Unit 2

Gauss's Law

In Unit 1 every field took an integral over the source, piece by piece. Gauss's law turns the problem inside out: instead of adding up the source, it counts the field lines coming out of a closed surface and says that number is fixed by the charge inside. When the charge is arranged symmetrically — a sphere, a long wire, a flat plate, which is to say most of the shapes engineers build — this gives the field in two lines of algebra. It also tells you, with no calculation at all, why the inside of a metal box is shielded, why charge sits on the outside of a conductor, and why coaxial cables do not radiate their field into neighbours. It is the first of Maxwell's four equations. The lab lets you drag a closed surface around charges and watch the flux change only when a charge crosses the boundary.

1 · Electric flux

Hold a hoop in a stream of water. How much water flows through it per second depends on three things: how fast the water moves, how big the hoop is, and how it is tilted. Face-on to the flow, maximum; edge-on, nothing. Electric flux is the same idea for field lines: it measures “how much field” passes through a surface. Nothing actually flows — think of it as the number of field lines piercing the surface.

A flat surface of area \(A\) in a uniform field. Its orientation is described by the unit normal \(\hat{\mathbf n}\), a unit vector perpendicular to the surface. The field lines that cross it are the same ones that would cross a face-on surface of area \(A\cos\theta\) (dashed), where \(\theta\) is the angle between \(\mathbf E\) and \(\hat{\mathbf n}\).

\[ \Phi_E = EA\cos\theta = \mathbf E\cdot\mathbf A,\qquad \mathbf A = A\,\hat{\mathbf n} \] \(\Phi_E\) (capital phi) is the electric flux, unit N·m²/C. \(\mathbf A\) is the area vector: magnitude \(A\), direction \(\hat{\mathbf n}\). Face-on (\(\theta = 0\)): \(\Phi = EA\). Edge-on (\(\theta = 90^\circ\)): \(\Phi = 0\). If \(\theta > 90^\circ\) the field goes through “backwards” and the flux is negative.

If the field varies over the surface, or the surface is curved, chop it into tiny patches \(d\mathbf A\), each flat and small enough that \(\mathbf E\) is constant on it, and add the patches up:

\[ \Phi_E = \int_S \mathbf E\cdot d\mathbf A \qquad\qquad \Phi_E = \oint_S \mathbf E\cdot d\mathbf A \ \ (\text{closed surface}) \] This is a surface integral: the sum of \(\mathbf E\cdot d\mathbf A\) over every patch. The circle on \(\oint\) says the surface is closed — a balloon, a box, a can — with an inside and an outside. For closed surfaces \(\hat{\mathbf n}\) always points outward, so field lines leaving count as positive flux and field lines entering count as negative.

You will rarely integrate

In this course the surface integral almost always collapses to “\(E\) times an area”, because we choose surfaces on which \(\mathbf E\) is either perpendicular to the surface with constant magnitude (flux \(= EA\)) or parallel to it (flux \(= 0\)). Section 3 is about choosing such surfaces.

Common mistake

Using the angle between \(\mathbf E\) and the surface instead of the angle between \(\mathbf E\) and the normal. If a problem says “the field makes \(30^\circ\) with the plane of the loop”, then \(\theta = 60^\circ\) and \(\Phi = EA\cos60^\circ\). Draw the normal every time.

2 · Gauss's law

Put a point charge \(q\) at the centre of an imaginary sphere of radius \(r\). On the sphere the field has magnitude \(kq/r^2\) everywhere and points straight out — parallel to \(\hat{\mathbf n}\). So the flux is simply field times area:

\[ \Phi_E = \frac{kq}{r^2}\cdot 4\pi r^2 = 4\pi k q = \frac{q}{\varepsilon_0} \] The \(r^2\) cancels! A bigger sphere has a weaker field but exactly proportionally more area. That cancellation is the inverse-square law at work, and it is also why \(k\) was written as \(1/(4\pi\varepsilon_0)\): the \(4\pi\) of a sphere's area cancels.

