Gauss's Law and Electric Flux

The flux of E out of any closed surface is the charge inside divided by ε₀. That is always true and usually useless, because it gives one number for the whole surface. Where the charge has a symmetry, though, a surface can be chosen that E crosses at right angles and with the same size everywhere it crosses, and then one line of algebra gives E. Pick a symmetry, size the Gaussian surface, and read off the charge it encloses, the flux and the field.

Gauss's law: flux out of any closed surface	$\oint\vec E\cdot d\vec A=\frac{Q_{\mathrm{enc}}}{\varepsilon_0}$
Flux through a flat area in a uniform field	$\Phi=\vec E\cdot\vec A=EA\cos\theta$

Worked example — Gauss's Law and Electric Flux

Charge enclosed	1.25×10⁻⁹ C
Area of the Gaussian sphere	0.03142 m²
Flux out of the surface	141.2 N·m²/C
Field on the surface	4,494 N/C

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Gauss's Law and Electric Flux

The electric flux through a surface is Φ = ∫E·dA, the field's component across the surface added over its area. Gauss's law says that for any closed surface the flux out of it is the charge inside divided by ε₀: ∮E·dA = Q_enc/ε₀.

What each symbol means

dA: a piece of the surface, as a vector: its area, pointing out of the closed surface
Q_enc: the charge inside the closed surface; charge outside contributes no net flux
ε₀: the permittivity of free space, 8.854 × 10⁻¹² C²/(N·m²)

Where the formula comes from

Flux is field times area, across the surface. For a flat area A in a uniform field E at angle θ to its normal, only E cos θ crosses it.
$\Phi=\int\vec E\cdot d\vec A,\qquad\Phi=EA\cos\theta$
A point charge's flux does not depend on the surface. Through a sphere of any radius about q it is kq/r² times 4πr², which is q/ε₀ with the r cancelled; a surface of any other shape is crossed by the same field lines, and adding charges adds fluxes.
$\oint\frac{kq}{r^{2}}\,dA=\frac{kq}{r^{2}}\cdot4\pi r^{2}=4\pi kq=\frac{q}{\varepsilon_0}$
Choose the surface so that E comes out of the integral. Symmetry makes E the same size all over the surface and perpendicular to it, or parallel where it is not, so the flux is E times the area it crosses.
$\oint\vec E\cdot d\vec A=E\oint dA=EA$
The three surfaces. A sphere for spherical charge, a cylinder of length L for a line or a long cylinder (its flat ends carry no flux), a pillbox for a plane (only its two faces do).
$4\pi r^{2},\qquad 2\pi rL,\qquad 2A$
Set E·A equal to Q_enc/ε₀ and solve. The three results are the point charge, the line and the sheet.
$E=\frac{Q}{4\pi\varepsilon_0r^{2}},\qquad E=\frac{\lambda}{2\pi\varepsilon_0r},\qquad E=\frac{\sigma}{2\varepsilon_0}$

Assumes: Charges at rest. The law itself holds for any closed surface and any arrangement; what needs symmetry is getting E out of it. The cylinder and the plane are infinitely long and wide; a real rod or plate behaves so only far from its ends and edges.

Problems solved in full

Flux through a flat surface 2 steps
A uniform field of 2.0 × 10⁴ N/C crosses a square of side 20 cm at 60° to the square's normal. Find the flux through the square.
1. Only the component of E along the normal crosses the surface:
  
  $\Phi=EA\cos\theta$
2. With A = (0.20 m)² = 0.040 m²:
  
  $\Phi=(2{.}0\times10^{4})(0{.}040)\cos60^{\circ}=400\ \mathrm{N\,m^{2}/C}$
Answer

400 N·m²/C. Tilt the square until the field skims along it (θ = 90°) and the flux is zero.

$\Phi=400\ \mathrm{N\,m^{2}/C}$
A charge at the centre of a cube 2 steps
A charge of 10 nC sits at the centre of a cube with sides of 20 cm. Find the flux through one face.

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1. The flux out of the whole cube is q/ε₀, whatever its shape:
  
  $\Phi=\frac{q}{\varepsilon_0}=\frac{1{.}0\times10^{-8}}{8{.}854\times10^{-12}}\approx1129\ \mathrm{N\,m^{2}/C}$
2. By symmetry the six faces share it equally:
  
  $\Phi_{1}=\frac{1}{6}\cdot\frac{q}{\varepsilon_0}\approx188{.}2\ \mathrm{N\,m^{2}/C}$
Answer

188.2 N·m²/C per face; the side of the cube does not enter.

$\Phi_{1}=\frac{q}{6\varepsilon_0}\approx188{.}2\ \mathrm{N\,m^{2}/C}$
A charge at a corner of a cube 2 steps
The same 10 nC sits at a corner of the cube. Find the flux through each face and through the cube.

