Part 3 · Chapter 23

Gauss' Law

Coulomb's law can find the field of any charge — if you are willing to integrate. Gauss' law offers a shortcut so powerful it feels like cheating: when a charge distribution is symmetric, you can read its field off a single, well-chosen surface without doing a single integral. It is the same physics seen from a higher vantage point.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • Electric flux as the amount of field "piercing" a surface, \(\Phi = \displaystyle\int \vec{E}\cdot d\vec{A}\), measured in N·m²/C.
  • The idea of a Gaussian surface — an imaginary closed surface you choose to exploit symmetry.
  • Gauss' law in its clean form \(\varepsilon_{0}\displaystyle\oint \vec{E}\cdot d\vec{A} = q_{\text{enc}}\), and why only the enclosed charge matters.
  • How Gauss' law and Coulomb's law are two faces of the same physics — each follows from the other.
  • The field of a charged isolated conductor: zero inside, and \(E = \sigma/\varepsilon_{0}\) just outside its surface.
  • Field by symmetry: a line of charge \(E = \dfrac{\lambda}{2\pi\varepsilon_{0}r}\), a sheet of charge \(E = \dfrac{\sigma}{2\varepsilon_{0}}\), and a spherical shell or ball of charge.
Section 23-1

What Is Physics?

In the last chapter we built electric fields one charge at a time, then summed or integrated. That always works in principle, but the integrals can be brutal — and for a long charged rod or a large charged plate they are genuinely hard. Yet the answers, when you finally grind them out, are often startlingly simple. A long line of charge gives a field that falls off as \(1/r\); a large flat sheet gives a field that does not fall off at all. Such clean results are a hint that there is an easier road.

That road is Gauss' law, the work of Carl Friedrich Gauss. It relates the electric field on a closed surface to the net charge sitting inside that surface — and nothing else. When the charge distribution has symmetry (a sphere, an infinite line, an infinite plane), Gauss' law lets us extract the field with almost no calculation. Equally important, it is one of the four Maxwell equations, the compact statements that contain all of classical electromagnetism. Learning to wield it is learning to think about fields the way physicists actually do.

Section 23-2

Electric Flux

Picture a uniform field \(\vec{E}\) and a flat loop of area \(A\) held up in it. The electric flux \(\Phi\) measures how much of the field passes through the loop. If we represent the area itself as a vector \(\vec{A}\) — magnitude \(A\), direction along the normal to the surface — then the flux is their dot product. Hold the loop face-on to the field and the flux is largest; turn it edge-on and the flux drops to zero, because nothing "pierces through."

Flux of a uniform field through a flat area
\[ \Phi = \vec{E}\cdot\vec{A} = EA\cos\theta \]
Here \(\theta\) is the angle between the field and the area's normal. The SI unit of flux is the newton·metre²/coulomb (N·m²/C). Flux is a scalar, and it can be positive, negative, or zero depending on the orientation.

When the field varies or the surface curves, we cut the surface into patches \(d\vec{A}\) small enough that \(\vec{E}\) is uniform over each, compute \(\vec{E}\cdot d\vec{A}\) for every patch, and add them up — an integral:

Flux through any surface
\[ \Phi = \int \vec{E}\cdot d\vec{A} \]
Read this as "sum the field component along the local normal, weighted by patch area, over the whole surface." It is the central object of the chapter — everything that follows is a statement about this integral.
The rain analogy. Think of the field as a steady rain and the loop as a wire hoop. The flux is the volume of rain per second passing through the hoop. Hold the hoop flat to the falling rain and lots passes through; tilt it edge-on and almost none does. The field lines play the role of the raindrops, and "flux" is just a count of how many thread the surface.
Section 23-3

Flux Through a Closed Surface

Gauss' law is about closed surfaces — surfaces that completely enclose a volume, like a balloon or a tin can. For a closed surface we adopt one firm convention: the area vector \(d\vec{A}\) always points outward. With that rule, field lines leaving the surface count as positive flux and lines entering count as negative flux. The total, written with a circle on the integral sign, is the net flux:

Net flux through a closed surface
\[ \Phi = \oint \vec{E}\cdot d\vec{A} \]
The little circle means the integral runs over the entire closed surface. If as many lines leave as enter, the net flux is zero — even if the field itself is large everywhere on the surface.
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What net flux tells you
Net outward flux is positive only if the surface encloses net positive charge.

