physics2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Derivation

Gauss's Law (Integral Form) from Coulomb

D-040 Home PU-102 Threads fields · symmetry Depends on Electric Field from Coulomb's Law, solid-angle-and-inverse-square-flux
Statement

For an electrostatic field built by superposing Coulomb fields of point charges, the outward electric flux through any closed surface \( S \) equals the total charge enclosed by \( S \) divided by \( \varepsilon_0 \): \( \displaystyle \oint_S \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \). The surface may have any shape and size; only the charge inside it contributes, and each interior charge contributes the same amount regardless of where inside it sits.

Why it matters

This is the bridge from the pointwise, action-at-a-distance Coulomb law to a global statement about an entire surface at once. It packages "inverse-square with the right coefficient" into a single conservation-of-flux law that no longer references distances or angles explicitly.

Because the result is independent of surface shape, it becomes the practical engine for computing fields of symmetric charge distributions, and — once localised via the divergence theorem — it is Maxwell's first equation. Deriving the integral form straight from Coulomb shows that the \( 1/r^2 \) exponent is not incidental: it is exactly the power that makes flux depend only on enclosed charge.

Assumptions
The field is the Coulomb (electrostatic) field.If \( \vec{E} \) carries any non-Coulomb part (e.g. an induced field from a changing \( \vec{B} \)), that part has its own flux and the clean \( Q_{\text{enc}}/\varepsilon_0 \) bookkeeping from the point-charge form no longer follows. Superposition holds, so the total field is a sum of point-charge fields.If fields did not add linearly, total flux could not be computed as a sum of single-charge fluxes and the derivation would not close. The closed surface \( S \) is piecewise smooth and passes through no point charge.A charge lying exactly on \( S \) subtends a solid angle of \( 2\pi \), not \( 4\pi \) or \( 0 \); the flux integral becomes ambiguous at that point until \( S \) is deformed off the charge. Charges are at rest and the medium is vacuum (or is folded into \( \vec{E} \) directly).In a polarisable medium the bound charge contributes to \( Q_{\text{enc}} \); ignoring it makes the naive \( \oint \vec{E}\cdot d\vec{A} = q_{\text{free}}/\varepsilon_0 \) statement wrong unless the \( \vec{D} \) field is used instead.
Derivation
1
\[ \vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r^2}\,\hat{r} \]
Start from the Coulomb field of a single point charge \( q \) at the origin (prior result: electric field from Coulomb's law). A
2
\[ d\Phi = \vec{E}\cdot d\vec{A} = \frac{q}{4\pi\varepsilon_0}\,\frac{\hat{r}\cdot d\vec{A}}{r^2} \]
Form the flux of \( \vec{E} \) through an area element \( d\vec{A} \) of the closed surface and pull the constants out. A
3
\[ d\Omega = \frac{\hat{r}\cdot d\vec{A}}{r^2} \]
Recognise the geometric factor as the solid angle subtended at the charge by \( d\vec{A} \) (prior result: solid angle and inverse-square flux). The dot product projects \( d\vec{A} \) onto the sphere of radius \( r \). B
4
\[ d\Phi = \frac{q}{4\pi\varepsilon_0}\,d\Omega \]
Substitute the solid angle. The \( r^2 \) of the inverse-square field cancels the \( r^2 \) in the projected area, so flux depends only on the angular size of the element — the crux of the whole argument. B
5
\[ \oint_S d\Omega = \begin{cases} 4\pi, & q \text{ inside } S \\[2pt] 0, & q \text{ outside } S \end{cases} \]
Integrate the solid angle over the whole closed surface. A charge inside sees the surface wrap completely around it (\( 4\pi \)). For a charge outside, every ray from it crosses \( S \) an even number of times; the crossings pair into equal-and-opposite \( \pm d\Omega \) contributions that cancel. C
6
\[ \Phi = \oint_S \vec{E}\cdot d\vec{A} = \frac{q}{4\pi\varepsilon_0}\oint_S d\Omega = \frac{q}{\varepsilon_0}\quad(q\ \text{inside}) \]
For an enclosed charge insert \( \oint d\Omega = 4\pi \); the \( 4\pi \) cancels the \( 4\pi \) in Coulomb's constant, leaving \( q/\varepsilon_0 \). An excluded charge contributes zero. B
7
\[ \oint_S \Big(\sum_i \vec{E}_i\Big)\cdot d\vec{A} = \sum_i \frac{q_i^{\text{in}}}{\varepsilon_0} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]
Apply superposition: the total field is the sum of point-charge fields and flux is linear, so sum the single-charge results. Only interior charges survive, giving the total enclosed charge. B
8
\[ \oint_S \vec{E}\cdot d\vec{A} = \frac{1}{\varepsilon_0}\int_V \rho\, dV \]
For a continuous distribution replace the discrete sum by an integral of the charge density \( \rho \) over the enclosed volume \( V \); superposition passes to the integral by linearity of flux. C
Result
\[ \oint_S \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]

