by step:

Task

A spherical surface of r=0.10 m encloses a single point charge q=+2.0 nC at its centre in vacuum. Find total flux, field magnitude and direction at the surface.

Solution path

Approach: $\Phi_E=\frac q{\varepsilon_0}$ .
The central source also permits $E_r4\pi r^2=\Phi_E$ .
Thus $\Phi_E\approx225.882\,\mathrm{N\,m^2\,C^{-1}}$ and $E_r=\frac{q}{4\pi\varepsilon_0r^2}\approx1797.51\,\mathrm{N\,C^{-1}}$ , radially outwards.

Check: field times sphere area reproduces flux. Without the centred-source assumption, only total flux is fixed.

Imagine a boundary around the charges

Prior knowledge: electric field is a vector. Its dot product with the outward surface normal selects the normal component.
A Gaussian surface is selected for calculation. It adds no conductor and does not change the original field.
Separate total flux from local field. The law answers the first question without symmetry.

[1]

What does Gauss’s law state, and what are the units of flux?

\Phi_E=\oint_{\partial V}\mathbf E\cdot\hat{\mathbf n}\,dA=\frac{Q_{\mathrm{tot,in}}}{\varepsilon_0}

Normal points outward; enclosed positive and negative charges are summed algebraically.[1]

How do you calculate electric flux through a surface?

d\Phi_E=E\cos\theta\,dA

θ is the angle between local field and outward normal, not between field and the surface.[1]

Explore further: solid angle and local form

\begin{aligned}d\Phi_E&=\frac{q}{4\pi\varepsilon_0}\,d\Omega\\\oint d\Omega&=4\pi\quad(\mathbf r_q\in V)\end{aligned}

For a point source, signed solid-angle projection connects inverse-square field to surface-area scaling.[1]

\nabla\cdot\mathbf E=\frac{\rho_{\mathrm{tot}}}{\varepsilon_0}

The local form follows from the divergence theorem for suitably smooth fields; ideal point sources require a distributional interpretation.[1]

How do you choose a Gaussian surface to calculate the field?

\begin{aligned}E_r(r)\,4\pi r^2&=\frac{Q_{\mathrm{in}}(r)}{\varepsilon_0}\\E_r(r)&=\frac{Q_{\mathrm{in}}(r)}{4\pi\varepsilon_0r^2}\end{aligned}

$E_r$ here is a signed radial component. Enclosed charge may depend on radius.[2]

Interior field of a uniformly charged insulating sphere

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Sphere and cube: distinguish charge, field and flux

Flux through a cube without field symmetry

Given: A cube of side 0.120 m encloses fixed point charges +3.00 nC and −0.500 nC at different interior locations. No charge lies on a face; external sources may be present. Vacuum model.
Find: Find total flux and state whether this determines E on every face.

Approach: Total flux requires only algebraic enclosed charge; cube geometry without symmetry does not supply local field values.
$\begin{aligned}Q_{\mathrm{in}}&=\sum_iq_i\\\Phi_E&=\frac{Q_{\mathrm{in}}}{\varepsilon_0}\end{aligned}$
Sum interior sources with signs.
$Q_{\mathrm{in}}=(3.00-0.500)\,\mathrm{nC}=2.50\times10^{-9}\,\mathrm C$
Total flux is independent of cube shape and side length.
$\Phi_E=\frac{Q_{\mathrm{in}}}{\varepsilon_0}\approx282.35227\,\mathrm{N\,m^2/C}$

Result: $\Phi_E$ ≈282.352 N·m²/C. Without further symmetry this determines neither local field nor equal flux through the six faces.

Check: A larger closed surface enclosing the same charges has the same total flux. External sources do not change the sum even though they can change its distribution among faces.

Uniformly charged insulating model sphere

Given: A nonconducting sphere has fixed uniform volume charge density, radius $R=0.0800\,\mathrm m$ and total charge $Q=-6.00\,\mathrm{nC}$ . Vacuum environment and no other sources. Find fields at $r=0.0400\,\mathrm m$ and $r=0.160\,\mathrm m$ .
Find: Find enclosed charge and signed radial fields.