Wrap an imaginary surface around empty space and every line that enters also leaves, so the net flux is zero. Wrap it around a positive charge and lines stream outward without returning — net positive flux. Wrap it around a negative charge and lines dive inward — net negative flux. The net flux is a charge detector.

Section 23-4

Gauss' Law

The qualitative observation above becomes exact and quantitative. Gauss' law states that the net electric flux through any closed surface is proportional to the net charge enclosed by that surface — and is utterly independent of the surface's shape, of how the charge is arranged inside, and of any charge sitting outside.

Gauss' law
\[ \varepsilon_{0}\oint \vec{E}\cdot d\vec{A} = q_{\text{enc}} \qquad\Longleftrightarrow\qquad \Phi = \frac{q_{\text{enc}}}{\varepsilon_{0}} \]
Here \(\varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{C^{2}/N\cdot m^{2}}\) is the permittivity constant and \(q_{\text{enc}}\) is the algebraic sum (signs included) of all charge inside the surface. Charge outside contributes exactly zero net flux: its lines enter the surface and leave again.
Two surprises worth dwelling on. First, the field \(\vec{E}\) appearing in the integral is the total field — produced by all charges, inside and out — yet only the enclosed charge survives on the right-hand side. The outside charges shape \(\vec{E}\) point by point but contribute nothing to the net flux. Second, Gauss' law holds for any closed surface whatsoever; we get to choose a convenient one, and that freedom is the whole secret of the method.
Section 23-5

Gauss' Law and Coulomb's Law

Gauss' law is not a new postulate competing with Coulomb's law — the two are equivalent. To see it, surround a single point charge \(q\) with an imaginary sphere of radius \(r\) centred on it. By symmetry the field has the same magnitude everywhere on the sphere and points radially outward, parallel to \(d\vec{A}\) at every patch. The flux integral then collapses:

Recovering Coulomb's law from Gauss' law
\[ \varepsilon_{0}\oint \vec{E}\cdot d\vec{A} = \varepsilon_{0}\,E\!\oint dA = \varepsilon_{0}\,E\,(4\pi r^{2}) = q \;\Longrightarrow\; E = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}} \]
Out drops the familiar inverse-square field of a point charge. Run the argument backwards and Coulomb's law plus superposition reproduces Gauss' law. They carry the same content; Gauss' law simply packages it in a form built for symmetry.
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The strategy in one sentence
Choose a Gaussian surface on which E is constant in magnitude and either parallel or perpendicular to dA — then the integral becomes E times an area.

Symmetry does the heavy lifting. On the right surface, \(\oint \vec{E}\cdot d\vec{A}\) reduces to \(EA\) over the part where the field pierces through, and to zero where the field skims along the surface. Set \(\varepsilon_{0}EA = q_{\text{enc}}\) and solve for \(E\). No integration tables required.

Section 23-6

A Charged Isolated Conductor

Gauss' law settles, almost instantly, a question that would otherwise be hard: where does the charge on a metal object go, and what is the field around it? In electrostatic equilibrium the charges in a conductor have stopped moving, which means the field inside the conducting material must be exactly zero — any leftover field would keep pushing the free electrons around. Draw a Gaussian surface just beneath the outer skin: since \(\vec{E} = 0\) on it, the enclosed charge is zero. All the excess charge must therefore reside on the outer surface.