Reading. The net field lines threading out through any closed surface count only the charge inside it, in units of \( 1/\varepsilon_0 \). Deforming the surface, or moving external charges around, leaves the flux untouched — external charges and the shape of \( S \) cancel out exactly.

Units check. \( [\vec{E}]\,[\,d\vec{A}\,] = (\mathrm{V\,m^{-1}})(\mathrm{m^2}) = \mathrm{V\,m} \). On the right, \( [Q]/[\varepsilon_0] = \mathrm{C}/(\mathrm{C^2\,N^{-1}\,m^{-2}}) = \mathrm{N\,m^2\,C^{-1}} \). Since \( 1\,\mathrm{V} = 1\,\mathrm{N\,m\,C^{-1}} \), the right side is also \( \mathrm{V\,m} \). Both sides carry units of \( \mathrm{V\,m} \equiv \mathrm{N\,m^2\,C^{-1}} \).

Limiting cases
  • Single charge, spherical surface centred on it: \( \vec{E} \parallel d\vec{A} \) with \( E \) constant, so \( \Phi = E\cdot 4\pi r^2 = q/\varepsilon_0 \), recovering \( E = q/(4\pi\varepsilon_0 r^2) \) — Coulomb back out.
  • Charge just outside \( S \): flux is exactly zero however close it sits — a discontinuous jump to \( q/\varepsilon_0 \) the instant it crosses the surface.
  • No enclosed charge (empty region): \( \oint \vec{E}\cdot d\vec{A} = 0 \) even where \( \vec{E}\neq 0 \) from distant sources.
  • Shrinking \( S \) to a point around charge density \( \rho \): dividing by the enclosed volume gives \( \nabla\cdot\vec{E} = \rho/\varepsilon_0 \), the differential form.
Breaks when
  • Non-Coulomb fields are present. A time-varying magnetic field induces an \( \vec{E} \) whose flux the point-charge argument above cannot supply; the static \( 1/r^2 \) form no longer gives the field, and the full Maxwell system is required.
  • The force law departs from inverse-square. If the field went as \( 1/r^{2+\delta} \) (a finite photon mass, or a modified law), the \( r^2 \) cancellation in Step 4 fails, \( \oint d\Omega \) no longer decouples, and flux would depend on the size and shape of \( S \). Precision Gauss's-law "null" experiments bound \( \delta \) this way.
  • A point charge lies on the surface. The solid angle it subtends is \( 2\pi \), giving \( \Phi = q/(2\varepsilon_0) \); the clean law is ambiguous until \( S \) is deformed off the charge.
  • Inside linear dielectrics if bound charge is forgotten. \( \oint \vec{E}\cdot d\vec{A} = (q_{\text{free}}+q_{\text{bound}})/\varepsilon_0 \); using only free charge here is wrong — that role belongs to \( \vec{D} \).
Failure modes
  • "External charges add to the flux." They do change \( \vec{E} \) on the surface, but their net flux is exactly zero; students wrongly fold them into \( Q_{\text{enc}} \).
  • Confusing flux with field. Reading \( \Phi = 0 \) as \( \vec{E} = 0 \). A neutral region can host strong fields with zero net flux.
  • Using \( \Phi = EA \) without symmetry. Pulling \( E \) out of \( \oint \vec{E}\cdot d\vec{A} \) is legal only when \( E \) is constant and \( \vec{E}\parallel d\vec{A} \) over the surface; done blindly it yields nonsense fields.
  • Dropping the dot product. Taking \( E\,dA \) instead of \( \vec{E}\cdot d\vec{A} \) ignores the projection \( \cos\theta \) and over-counts flux through tilted elements.
  • Miscounting enclosed charge. Placing the Gaussian surface so a charge sits on it, or mishandling a charge straddling the boundary, corrupts \( Q_{\text{enc}} \).
  • Assuming the law gives \( \vec{E} \) directly. Gauss's law fixes only the surface integral; without symmetry it does not determine \( \vec{E} \) at a point.
Discussion

The single fact doing all the work is the exact cancellation between the \( 1/r^2 \) of the Coulomb field and the \( r^2 \) growth of area on a sphere. That is why flux reduces to solid angle, and solid angle is a purely topological count: does the surface enclose the source or not? The strength of the law — its indifference to surface shape and to external charges — is inherited entirely from this cancellation and would be destroyed by any other power law.