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1. Eight such cubes, meeting at the corner, would enclose the charge between them, and by symmetry each takes an equal share:
  
  $\Phi=\frac{1}{8}\cdot\frac{q}{\varepsilon_0}\approx141{.}2\ \mathrm{N\,m^{2}/C}$
2. The three faces that meet at the corner contain the charge's own field lines, which run along them, so they carry nothing; the other three share the cube's flux equally:
  
  $\Phi_{1}=0,\qquad\Phi_{2}=\frac{1}{3}\cdot\frac{q}{8\varepsilon_0}=\frac{q}{24\varepsilon_0}\approx47{.}06\ \mathrm{N\,m^{2}/C}$
Answer

141.2 N·m²/C through the cube: 47.06 through each of the three far faces and none through the three near ones.

$\Phi=\frac{q}{8\varepsilon_0}\approx141{.}2\ \mathrm{N\,m^{2}/C}$
Inside and outside a charged ball 4 steps
A ball of radius 10 cm carries 10 nC spread evenly through it. Find E at 5 cm and at 20 cm from its centre.

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1. Inside, a Gaussian sphere of radius r encloses the fraction (r/R)³ of the charge:
  
  $Q_{\mathrm{enc}}=Q\,\frac{r^{3}}{R^{3}}$
2. Gauss's law over that sphere:
  
  $E\cdot4\pi r^{2}=\frac{Qr^{3}}{\varepsilon_0R^{3}}\ \Rightarrow\ E=\frac{kQr}{R^{3}}$
3. At r = 0.05 m:
  
  $E=\frac{(8{.}988\times10^{9})(1{.}0\times10^{-8})(0{.}05)}{(0{.}10)^{3}}\approx4494\ \mathrm{N/C}$
4. Outside, the sphere encloses all of Q and the field is a point charge's:
  
  $E=\frac{kQ}{r^{2}}=\frac{(8{.}988\times10^{9})(1{.}0\times10^{-8})}{(0{.}20)^{2}}\approx2247\ \mathrm{N/C}$
Answer

4,494 N/C at 5 cm and 2,247 N/C at 20 cm. The surface field is 8,988 N/C: halfway in it is halved, because the field inside grows in proportion to r, and twice as far out it is quartered.

$E(5\ \mathrm{cm})\approx4494\ \mathrm{N/C},\qquad E(20\ \mathrm{cm})\approx2247\ \mathrm{N/C}$
A charge inside a conducting shell 3 steps
A charge of 10 nC sits at the centre of a metal shell with inner radius 10 cm and outer radius 15 cm, which carries 5 nC of its own. Find the charge on each wall of the shell, and E at 5 cm, 12 cm and 20 cm.

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1. In the metal E = 0, so a Gaussian sphere inside the metal encloses no net charge. The inner wall must hold −q, and the rest of the shell's charge goes to the outer wall:
  
  $Q_{a}=-q=-10\ \mathrm{nC},\qquad Q_{b}=q+Q_s=15\ \mathrm{nC}$
2. In the cavity, at 5 cm, only q is enclosed:
  
  $E=\frac{kq}{r^{2}}=\frac{(8{.}988\times10^{9})(1{.}0\times10^{-8})}{(0{.}05)^{2}}\approx3{.}595\times10^{4}\ \mathrm{N/C}$
3. In the metal, at 12 cm, E = 0. Outside, at 20 cm, the sphere encloses q + Qs:
  
  $E=\frac{k\,(q+Q_s)}{r^{2}}=\frac{(8{.}988\times10^{9})(1{.}5\times10^{-8})}{(0{.}20)^{2}}\approx3370\ \mathrm{N/C}$
Answer

−10 nC on the inner wall, 15 nC on the outer; E = 3.595 × 10⁴ N/C at 5 cm, 0 at 12 cm, 3,370 N/C at 20 cm.

$E=3{.}595\times10^{4},\ 0,\ 3370\ \mathrm{N/C}$
Inside a charged rod 3 steps
A long cylinder of radius 10 cm carries 10 nC per metre spread evenly through its volume. Find E at 5 cm from its axis.

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1. Take a Gaussian cylinder of radius r and length L. Only its curved side carries flux, and it encloses the fraction (r/R)² of the charge in that length:
  
  $E\cdot2\pi rL=\frac{\lambda L}{\varepsilon_0}\cdot\frac{r^{2}}{R^{2}}$
2. Solve for E:
  
  $E=\frac{\lambda r}{2\pi\varepsilon_0R^{2}}=\frac{2k\lambda r}{R^{2}}$
3. At r = 0.05 m:
  
  $E=\frac{2\,(8{.}988\times10^{9})(1{.}0\times10^{-8})(0{.}05)}{(0{.}10)^{2}}\approx898{.}8\ \mathrm{N/C}$
Answer

898.8 N/C, a quarter of the 3,595 N/C an infinitely thin line with the same λ would give at 5 cm.

$E=\frac{2k\lambda r}{R^{2}}\approx898{.}8\ \mathrm{N/C}$
A coaxial cable 2 steps
A coaxial cable carries +10 nC per metre on a wire of radius 2.0 cm and −10 nC per metre on a thin outer conductor of radius 10 cm. Find E at 5.0 cm and at 15 cm from the axis.