Field just outside a charged conductor
\[ E_{\text{inside}} = 0, \qquad E_{\text{just outside}} = \frac{\sigma}{\varepsilon_{0}} \]
The field at the surface is perpendicular to it (a tangential component would drive surface currents) and has magnitude \(\sigma/\varepsilon_{0}\), where \(\sigma\) is the local surface charge density. Note this is twice the field of an isolated nonconducting sheet of the same \(\sigma\) — the conductor's charge lives entirely on one side.
The cavity and the cage. If a conductor has an empty internal cavity with no charge in it, the field inside that cavity is zero, and no charge appears on the cavity wall. This is why a metal enclosure shields its interior from outside fields — the principle of the Faraday cage that protects you inside a car during a lightning strike, and that keeps sensitive electronics quiet.
Section 23-7

Applying Gauss' Law: Cylindrical Symmetry

For an infinitely long line (or thin rod) carrying uniform linear charge density \(\lambda\), the natural Gaussian surface is a coaxial cylinder of radius \(r\) and length \(L\). By symmetry the field points straight out, perpendicular to the line, with the same magnitude all around. On the two flat end caps the field skims along the surface and contributes no flux; only the curved side, of area \(2\pi r L\), matters. The charge enclosed is \(\lambda L\).

Field of an infinite line of charge
\[ \varepsilon_{0}\,E\,(2\pi r L) = \lambda L \;\Longrightarrow\; E = \frac{\lambda}{2\pi\varepsilon_{0}\,r} \]
The length \(L\) cancels, as it must for an infinite line, and the field falls off as \(1/r\) — slower than the \(1/r^{2}\) of a point charge, because charge stretches away on both sides. Compare the effort: Chapter 22 reached this only after a real integral; here it is three lines.
When "infinite" is good enough. No real wire is infinite, but the formula is accurate whenever you are much closer to the wire than its length and far from its ends. That covers most practical cases — coaxial cables, charged filaments, the field near a long charged rod in the lab.
Section 23-8

Applying Gauss' Law: Planar Symmetry

For an infinite nonconducting sheet with uniform surface charge density \(\sigma\), symmetry says the field points straight away from the sheet on both sides, with the same magnitude everywhere. The Gaussian surface of choice is a small cylinder ("pillbox") piercing the sheet, with its two caps (each of area \(A\)) parallel to the sheet. The field pierces both caps but skims the curved side, so the flux is \(2EA\); the enclosed charge is \(\sigma A\).

Field of an infinite sheet of charge
\[ \varepsilon_{0}\,(2EA) = \sigma A \;\Longrightarrow\; E = \frac{\sigma}{2\varepsilon_{0}} \]
The distance from the sheet has vanished entirely: the field of an infinite sheet is uniform. This matches the large-disk limit found by integration in Chapter 22, now obtained in a single step. The factor of two comes from the field emerging on both faces of the sheet.
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Sheet vs. conductor — mind the factor of 2
Nonconducting sheet: E = σ/2ε₀. Conductor's surface: E = σ/ε₀.

A lone charged sheet sends field out both sides, so each side gets \(\sigma/2\varepsilon_{0}\). A conductor confines all its charge to one outward face, so that face gets the full \(\sigma/\varepsilon_{0}\). Two oppositely charged conducting plates (a capacitor) produce a uniform \(\sigma/\varepsilon_{0}\) between them and nearly zero outside — the workhorse of the next chapters.

Section 23-9

Applying Gauss' Law: Spherical Symmetry

For any spherically symmetric charge distribution, a concentric spherical Gaussian surface of radius \(r\) reduces the flux to \(E(4\pi r^{2})\). Two classic results follow, summarised in the shell theorems.

Table 23-1 · Field by Gauss' law for symmetric distributions
DistributionField magnitudeFalls off as
Point charge / outside any sphere\(E = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q}{r^{2}}\)1/r²
Infinite line of charge\(E = \dfrac{\lambda}{2\pi\varepsilon_{0}r}\)1/r
Infinite sheet of charge\(E = \dfrac{\sigma}{2\varepsilon_{0}}\)uniform
Inside a thin charged shell\(E = 0\)—
Inside a uniform ball of charge\(E = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{qr}{R^{3}}\)∝ r
The two shell theorems
\[ E_{\text{outside}} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}}\;(r\ge R), \qquad E_{\text{inside shell}} = 0\;(r\lt R) \]
A uniform shell acts, from outside, exactly like a point charge at its centre; inside the shell the enclosed charge is zero, so the field vanishes. For a solid uniform ball, the inside field grows linearly with \(r\) because only the charge within radius \(r\) counts.
Why "enclosed" is the only word that matters. Standing inside a charged shell you feel no field at all, even though charge surrounds you on every side — the pulls from all directions cancel perfectly. Gauss' law sees this at a glance: enclose zero charge, get zero flux, get zero field. The same logic governs gravity inside a hollow planet.
Worked Examples

Putting It to Work

1 Flux through a cube in a uniform field

Problem. A uniform field \(\vec{E}\) points along the \(x\)-axis. A cube of edge \(L\) sits with its faces parallel to the coordinate planes. What is the net flux through the cube?