Now picture field lines instead of formulas. The number of lines leaving \(q\) is fixed by \(q\). Any closed surface around \(q\) — a sphere, a lumpy potato, a box — is crossed by all of those lines, so it has the same flux. And a charge outside the surface sends every line that enters out again somewhere else: zero net flux. Superposition then gives the general law:

\[ \oint_S \mathbf E\cdot d\mathbf A = \frac{Q_{\text{enc}}}{\varepsilon_0} \] Gauss's law. The net outward flux through any closed surface \(S\) (called a Gaussian surface) equals the total charge enclosed by it, \(Q_{\text{enc}}\), divided by \(\varepsilon_0\). The surface is imaginary — a mathematical bubble you choose, not a physical object.

Common mistakes

“Zero flux means zero field.” No. A charge outside the surface makes a non-zero field all over it, but its flux in and out cancels. Zero net flux only says the enclosed charge is zero.
“\(\mathbf E\) in Gauss's law is only the field of the enclosed charge.” No. \(\mathbf E\) is the total field from all charges, inside and out. It is the flux, not the field, that depends only on the inside.
Including charge outside \(S\) in \(Q_{\text{enc}}\), or including all of a charged object when only part of it is inside \(S\).

EE connection

Gauss's law is the first of Maxwell's equations, the four laws that contain all of electromagnetism — circuits, motors, antennas, optics, fibre and Wi-Fi (Unit 9). In the electromagnetic field solvers used to design chips, packages and antennas, this equation is literally one of the equations being solved on a mesh of millions of tiny cells.

3 · Choosing a Gaussian surface

Gauss's law is true for every closed surface, but it only hands you \(E\) when you can pull \(E\) out of the integral. That needs a surface that matches the symmetry of the charge. There are exactly three symmetries that work in practice:

The three standard Gaussian surfaces. On each, the field is either perpendicular to a part of the surface with the same magnitude everywhere on that part (the sphere; the cylinder's curved side; the pillbox's two ends) or parallel to it (cylinder caps, pillbox side), where it contributes no flux.

Symmetry	Charge looks like	\(\mathbf E\) points	Gaussian surface	Flux
Spherical	point, ball, spherical shell	radially from a centre	concentric sphere, radius \(r\)	\(E\cdot4\pi r^2\)
Cylindrical	long line, wire, rod, coax	radially from an axis	coaxial cylinder, radius \(r\), length \(\ell\)	\(E\cdot2\pi r\ell\)
Planar	large flat sheet or slab	perpendicular to the plane	pillbox through the plane	\(E\cdot2A\) (or \(EA\) at a conductor)

The five-step recipe

Symmetry. Decide which of the three it is, and argue which way \(\mathbf E\) points and what it can depend on (only \(r\), say).
Surface. Choose the matching Gaussian surface passing through the point where you want \(E\).
Left side. Write the flux: \(E\) times the area of the parts where \(\mathbf E\perp\) surface; zero on parts where \(\mathbf E\parallel\) surface.
Right side. Find \(Q_{\text{enc}}\) — only the charge inside the surface. For a partly enclosed object, integrate the density over the enclosed part.
Solve for \(E\), then check the limits and direction.

Interactive lab Gauss’s law: flux through any surface Open it in Interactive Laboratories →

4 · Spheres: a uniform ball and a thin shell

Take an insulating ball of radius \(R\) with charge \(Q\) spread uniformly through its volume, so the volume charge density is \(\rho = Q\big/\tfrac43\pi R^3\). Spherical symmetry says \(\mathbf E\) is radial and depends only on the distance \(r\) from the centre. Use a concentric sphere of radius \(r\).

Outside (\(r \ge R\)): the whole charge is enclosed.

\[ E\cdot4\pi r^2 = \frac{Q}{\varepsilon_0}\quad\Longrightarrow\quad E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{kQ}{r^2} \] From outside, a spherically symmetric charge is indistinguishable from a point charge at its centre. (Newton proved the same thing for gravity with much more effort — this is why planets act like point masses.)