Read the other way, Gauss's law plus symmetry replaces vector integration with algebra. For a distribution with spherical, cylindrical, or planar symmetry, a Gaussian surface aligned with the symmetry makes \( \vec{E} \) constant and normal over it, so \( \oint \vec{E}\cdot d\vec{A} = EA \) and \( E = Q_{\text{enc}}/(\varepsilon_0 A) \). This is why Gauss's law is a computational tool as much as a physical statement, but the shortcut is available only when symmetry is present.

Applying the divergence theorem, \( \oint_S \vec{E}\cdot d\vec{A} = \int_V \nabla\cdot\vec{E}\, dV \), and equating integrands with \( Q_{\text{enc}} = \int_V \rho\, dV \) for arbitrary \( V \), yields \( \nabla\cdot\vec{E} = \rho/\varepsilon_0 \). The integral law is the global shadow of a local statement: charge is the source of the divergence of \( \vec{E} \). Historically this is the first of Maxwell's equations and it survives unchanged into the dynamical theory.

More deeply, the solid-angle argument is precisely the distributional identity \( \nabla\cdot(\hat{r}/r^2) = 4\pi\,\delta^3(\vec{r}) \): the point-charge field is divergence-free everywhere except at the source, where it carries a delta-function divergence of weight \( 4\pi \). Gauss's law is the integrated form of that identity, and its shape-independence is the same fact as the harmonicity of the \( 1/r \) potential away from the origin — the \( 1/r \) Coulomb potential is the Green's function of the Laplacian in three dimensions, so \( \nabla^2(1/r) = -4\pi\,\delta^3(\vec{r}) \).

Common misconceptions. Gauss's law is always true, but it is only useful for finding \( \vec{E} \) when symmetry lets you evaluate the integral. A vanishing net flux never implies a vanishing field, and a nonzero field on the surface never implies the enclosed charge is nonzero.

Worked examples
1
Flux through a cube from an interior charge. A point charge \( q = 5.0\,\mathrm{nC} \) sits somewhere strictly inside a cubical box. Find the total flux through the whole cube, and the flux through one face if the charge is at the cube's centre. A
2
\[ \Phi_{\text{total}} = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{q}{\varepsilon_0} \]
The whole charge is enclosed; the box shape is irrelevant. A
3
\[ \Phi_{\text{total}} = \frac{5.0\times10^{-9}\,\mathrm{C}}{8.854\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}}} = 5.6\times10^{2}\ \mathrm{N\,m^2\,C^{-1}} \]
Insert numbers with \( \varepsilon_0 = 8.854\times10^{-12} \). A
4
\[ \Phi_{\text{face}} = \frac{1}{6}\,\Phi_{\text{total}} = 94\ \mathrm{N\,m^2\,C^{-1}}\quad(\text{centre only}) \]
Only with the charge at the centre do the six faces share the flux equally by symmetry. B
\[ \Phi_{\text{total}} = 5.6\times10^{2}\ \mathrm{N\,m^2\,C^{-1}},\qquad \Phi_{\text{face,centre}} = 94\ \mathrm{N\,m^2\,C^{-1}} \]

Reading. The box shape is invisible to the total flux; only the per-face split needs the centring symmetry.

Units check. \( \mathrm{C}/(\mathrm{C^2\,N^{-1}\,m^{-2}}) = \mathrm{N\,m^2\,C^{-1}} \), correct for flux.