Approach: Uniform fixed volume charge and spherical symmetry give enclosed charge first, then the signed radial field component.
$\begin{aligned}Q_{\mathrm{in}}(r<R)&=Q\left(\frac rR\right)^3\\E_r(r)&=\frac{Q_{\mathrm{in}}(r)}{4\pi\varepsilon_0r^2}\end{aligned}$
Inside, enclosed charge grows with sphere volume.
$\begin{aligned}Q_{\mathrm{in}}(r)&=Q\left(\frac rR\right)^3\\Q_{\mathrm{in}}(0.0400\,\mathrm m)&=-0.750\,\mathrm{nC}\end{aligned}$
The interior Gauss solution is proportional to radius.
$\begin{aligned}E_r(r<R)&=\frac{Qr}{4\pi\varepsilon_0R^3}\\E_r(0.0400\,\mathrm m)&\approx-4212.915\,\mathrm{N/C}\end{aligned}$
Outside, all charge is enclosed.
$\begin{aligned}E_r(r\ge R)&=\frac Q{4\pi\varepsilon_0r^2}\\E_r(0.160\,\mathrm m)&\approx-2106.457\,\mathrm{N/C}\end{aligned}$

Result: Both fields point inward. At half the radius only one eighth of charge is enclosed; outside the inverse-square relation holds.

Flux with no enclosed charge

Given: A cube of side 4.00 cm lies in a uniform field $E=250\,\mathrm{N\,C^{-1}}$ along +x. It contains no charge; its faces are perpendicular to the coordinate axes.
Find: Find flux through the right and left faces and through the whole boundary.

Approach: each flat face has $\Phi=EA\cos\theta$ . Outward normals of the two x-faces point oppositely.
$\begin{aligned}\Phi_{+x}&=+EA\\\Phi_{-x}&=-EA\\\Phi_{\mathrm{other}}&=0\\\Phi_{\mathrm{closed}}&=\Phi_{+x}+\Phi_{-x}=0\end{aligned}$
Substitute: convert side length to metres before squaring.
$\begin{aligned}A&=(0.0400\,\mathrm m)^2=0.00160\,\mathrm{m^2}\\\Phi_{+x}&=250(0.00160)\,\mathrm{N\,m^2\,C^{-1}}=+0.400\,\mathrm{N\,m^2\,C^{-1}}\\\Phi_{-x}&=-0.400\,\mathrm{N\,m^2\,C^{-1}}\end{aligned}$

Result: Net flux is zero even though field magnitude is 250 N/C everywhere. The other four faces contribute zero because the field is tangent to them.

Check: Gauss independently gives $\frac{Q_{\mathrm{in}}}{\varepsilon_0}=0$ . Equal incoming and outgoing flux confirms the balance, not a vanishing field.

Which charge belongs in the flux laws for E and D?

Five flux and symmetry exercises

Calculate · Foundation

A closed surface encloses +4.00 nC and −7.00 nC. Find total flux.

Show hint

Add charges algebraically, not by magnitude.

Learning goal: Convert enclosed net charge to total flux.

Show solution and calculation

Model and approach: Add enclosed charges algebraically, not their magnitudes.
$\Phi_E=\frac{Q_++Q_-}{\varepsilon_0}$ $Q_{\mathrm{in}}$ =−3.00 nC.
$\Phi_E$ ≈−338.823 N·m²/C.
Negative sign means net inward flux.

Check: $\varepsilon_0\Phi_E=-3.00\,\mathrm{nC}$ .

\Phi_E=\frac{-3.00\times10^{-9}\,\mathrm C}{\varepsilon_0}\approx-338.823\,\mathrm{N\,m^2/C}

[1]

Interpret · Application

A central positive charge +1.00 nC lies inside concentric spheres of radii 0.0500 m and 0.100 m. Compare fields and total fluxes.

Show hint

Field scales as $\frac1{r^2}$ ; total flux counts charge.

Learning goal: Separate radius-dependent field from radius-independent total flux.

Show solution and calculation

Model and approach: The positive point charge is at the common centre, so field is radial.
$\begin{aligned}E(r)&=\frac{q}{4\pi\varepsilon_0r^2}\\\Phi_E&=\frac q{\varepsilon_0}\end{aligned}$ The larger sphere has one quarter of the field magnitude.
Both fluxes are ≈112.941 N·m²/C.
Field and flux are distinct quantities.