Inside (\(r < R\)): only the charge within radius \(r\) is enclosed, a fraction \(r^3/R^3\) of the total.

\[ Q_{\text{enc}} = \rho\cdot\tfrac43\pi r^3 = Q\frac{r^3}{R^3} \quad\Longrightarrow\quad E\cdot4\pi r^2 = \frac{Qr^3}{\varepsilon_0R^3} \quad\Longrightarrow\quad E = \frac{kQ}{R^3}\,r = \frac{\rho\,r}{3\varepsilon_0} \] The field grows linearly from zero at the centre to \(kQ/R^2\) at the surface, where it matches the outside formula — the field is continuous at \(r = R\) because there is no surface charge there.

A thin spherical shell of radius \(R\) with charge \(Q\) on its surface is different inside: a Gaussian sphere with \(r < R\) encloses nothing, so \(E = 0\) everywhere inside. Outside it is again \(kQ/r^2\). This is the shell theorem.

Field of a uniformly charged ball (accent) and of a thin shell with the same total charge (blue). Outside they are identical. Inside, the ball's field grows linearly; the shell's is zero. The shell's field jumps at \(r = R\) by \(\sigma/\varepsilon_0\), the signature of a surface charge.

Common mistake

Using the total \(Q\) for \(Q_{\text{enc}}\) at an inside point. Inside the ball, the charge outside your Gaussian sphere contributes zero net field there (the shell theorem, layer by layer), and the formula knows this only if you use \(Q_{\text{enc}} = Qr^3/R^3\). For a non-uniform density \(\rho(r)\), you must integrate: \(Q_{\text{enc}} = \int_0^r \rho(r')\,4\pi r'^2\,dr'\).

5 · The infinite line and the infinite sheet

Line of charge

A very long straight line with charge per length \(\lambda\). By symmetry \(\mathbf E\) points radially away from the line and depends only on the distance \(r\) from it. Use a coaxial cylinder of radius \(r\) and length \(\ell\). Through the end caps the flux is zero (\(\mathbf E\) lies along them); through the curved side it is \(E\cdot 2\pi r\ell\). The enclosed charge is \(\lambda\ell\):

\[ E\,(2\pi r\ell) = \frac{\lambda\ell}{\varepsilon_0}\quad\Longrightarrow\quad E = \frac{\lambda}{2\pi\varepsilon_0 r} \] The same answer Unit 1 needed a trig-substitution integral and a limit for. The length \(\ell\) cancels, as it must — it was our arbitrary choice.

Sheet of charge

A large flat sheet with charge per area \(\sigma\). By symmetry \(\mathbf E\) is perpendicular to the sheet, pointing away on both sides (for \(\sigma > 0\)), with the same magnitude at equal distances. Use a pillbox with end area \(A\) straddling the sheet. Flux leaves through both ends, none through the side:

\[ 2EA = \frac{\sigma A}{\varepsilon_0}\quad\Longrightarrow\quad E = \frac{\sigma}{2\varepsilon_0} \] Independent of distance, as found in Unit 1 from the disk. Real sheets have edges; this holds where you are much closer to the sheet than to its edges.

Two sheets: the parallel-plate arrangement

Two large parallel sheets with \(+\sigma\) and \(-\sigma\). Each makes \(\sigma/(2\varepsilon_0)\) on both of its sides. Between them, the two fields point the same way (away from \(+\), toward \(-\)) and add; outside they point opposite ways and cancel:

\[ E_{\text{between}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0},\qquad E_{\text{outside}} = 0 \] A uniform field confined to the gap. This is the parallel-plate capacitor of Unit 4, and the “uniform field between plates” used for electron deflection in Unit 1.

EE connection

Every capacitor in your circuits — from the ceramic chip on a PCB to the gate of each transistor — relies on this result: equal and opposite charges on two close plates trap a uniform field between them and almost none outside. The line result gives the field inside a coaxial cable (Worked example 4), which sets its capacitance per metre and helps set its characteristic impedance (the “50 \(\Omega\)” printed on the cable).

6 · Conductors in electrostatic equilibrium

A conductor has free electrons. If there were any field inside the metal, they would move. So once charges stop moving — electrostatic equilibrium, reached almost instantly in a metal — the following must be true. Each follows from Gauss's law with no integration.