1
Field of an infinite line charge via Gauss's law. A uniform line charge has linear density \( \lambda = 3.0\,\mu\mathrm{C\,m^{-1}} \). Find \( E \) at radial distance \( s = 0.20\,\mathrm{m} \) using a coaxial cylinder of length \( L \). B
2
\[ \oint \vec{E}\cdot d\vec{A} = E\,(2\pi s L) = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{\lambda L}{\varepsilon_0} \]
By cylindrical symmetry \( \vec{E} \) is radial and constant over the curved wall; the flat ends contribute nothing because \( \vec{E}\perp d\vec{A} \) there. B
3
\[ E = \frac{\lambda}{2\pi\varepsilon_0 s} \]
Cancel \( L \) and solve for \( E \) — symbols before numbers. B
4
\[ E = \frac{3.0\times10^{-6}}{2\pi(8.854\times10^{-12})(0.20)}\ \mathrm{V\,m^{-1}} = 2.7\times10^{5}\ \mathrm{V\,m^{-1}} \]
Substitute \( \lambda,\ \varepsilon_0,\ s \). A
\[ E = \frac{\lambda}{2\pi\varepsilon_0 s} = 2.7\times10^{5}\ \mathrm{V\,m^{-1}}\ \text{(radially outward)} \]

Reading. The cancellation of \( L \) shows the field of an infinite line falls as \( 1/s \), not \( 1/s^2 \) — a direct payoff of Gauss's law plus symmetry.

Units check. \( (\mathrm{C\,m^{-1}})/(\mathrm{C^2\,N^{-1}\,m^{-2}}\cdot\mathrm{m}) = \mathrm{N\,C^{-1}} = \mathrm{V\,m^{-1}} \).

Problems
  1. A charge \( q = 2.0\,\mathrm{nC} \) is placed at one corner of a cube. What flux passes through the cube?
    Solution A corner is shared by 8 cubes, so the charge subtends \( 1/8 \) of the full solid angle: \( \Phi = \tfrac{1}{8}\,\dfrac{q}{\varepsilon_0} = \dfrac{2.0\times10^{-9}}{8(8.854\times10^{-12})} = 28\ \mathrm{N\,m^2\,C^{-1}} \). The three faces meeting at the corner catch zero each; the far three share the \( 28 \).
  2. A spherical shell of radius \( R \) carries total charge \( Q \) spread uniformly over its surface. Using Gauss's law, find \( E \) for \( r < R \) and \( r > R \).
    Solution For \( r < R \): a concentric Gaussian sphere encloses no charge, so \( \oint \vec{E}\cdot d\vec{A} = 0 \Rightarrow E = 0 \). For \( r > R \): the enclosed charge is \( Q \), so \( E(4\pi r^2) = Q/\varepsilon_0 \Rightarrow E = \dfrac{Q}{4\pi\varepsilon_0 r^2} \), identical to a point charge \( Q \) at the centre.
  3. An infinite plane sheet has surface charge density \( \sigma = 1.5\,\mu\mathrm{C\,m^{-2}} \). Find \( E \) on either side.
    Solution Use a pillbox of face area \( A \) straddling the sheet. Flux exits both faces: \( 2EA = \sigma A/\varepsilon_0 \Rightarrow E = \dfrac{\sigma}{2\varepsilon_0} \). Numerically \( E = \dfrac{1.5\times10^{-6}}{2(8.854\times10^{-12})} = 8.5\times10^{4}\ \mathrm{V\,m^{-1}} \), directed away from the sheet and independent of distance.
  4. A solid insulating sphere of radius \( R = 0.10\,\mathrm{m} \) carries charge \( Q = 4.0\,\mathrm{nC} \) uniformly through its volume. Find \( E \) at \( r = 0.05\,\mathrm{m} \).
    Solution Enclosed charge scales with volume: \( Q_{\text{enc}} = Q\,(r/R)^3 \). Gauss: \( E(4\pi r^2) = Q(r/R)^3/\varepsilon_0 \Rightarrow E = \dfrac{Q\,r}{4\pi\varepsilon_0 R^3} \). At \( r = 0.05 \): \( E = \dfrac{(8.99\times10^{9})(4.0\times10^{-9})(0.05)}{(0.10)^3} = 1.8\times10^{3}\ \mathrm{V\,m^{-1}} \), using \( 1/(4\pi\varepsilon_0)=8.99\times10^{9} \). The field grows linearly with \( r \) inside.
  5. Two point charges \( +q \) and \( -q \) both sit inside a closed surface \( S \), and a third charge \( +2q \) sits just outside it. What is \( \oint_S \vec{E}\cdot d\vec{A} \)?
    Solution Only enclosed charge counts: \( Q_{\text{enc}} = +q + (-q) = 0 \), so \( \Phi = 0 \), regardless of the external \( +2q \), which contributes zero net flux even though it changes \( \vec{E} \) everywhere on \( S \). Note \( \vec{E}\neq 0 \) on \( S \); only the net flux vanishes.