Check: $E_1\approx3595.02\,\mathrm{N\,C^{-1}},\quad E_2\approx898.755\,\mathrm{N\,C^{-1}}$ .

\begin{aligned}\frac{E_2}{E_1}&=\frac14\\\Phi_2&=\Phi_1\end{aligned}

[2]

Interpret · Application

A uniform external field points along +x through an empty cube. Is E zero because $Q_{\mathrm{in}}$ =0?

Show hint

Compare entry and exit faces.

Learning goal: Explain zero total flux despite nonzero field.

Show solution and calculation

Model and approach: Use outward normals, opposite on the two x-faces.
$\Phi_E=\sum_{\mathrm{faces}}\mathbf E\cdot\hat{\mathbf n}\,A$ No.
The two x faces contribute −EA and +EA; the other four have tangential field and zero flux.
Total flux is zero despite nonzero field.

\Phi_E=-EA+EA=0

[1]

Interpret · Application

A spherical surface encloses one point charge q. The charge moves from the sphere centre to another interior point without crossing the surface. Compare total flux before and after. May field still be treated as uniform and normal over the sphere?

Show hint

Gauss counts enclosed charge; pulling E outside the integral needs additional symmetry.

Learning goal: Recognize flux invariance without a uniform surface field.

Show solution and calculation

Approach: enclosed charge remains q, so total flux remains $\frac q{\varepsilon_0}$ .
Off centre, distances and angles vary across the surface; using $E_r\,4\pi R^2$ with one magnitude is unjustified.

Check: the charge has not crossed the surface, so the flux balance still applies. Individual surface regions can change contributions while the total stays fixed.

\Phi_E^{\mathrm{before}}=\Phi_E^{\mathrm{after}}=\frac q{\varepsilon_0},\qquad\Phi_E=\oint\mathbf E\cdot\hat{\mathbf n}\,dA

[1][2]

Error analysis · Challenge

A closed mathematical surface encloses +2.00 nC free and −0.50 nC net bound charge. A solution uses only $\frac{2.00\,\mathrm{nC}}{\varepsilon_0}$ for ∮E·dA. Correct using ε₀=8.8541878×10⁻¹² F/m. The requested quantity is E flux, not D flux.

Show hint

The electric Gauss law with ε₀ uses total enclosed charge.

Learning goal: Distinguish total charge in E flux from free charge.

Show solution and calculation

Approach: $Q_{\mathrm{tot}}=Q_{\mathrm{free}}+Q_{\mathrm{bound}}=1.50\,\mathrm{nC}$ .
E flux is about 169.411 N·m²/C, not 225.882 N·m²/C.

\begin{aligned}Q_{\mathrm{tot}}&=Q_{\mathrm{free}}+Q_{\mathrm{bound}}=1.50\,\mathrm{nC}\\\Phi_E&=\frac{Q_{\mathrm{tot}}}{\varepsilon_0}\\&=\frac{1.50\cdot10^{-9}\,\mathrm C}{8.8541878\cdot10^{-12}\,\mathrm{F/m}}\\&\approx169.411\,\mathrm{N\,m^2/C}\end{aligned}

[1][4]

Distinguish electric and magnetic Gauss laws

What to take away

Closed flux counts enclosed net charge; external sources can still change local field.
Zero net flux does not mean zero field. Incoming and outgoing contributions may cancel.
Calculating local field requires additional symmetry of the source arrangement. The E-field balance counts free plus bound charge.

[1][2][4]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Clarified total flux, units and symmetry; distinguished zero end-cap flux from zero field and made existing examples and links to line charge, Coulomb and Maxwell accessible.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

Cite this reference page

Quanta. Gauss's Law for Electrostatics. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/gauss-gesetz-elektrisches-feld

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Gauss's Law for Electrostatics

What does Gauss’s law say about the electric field?+

When does Gauss’s law help me calculate an electric field?+

Why must the Gaussian surface be closed?+

How do I calculate the flux of a sphere around a point charge?+

Does zero total flux mean the electric field is zero everywhere?+

Retain Gauss's Law for Electrostatics for exams

Create a curated FSRS exam set for ∮ E·dA = Q_innen / ε₀: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Gauss’s law for the electric field

Variables & units

Gauss's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electric field and flux

Worked example

Applications

How do you calculate with Gauss's Law for Electrostatics?