Fact	Why
1. \(\mathbf E = 0\) everywhere inside the conducting material.	Otherwise free electrons would feel a force and move; that is not equilibrium. The charges rearrange until their own field cancels the applied one inside.
2. Any net charge sits on the surface.	Draw a Gaussian surface just inside the metal surface. \(\mathbf E = 0\) on it, so flux \(= 0\), so \(Q_{\text{enc}} = 0\). No charge inside; it must all be on the surface.
3. Just outside, \(\mathbf E\) is perpendicular to the surface, with \(E = \sigma/\varepsilon_0\).	A parallel component would push charges along the surface. For the magnitude, use a tiny pillbox with one end inside the metal (\(E = 0\)) and one just outside: \(EA = \sigma A/\varepsilon_0\).
4. An empty cavity inside a conductor has \(\mathbf E = 0\) and no charge on its walls.	Whatever happens outside, the metal shields the cavity. This is the Faraday cage.
5. A charge \(q\) inside a cavity induces \(-q\) on the cavity wall and \(+q\) extra on the outer surface (if the conductor is otherwise neutral).	A Gaussian surface inside the metal surrounding the cavity has zero flux, so it encloses zero net charge: the wall must hold \(-q\). Charge conservation puts the balancing \(+q\) on the outside.

A charge \(+q\) inside a neutral conducting shell. The field in the cavity pulls \(-q\) onto the inner wall; the shell is neutral, so \(+q\) appears on the outer surface. The metal itself is field-free, and from outside the shell looks like a point charge \(q\) — wherever in the cavity \(q\) happens to sit.

Common mistake: \(\sigma/\varepsilon_0\) or \(\sigma/2\varepsilon_0\)?

A lone charged sheet gives \(\sigma/(2\varepsilon_0)\) on each side. Just outside a conductor the field is \(\sigma/\varepsilon_0\) — twice as big — because the rest of the conductor's charges cancel the field inside and double it outside. In the pillbox derivation the difference shows up as flux through two ends (sheet) versus one end (conductor, whose inner end is in a zero field).

One more qualitative fact: on a conductor of irregular shape, charge crowds at sharp points, where the surface curves most, so the field just outside (\(\sigma/\varepsilon_0\)) is strongest there. Dry air breaks down and conducts at about \(3\times10^{6}\) V/m (N/C), so a sharp point sprays charge (corona discharge) long before a smooth surface does. Unit 3 shows why.

EE connection

Shielding is facts 4 and 5 put to work. The braided outer conductor of a coax cable, the metal can over an RF chip, a shielded room for sensitive measurements and the metal body of a car in a lightning storm are all Faraday cages. (For rapidly changing fields a shield must also be thick enough and its seams small compared with the wavelength, but the idea is Gauss's.) High-voltage hardware uses smooth, rounded terminals to avoid corona; a lightning rod uses a sharp point on purpose.

7 · Always true, not always useful

Gauss's law holds for every closed surface and every charge distribution. But it is one equation about an integral of \(\mathbf E\). To extract \(E\) itself you need \(E\) to be constant (or zero) on each piece of the surface, and that requires symmetry.

A finite rod: the flux through a cylinder around it is still \(Q_{\text{enc}}/\varepsilon_0\), but near the ends the field tilts and weakens, so \(E\) is not constant on the cylinder and cannot be pulled out. Use the Unit 1 integral.
A cube with a charge at its centre: the flux through each face is exactly \(q/(6\varepsilon_0)\) by symmetry — but the field on a face varies from point to point, so the cube does not give \(E\). (The flux was easy; the field was not.)
A dipole: any surface around both charges has zero flux, yet the field is certainly not zero. Gauss's law is true here and tells you nothing about \(E\).

Symmetry argument first

Before writing \(E\cdot(\text{area})\), say out loud why \(E\) is the same everywhere on that part of the surface and why it is perpendicular to it. If you cannot, Gauss's law will still be true, but you are about to write a false equation.