Solution. Only the two faces perpendicular to \(\vec{E}\) matter; the other four are skimmed by the field. The field enters the left face and leaves the right.

Inflow cancels outflow
\[ \Phi = \underbrace{(+EL^{2})}_{\text{right face}} + \underbrace{(-EL^{2})}_{\text{left face}} + 0 = 0 \]

Zero, as Gauss' law demands — the cube encloses no charge. A uniform field threads cleanly through, entering and leaving in equal measure.

2 Net flux from an enclosed charge

Problem. A charge \(q = +3.0\times10^{-6}\,\mathrm{C}\) sits somewhere inside a closed surface of irregular shape. Find the net flux through the surface.

Solution. Gauss' law cares only about the enclosed charge, not the shape or the position of \(q\) inside.

Φ = q_enc / ε₀
\[ \Phi = \frac{q}{\varepsilon_{0}} = \frac{3.0\times10^{-6}}{8.85\times10^{-12}} \approx 3.4\times10^{5}\,\mathrm{N\cdot m^{2}/C} \]

If a second charge \(-q\) were placed outside the surface, the answer would not change at all — outside charge contributes zero net flux.

3 Field near a long charged wire

Problem. A long straight wire carries a uniform linear charge density \(\lambda = 2.0\times10^{-6}\,\mathrm{C/m}\). Find the field magnitude at \(r = 0.10\,\mathrm{m}\) from the wire.

Solution. Use the cylindrical-symmetry result directly.

E = λ / 2πε₀r
\[ E = \frac{2.0\times10^{-6}}{2\pi(8.85\times10^{-12})(0.10)} \approx 3.6\times10^{5}\,\mathrm{N/C} \]

The field points radially away from the positively charged wire. Move twice as far out and it halves, the signature of the \(1/r\) law.

4 Field between two charged plates

Problem. Two large parallel conducting plates carry equal and opposite surface charge densities \(\pm\sigma\), with \(\sigma = 1.8\times10^{-7}\,\mathrm{C/m^{2}}\). Find the field between them.

Solution. Between the plates the two sheets' fields add; the conductor result \(\sigma/\varepsilon_{0}\) already accounts for the one-sided charge.

E = σ / ε₀ (uniform, between the plates)
\[ E = \frac{\sigma}{\varepsilon_{0}} = \frac{1.8\times10^{-7}}{8.85\times10^{-12}} \approx 2.0\times10^{4}\,\mathrm{N/C} \]

The field is uniform across the gap and points from the positive plate to the negative one. Outside the pair the fields nearly cancel — this is exactly the parallel-plate capacitor of Chapter 25.

5 Inside and outside a uniform ball of charge

Problem. A nonconducting sphere of radius \(R\) carries total charge \(q\) spread uniformly through its volume. Find the field at a radius \(r\lt R\) and at \(r\gt R\).

Solution. Use a concentric spherical Gaussian surface. Outside, it encloses all of \(q\); inside, it encloses only the fraction \(q\,(r/R)^{3}\) within radius \(r\).

ε₀E(4πr²) = q_enc
\[ E_{\text{in}} = \frac{1}{4\pi\varepsilon_{0}}\frac{q\,r}{R^{3}}\;(r\lt R), \qquad E_{\text{out}} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}}\;(r\gt R) \]

The field rises linearly from zero at the centre to a maximum at the surface, then falls off as \(1/r^{2}\) outside — the two pieces meet smoothly at \(r = R\), a good consistency check.

Review

Chapter Summary

✓ Electric flux

\(\Phi = \int \vec{E}\cdot d\vec{A}\), in N·m²/C — the amount of field threading a surface; for a flat patch \(EA\cos\theta\).