Imagine a boundary around the charges

What does Gauss’s law state, and what are the units of flux?

How do you calculate electric flux through a surface?

Explore further: solid angle and local form

How do you choose a Gaussian surface to calculate the field?

Interior field of a uniformly charged insulating sphere

Sphere and cube: distinguish charge, field and flux

Flux through a cube without field symmetry

Uniformly charged insulating model sphere

Flux with no enclosed charge

Which charge belongs in the flux laws for E and D?

Five flux and symmetry exercises

A closed surface encloses +4.00 nC and −7.00 nC. Find total flux.

A central positive charge +1.00 nC lies inside concentric spheres of radii 0.0500 m and 0.100 m. Compare fields and total fluxes.

A uniform external field points along +x through an empty cube. Is E zero because QinQ_{\mathrm{in}}Qin​=0?

A spherical surface encloses one point charge q. The charge moves from the sphere centre to another interior point without crossing the surface. Compare total flux before and after. May field still be treated as uniform and normal over the sphere?

A closed mathematical surface encloses +2.00 nC free and −0.50 nC net bound charge. A solution uses only 2.00 nCε0\frac{2.00\,\mathrm{nC}}{\varepsilon_0}ε0​2.00nC​ for ∮E·dA. Correct using ε0=8.8541878×10−12 F/m. The requested quantity is E flux, not D flux.

Distinguish electric and magnetic Gauss laws

What to take away

Sources for this explanation

Frequently asked questions about Gauss's Law for Electrostatics

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Gauss's Law for Electrostatics for exams

Gauss’s law for the electric field

Variables & units

Gauss's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electric field and flux

Worked example

Applications

How do you calculate with Gauss's Law for Electrostatics?

Imagine a boundary around the charges

What does Gauss’s law state, and what are the units of flux?

How do you calculate electric flux through a surface?

Explore further: solid angle and local form

How do you choose a Gaussian surface to calculate the field?

Interior field of a uniformly charged insulating sphere

Sphere and cube: distinguish charge, field and flux

Flux through a cube without field symmetry

Uniformly charged insulating model sphere

Flux with no enclosed charge

Which charge belongs in the flux laws for E and D?

Five flux and symmetry exercises

A closed surface encloses +4.00 nC and −7.00 nC. Find total flux.

A central positive charge +1.00 nC lies inside concentric spheres of radii 0.0500 m and 0.100 m. Compare fields and total fluxes.

A uniform external field points along +x through an empty cube. Is E zero because QinQ_{\mathrm{in}}Qin​=0?

A spherical surface encloses one point charge q. The charge moves from the sphere centre to another interior point without crossing the surface. Compare total flux before and after. May field still be treated as uniform and normal over the sphere?

A closed mathematical surface encloses +2.00 nC free and −0.50 nC net bound charge. A solution uses only 2.00 nCε0\frac{2.00\,\mathrm{nC}}{\varepsilon_0}ε0​2.00nC​ for ∮E·dA. Correct using ε0=8.8541878×10−12 F/m. The requested quantity is E flux, not D flux.

Distinguish electric and magnetic Gauss laws

What to take away

Sources for this explanation

Frequently asked questions about Gauss's Law for Electrostatics

Subject references and further reading

Related formulas

A uniform external field points along +x through an empty cube. Is E zero because $Q_{\mathrm{in}}$ =0?

A closed mathematical surface encloses +2.00 nC free and −0.50 nC net bound charge. A solution uses only $\frac{2.00\,\mathrm{nC}}{\varepsilon_0}$ for ∮E·dA. Correct using ε₀=8.8541878×10⁻¹² F/m. The requested quantity is E flux, not D flux.

A uniform external field points along +x through an empty cube. Is E zero because $Q_{\mathrm{in}}$ =0?

A closed mathematical surface encloses +2.00 nC free and −0.50 nC net bound charge. A solution uses only $\frac{2.00\,\mathrm{nC}}{\varepsilon_0}$ for ∮E·dA. Correct using ε₀=8.8541878×10⁻¹² F/m. The requested quantity is E flux, not D flux.