A deeper look. Gauss's law is equivalent to the inverse-square law: it works only because the \(r^2\) in Coulomb's law cancels the \(r^2\) in a sphere's area. Physicists test it to extraordinary precision for this reason. Shrinking the Gaussian surface to a point turns it into a local statement, the differential form \(\nabla\cdot\mathbf E = \rho/\varepsilon_0\): field lines spread out (“diverge”) only from places where there is charge. That is how it appears in Maxwell's equations and in field-solver software.

8 · Worked examples

1. Flux through a tilted square and a closed box

A uniform field \(E = 2.0\times10^{3}\) N/C passes through a 10 cm \(\times\) 10 cm square whose normal makes \(60^\circ\) with \(\mathbf E\). (a) Find the flux. (b) What is the net flux through a closed cube of the same size placed in this field?

\[ \text{(a)}\ \Phi = EA\cos\theta = (2.0\times10^{3})(0.010)(0.500) = 10\ \text{N}\cdot\text{m}^2/\text{C} \]

(b) Zero. There is no charge inside the cube, and indeed every field line that enters through one face leaves through another: inward flux (negative) and outward flux (positive) cancel. This holds for any orientation of the cube.

2. A charge in a cube

A point charge \(q = 5.0\) nC sits at the centre of a cube of side 20 cm. Find the flux through the whole cube and through one face. What if the charge were moved just outside the cube?

\[ \Phi_{\text{total}} = \frac{q}{\varepsilon_0} = \frac{5.0\times10^{-9}}{8.85\times10^{-12}} = 565\ \text{N}\cdot\text{m}^2/\text{C},\qquad \Phi_{\text{face}} = \frac{565}{6} = 94.2\ \text{N}\cdot\text{m}^2/\text{C} \]

The cube's size does not matter. The one-face result needs the charge to be at the centre so all six faces are equivalent. If the charge is moved just outside, \(Q_{\text{enc}} = 0\), so the total flux is zero — even though the field on the nearest face is now very strong.

3. A uniformly charged insulating ball

An insulating ball of radius 10 cm has 8.0 nC spread uniformly through its volume. Find \(\rho\) and the field at \(r = 5.0\), 10 and 20 cm from the centre.

\[ \rho = \frac{Q}{\tfrac43\pi R^3} = \frac{8.0\times10^{-9}}{4.19\times10^{-3}} = 1.91\times10^{-6}\ \text{C/m}^3 \] \[ r = 0.05:\ E = \frac{kQr}{R^3} = \frac{(8.99\times10^{9})(8.0\times10^{-9})(0.05)}{(0.10)^3} = 3.60\times10^{3}\ \text{N/C} \] \[ r = 0.10:\ E = \frac{kQ}{R^2} = \frac{71.9}{0.010} = 7.19\times10^{3}\ \text{N/C} \] \[ r = 0.20:\ E = \frac{kQ}{r^2} = \frac{71.9}{0.040} = 1.80\times10^{3}\ \text{N/C} \]

All radially outward. Check via \(\rho r/(3\varepsilon_0) = (1.91\times10^{-6})(0.05)/(2.655\times10^{-11}) = 3.60\times10^{3}\) N/C ✓. Note the pattern: halfway in, half the surface field (linear); twice as far out, a quarter (inverse square).

4. EE: the field inside a coaxial cable

A coax has a solid inner conductor of radius \(a = 0.50\) mm carrying \(+\lambda = 2.0\) nC/m and a thin outer conducting sleeve of inner radius \(b = 2.0\) mm carrying \(-\lambda\). Find \(E\) (a) inside the inner conductor, (b) at \(r = 1.0\) mm, (c) outside the cable. (d) Where is \(E\) greatest, and why does that matter?

(a) \(r < a\): inside a conductor in equilibrium, \(E = 0\). The inner conductor's charge is on its surface.