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1. Between the conductors the Gaussian cylinder encloses only the wire's charge:
  
  $E\cdot2\pi rL=\frac{\lambda L}{\varepsilon_0}\ \Rightarrow\ E=\frac{2k\lambda}{r}=\frac{2\,(8{.}988\times10^{9})(1{.}0\times10^{-8})}{0{.}05}\approx3595\ \mathrm{N/C}$
2. Outside the cable it encloses both, which cancel:
  
  $E\cdot2\pi rL=\frac{(\lambda-\lambda)L}{\varepsilon_0}=0\ \Rightarrow\ E=0$
Answer

3,595 N/C at 5.0 cm, and zero at 15 cm. Equal and opposite charges on the two conductors keep the field inside the cable.

$E(5\ \mathrm{cm})\approx3595\ \mathrm{N/C},\qquad E(15\ \mathrm{cm})=0$
A sheet and two plates 2 steps
Find the field of a large sheet carrying 10 nC per square metre, and the field between two large plates 10 cm apart carrying +10 and −10 nC per square metre.

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1. A pillbox crossing the sheet has two faces of area A, both with E out of them, and encloses σA:
  
  $2EA=\frac{\sigma A}{\varepsilon_0}\ \Rightarrow\ E=\frac{\sigma}{2\varepsilon_0}=\frac{1{.}0\times10^{-8}}{2\,(8{.}854\times10^{-12})}\approx564{.}7\ \mathrm{N/C}$
2. Between the plates the two sheets' fields point the same way, from + to −, and add; outside they point opposite ways and cancel:
  
  $E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}\approx1129\ \mathrm{N/C}$
Answer

564.7 N/C for the sheet; 1,129 N/C between the plates, and zero outside. Neither depends on the distance from the plates.

$E=\frac{\sigma}{\varepsilon_0}\approx1129\ \mathrm{N/C}$
A sphere whose charge is not spread evenly 3 steps
A ball of radius R = 10 cm carries charge with density ρ = ρ₀r/R, where ρ₀ = 2.0 μC/m³. Find its total charge and E at r = 5.0 cm.
Show solution
1. The charge inside radius r, adding thin shells of area 4πr′² and thickness dr′:
  
  $Q_{\mathrm{enc}}=\int_{0}^{r}\rho_0\frac{r'}{R}\,4\pi r'^{2}\,dr'=\frac{\pi\rho_0r^{4}}{R}$
2. Gauss's law over a sphere of radius r:
  
  $E\cdot4\pi r^{2}=\frac{\pi\rho_0r^{4}}{\varepsilon_0R}\ \Rightarrow\ E=\frac{\rho_0r^{2}}{4\varepsilon_0R}$
3. At r = 0.050 m, and the total charge at r = R:
  
  $E=\frac{(2{.}0\times10^{-6})(0{.}050)^{2}}{4\,(8{.}854\times10^{-12})(0{.}10)}\approx1412\ \mathrm{N/C},\qquad Q=\pi\rho_0R^{3}\approx6{.}283\times10^{-9}\ \mathrm{C}$
Answer

Q = 6.283 nC; E = 1,412 N/C at 5.0 cm. At r = R the inside formula gives 5,647 N/C, the same as kQ/R², as it must.

$Q\approx6{.}283\ \mathrm{nC},\qquad E\approx1412\ \mathrm{N/C}$

Supplementary problems, answers only

A charge of 3.0 μC sits inside a closed surface of any shape. Find the flux out of the surface.

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3.388 × 10⁵ N·m²/C.

$\Phi=\frac{q}{\varepsilon_0}\approx3{.}388\times10^{5}\ \mathrm{N\,m^{2}/C}$
A uniform field of 1,000 N/C is parallel to the axis of a hemisphere of radius 10 cm. Find the flux through its curved surface.

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31.42 N·m²/C: the same field lines cross the flat disk that closes it, πR²E.

$\Phi=\pi R^{2}E\approx31{.}42\ \mathrm{N\,m^{2}/C}$
A metal sphere of radius 5.0 cm carries 4.0 nC. Find E at 3.0 cm and at 10 cm from its centre.

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Zero at 3.0 cm (inside the metal), 3,595 N/C at 10 cm.