✓ Gauss' law

\(\varepsilon_{0}\oint \vec{E}\cdot d\vec{A} = q_{\text{enc}}\); only enclosed charge matters, for any closed surface.

✓ Gaussian surface

Choose one matching the symmetry so \(\oint \vec{E}\cdot d\vec{A}\) becomes a simple \(EA\).

✓ Equivalent to Coulomb

A spherical surface around a point charge gives back \(E = kq/r^{2}\); the two laws are one.

✓ Conductor

Field inside is zero; charge sits on the outer surface; just outside \(E = \sigma/\varepsilon_{0}\).

✓ Line of charge

Cylindrical symmetry gives \(E = \dfrac{\lambda}{2\pi\varepsilon_{0}r}\), falling as \(1/r\).

✓ Sheet of charge

Planar symmetry gives a uniform \(E = \dfrac{\sigma}{2\varepsilon_{0}}\) (nonconductor).

✓ Spherical shell & ball

Outside: like a point charge. Inside a shell: \(E = 0\). Inside a uniform ball: \(E \propto r\).

✓ A Maxwell equation

Gauss' law for the electric field is one of the four equations summarising all of electromagnetism.

Practice

Problems

Take \(k = 8.99\times10^{9}\,\mathrm{N\cdot m^{2}/C^{2}}\), \(\varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{C^{2}/N\cdot m^{2}}\), and \(e = 1.602\times10^{-19}\,\mathrm{C}\). For each problem, first identify the symmetry, then choose a Gaussian surface on which \(E\) is constant and either parallel or perpendicular to \(d\vec{A}\).

  1. A flat square of side 0.20 m sits in a uniform field of 450 N/C with its normal at \(30^{\circ}\) to the field. Find the flux through it.
  2. A point charge \(q = 5.0\,\mu\mathrm{C}\) is at the centre of a cube. Find the flux through the whole cube, and through one face.
  3. A closed surface encloses charges \(+8.0\,\mathrm{nC}\), \(-3.0\,\mathrm{nC}\), and \(+2.0\,\mathrm{nC}\). Find the net flux through it.
  4. Why does a charge placed outside a closed surface contribute zero net flux? Argue from field lines.
  5. A long wire has linear charge density \(\lambda = 4.5\times10^{-6}\,\mathrm{C/m}\). Find the field at 8.0 cm from the wire.
  6. At what distance from the wire of Problem 5 does the field fall to \(1.0\times10^{5}\,\mathrm{N/C}\)?
  7. An infinite nonconducting sheet has \(\sigma = 6.0\times10^{-6}\,\mathrm{C/m^{2}}\). Find the field on either side.
  8. Two infinite parallel nonconducting sheets carry \(+\sigma\) and \(-\sigma\). Find the field between them and outside them by superposing the single-sheet fields.
  9. A large conducting plate carries a surface charge density \(\sigma = 2.0\times10^{-6}\,\mathrm{C/m^{2}}\). Find the field just outside its surface.
  10. A thin spherical shell of radius 10 cm carries \(q = 6.0\,\mathrm{nC}\). Find the field at \(r = 5.0\,\mathrm{cm}\) and at \(r = 20\,\mathrm{cm}\).
  11. A solid nonconducting sphere of radius \(R = 8.0\,\mathrm{cm}\) carries \(q = 12\,\mathrm{nC}\) spread uniformly through its volume. Find the field at \(r = 4.0\,\mathrm{cm}\).
  12. For the sphere of Problem 11, find the field at the surface and confirm the inside and outside formulas agree there.
  13. A conducting sphere of radius \(R\) carries charge \(Q\). Sketch \(E(r)\) for all \(r\) and explain the jump at \(r = R\).
  14. Starting from Gauss' law, derive Coulomb's law for the field of a point charge, stating clearly where symmetry is used.
Tip: three habits make Gauss' law effortless. First, name the symmetry (spherical, cylindrical, or planar) before drawing anything — it dictates the Gaussian surface and the direction of \(\vec{E}\). Second, split the chosen surface into a part the field pierces (contributes \(EA\)) and a part it skims (contributes zero); never integrate what symmetry already settles. Third, on the right-hand side count only the charge actually enclosed — charge outside the surface, however large, never appears.