(b) Gaussian cylinder with \(a < r < b\) encloses \(\lambda\ell\):

\[ E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r} = \frac{2(8.99\times10^{9})(2.0\times10^{-9})}{1.0\times10^{-3}} = 3.60\times10^{4}\ \text{N/C, radially outward} \]

(c) A cylinder outside the whole cable encloses \(\lambda\ell - \lambda\ell = 0\). With cylindrical symmetry, \(E\cdot2\pi r\ell = 0\), so \(E = 0\). The field is entirely confined between the conductors.

(d) \(E\propto 1/r\), so it is largest at the surface of the inner conductor: \(E(a) = 35.96/(5.0\times10^{-4}) = 7.19\times10^{4}\) N/C. Insulation breaks down first there, which is what limits a cable's maximum voltage. And because the field stays inside, a coax neither leaks signal to nearby circuits nor picks up their fields — the reason it carries radio, video and test-instrument signals.

5. Two oppositely charged plates

Two large parallel plates carry \(\pm3.0\ \mu\text{C/m}^2\). Find the field between and outside them, and the force on an electron between them.

\[ E_{\text{between}} = \frac{\sigma}{\varepsilon_0} = \frac{3.0\times10^{-6}}{8.85\times10^{-12}} = 3.39\times10^{5}\ \text{N/C},\qquad E_{\text{outside}} = 0 \] \[ F = eE = (1.60\times10^{-19})(3.39\times10^{5}) = 5.42\times10^{-14}\ \text{N, toward the positive plate} \]

The field points from the \(+\) plate to the \(-\) plate; the electron is pushed the opposite way. The force is the same anywhere in the gap (uniform field) — move the electron closer to a plate and nothing changes, unlike near a point charge.

6. Point charge inside a charged conducting shell

A point charge \(+3.0\) nC sits at the centre of a thick conducting spherical shell with inner radius 10 cm and outer radius 15 cm. The shell carries net charge \(-5.0\) nC. Find the charge on each surface of the shell and \(E\) at \(r = 5\), 12 and 30 cm.

Surfaces: a Gaussian sphere inside the metal (\(r = 12\) cm) has \(E = 0\), so \(Q_{\text{enc}} = 0\): the inner surface holds \(-3.0\) nC. The shell's total is \(-5.0\) nC, so the outer surface holds \(-5.0 - (-3.0) =\) \(-2.0\) nC.

\[ r = 0.05:\ E = \frac{k(3.0\times10^{-9})}{(0.05)^2} = 1.08\times10^{4}\ \text{N/C (outward)} \] \[ r = 0.12:\ E = 0\ \ (\text{inside the metal}) \] \[ r = 0.30:\ Q_{\text{enc}} = 3.0 - 5.0 = -2.0\ \text{nC},\quad E = \frac{k(2.0\times10^{-9})}{(0.30)^2} = 200\ \text{N/C (inward)} \]

Check with fact 3: on the outer surface, \(\sigma = -2.0\times10^{-9}/(4\pi(0.15)^2) = -7.07\times10^{-9}\) C/m², so just outside \(|E| = |\sigma|/\varepsilon_0 \approx 799\) N/C; and \(kQ/r^2\) at \(r = 0.15\) m gives \(799\) N/C ✓.

Checklist

I can compute flux \(\Phi = EA\cos\theta\) using the angle to the normal, and I know outward flux is positive for a closed surface.
I can state Gauss's law and explain why charges outside a surface contribute no net flux (but do contribute field).
I identify spherical, cylindrical or planar symmetry and pick the matching sphere, cylinder or pillbox through the field point.
I can derive \(E\) inside and outside a uniform ball (\(kQr/R^3\), \(kQ/r^2\)) and a thin shell (0, \(kQ/r^2\)).
I can derive \(\lambda/(2\pi\varepsilon_0r)\) for a line and \(\sigma/(2\varepsilon_0)\) for a sheet, and \(\sigma/\varepsilon_0\) between parallel plates.
I know the conductor facts: \(E = 0\) inside, charge on the surface, \(E = \sigma/\varepsilon_0\) perpendicular just outside, cavities shielded.
I can find the induced charges on the surfaces of a conducting shell around a charge.
I know Gauss's law is always true but gives \(E\) only with symmetry.