$E(3\ \mathrm{cm})=0,\qquad E(10\ \mathrm{cm})=\frac{kQ}{r^{2}}\approx3595\ \mathrm{N/C}$
A long cylinder of radius 2.0 cm carries 5.0 nC per metre spread evenly through its volume. Find E at 1.0 cm and at 4.0 cm from the axis.

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2,247 N/C at both: half the radius inside gives half the surface field, twice the radius outside gives half too.

$E=\frac{2k\lambda r}{R^{2}}=\frac{2k\lambda}{r'}\approx2247\ \mathrm{N/C}$
A large slab 4.0 cm thick carries ρ = 1.0 μC/m³ spread evenly. Find E 1.0 cm from its mid-plane and outside it.

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1,129 N/C at 1.0 cm; 2,259 N/C everywhere outside.

$E=\frac{\rho x}{\varepsilon_0}\approx1129\ \mathrm{N/C},\qquad E=\frac{\rho a}{\varepsilon_0}\approx2259\ \mathrm{N/C}$
A point charge q sits at the centre of one face of a cube. Find the flux through the cube.

Show answer

q/(2ε₀): a second, identical cube on the other side of the face would take the other half.

$\Phi=\frac{q}{2\varepsilon_0}$

From outside, a charged ball and a point charge cannot be told apart

Press A charged ball and drag the Gaussian sphere out past R = 10 cm. Beyond the surface the field lies exactly on kQ/r², the field of all 10 nC at the centre: a sphere of radius r outside the ball encloses all of Q wherever its pieces are, and symmetry spreads the flux evenly over 4πr². Switch to the shell and the outside is the same curve again. All the difference is inside. The shell's field is zero everywhere within it, and the ball's rises in a straight line, E = kQr/R³, from nothing at the centre to its largest value, 8,988 N/C, at the surface. A Gaussian sphere inside the ball encloses only the fraction (r/R)³ of the charge and spreads it over an area that grows as r², which leaves one power of r.

Move the charge to a corner and only an eighth of its flux goes through the cube

Press Charge at the centre of a cube: each face carries 188.2 N·m²/C, a sixth of q/ε₀ = 1,129 N·m²/C. Slide t towards the corner. The three near faces gain and the three far faces lose, and the total stays at 1,129 until the very last step: at t = 0.99 the near faces carry 328.8 each and the far faces 47.7. At the corner itself the near faces carry nothing, because the field there runs along them, and each far face carries 47.06, a twenty-fourth of q/ε₀. Eight cubes meet at that corner and would share the flux equally. Past the corner the total is zero, with flux going in through three faces and out through the other three.

The empty inside of a charged shell is the strictest test of Coulomb's law

Press A charged shell: inside, the field is zero at every radius. Gauss's law gets that in one line, but only because Coulomb's force falls exactly as 1/r². Make it 1/r^(2 + q) with any q other than zero and the near side of the shell no longer exactly balances the far side, so a field appears inside. A null result inside a shell is therefore a measurement of q. In 1936 Plimpton and Lawton put an alternating potential of more than 3,000 V on an outer metal globe and looked, with a detector that could see about 10⁻⁶ V, for any potential difference between it and a globe inside; there was none, and they put q below 2 × 10⁻⁹. Williams, Faller and Hill's high-frequency test of 1971 gave q = (2.7 ± 3.1) × 10⁻¹⁶.

Learning path

Where fields come from

References (4)

Gauss's law and flux: the flux through a closed surface is the same “regardless of the shape or size of the surface, as long as the surface encloses the same amount of charge”, and a charge outside contributes zero: S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 6.2, “Explaining Gauss's Law”.
The symmetric cases worked as here: a uniformly charged sphere inside and out (Example 6.6, E growing linearly inside), a non-uniformly charged sphere (6.7), a cylindrical shell (6.8), and the infinite plane σ/(2ε₀): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 6.3, “Applying Gauss's Law”.
Insight block 3: concentric metal globes, an alternating potential above 3,000 V on the outer one, a detector inside sensitive to about 10⁻⁶ V, and the exponent 2 confirmed to within 2 × 10⁻⁹: S. J. Plimpton and W. E. Lawton, “A Very Accurate Test of Coulomb's Law of Force Between Charges”, Physical Review 50(11), 1936, pp. 1066-1071. DOI 10.1103/PhysRev.50.1066
Insight block 3: the 1971 high-frequency test, q = (2.7 ± 3.1) × 10⁻¹⁶ for a force of the form 1/r^(2+q): E. R. Williams, J. E. Faller and H. A. Hill, “New Experimental Test of Coulomb's Law: A Laboratory Upper Limit on the Photon Rest Mass”, Physical Review Letters 26(12), 1971, pp. 721-724. DOI 10.1103/PhysRevLett.26.721

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.