Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Gauss's law is used in two ways in practice. The first is to compute the field of a symmetric source in a few lines (Problems 3.2.1–3.2.5). The second, together with the fact that the electric field vanishes inside a conductor in electrostatic equilibrium, is to say where the charge on a conductor sits and what the conductor shields. The two applications of this section use these ideas:

Faraday cage: no field inside a closed conductor (shielded rooms, cable screens, the car in a thunderstorm).
Coaxial cable: $E \propto 1/ ρ$ between the conductors, nothing outside, and the field peaks at the inner conductor.

Three basic results about conductors

A conductor contains charges that are free to move: the conduction electrons of a metal. When a conductor is placed in a field or given a charge, these free charges move until they reach a configuration in which no force acts on them any more. This state, reached in a very short time, is electrostatic equilibrium. Three results follow, the last two from Gauss's law.

(i) $E = 0$ inside the conducting material. If there were a field, the free charges would feel the force $q E$ and would still be moving, contrary to the assumption of equilibrium. Figure 3.2.11 shows how this comes about: the free charges pile up on the surfaces until their own field cancels the applied one inside the metal.
(ii) Any net charge resides on the surface. Since $E = 0$ , hence $D = 0$ , at every point inside the material, $ρ_{v} = \nabla \cdot D = 0$ there. Equivalently, a Gaussian surface drawn anywhere inside the metal, however close to the outer surface, has zero flux and encloses zero net charge. Excess charge can only sit on the surface, as a surface charge density $ρ_{S}$ .
(iii) Just outside, $E$ is normal to the surface and $D_{n} = ρ_{S}$ . A tangential component of $E$ at the surface would drive the surface charges along the surface, so in equilibrium the field just outside is normal to it. Now take a small pillbox of face area $Δ S$ straddling the surface, one face inside the metal and the other just outside (Figure 3.2.12), and let its height shrink to zero. The inner face has $D = 0$ and the side contributes nothing in the limit, so only the outer face counts:

$\Rightarrow \oint D \cdot d S = D_{n} Δ S = ρ_{S} Δ S D_{n} = ρ_{S}, E_{n} = \frac{ρ _{S}}{ε _{0}} .$

How a conductor reaches equilibrium (a metal slab in a uniform field \vec E_0). (a) At the first instant the field penetrates the metal and drives the free electrons against \vec E_0. (b) Electrons accumulate on the left face and leave positive charge on the right face, with \rho_S=\mp\varepsilon_0E_0. Their field cancels \vec E_0 inside, while outside the field is unchanged. (c) For any conductor in equilibrium, a Gaussian surface drawn inside the metal has \vec D=0 on it and therefore encloses no net charge: excess charge lives on the surface. — **Figure 3.2.11.** How a conductor reaches equilibrium (a metal slab in a uniform field $E_{0}$ ). (a) At the first instant the field penetrates the metal and drives the free electrons against $E_{0}$ . (b) Electrons accumulate on the left face and leave positive charge on the right face, with $ρ_{S} = \mp ε_{0} E_{0}$ . Their field cancels $E_{0}$ inside, while outside the field is unchanged. (c) For any conductor in equilibrium, a Gaussian surface drawn inside the metal has $D = 0$ on it and therefore encloses no net charge: excess charge lives on the surface.

A pillbox straddling the surface of a conductor in equilibrium. Inside the metal \vec D=0, and just outside \vec D is normal to the surface. With the height \Delta h\to0 only the outer face contributes, so D_n=\rho_S: the field just outside a conductor is fixed by the local surface charge. — **Figure 3.2.12.** A pillbox straddling the surface of a conductor in equilibrium. Inside the metal $D = 0$ , and just outside $D$ is normal to the surface. With the height $Δ h \to 0$ only the outer face contributes, so $D_{n} = ρ_{S}$ : the field just outside a conductor is fixed by the local surface charge.

In plain words

In a conductor the charges are free, so they keep moving until they feel no force: no field inside, all excess charge on the skin, and just outside a field that stands perpendicular to the surface with strength $ρ_{S} / ε_{0}$ .

Quick check

$5 nC$ is placed on a solid copper sphere. Where is the charge, and what is $E$ at the centre?

All $5 nC$ sit on the outer surface (spread uniformly, by symmetry), and $E = 0$ everywhere in the copper, including the centre.

Application 3.2.1 · Faraday cage

In a conductor in equilibrium the free charges move until $E = 0$ everywhere in the metal, otherwise they would still be moving. M. Faraday showed experimentally in 1836 that a closed conducting shell therefore screens static fields completely from its interior.

When an external field $E_{0}$ is switched on (Figure 3.2.13), the free charges move and pile up on the outer faces of the shell, negative on the side the field comes from and positive on the other side, until their own field cancels $E_{0}$ everywhere inside the metal. Now apply Gauss's law on a closed surface $S$ drawn inside the metal and surrounding the cavity. On $S$ we have $D = 0$ , so

\oint_{S} D \cdot d S = 0 ⟹ Q_{enc} = 0 :

if the cavity contains no charge, the total charge on the cavity wall is zero. In fact the wall carries no charge at all and the field in the empty cavity vanishes identically (see the box below). Whatever happens outside, a sensitive circuit inside the cage feels no static field. Figure 3.2.14 shows the computed field lines around a long conducting tube placed across a uniform field: the lines bend so as to meet the metal at right angles, end on the induced charges, and not a single line enters.

Faraday cage. Charges induced on the outer faces of a closed conducting shell cancel the external field \vec E_0 in the metal. A Gaussian surface S inside the metal encloses zero charge, and the cavity is free of field. — **Figure 3.2.13.** Faraday cage. Charges induced on the outer faces of a closed conducting shell cancel the external field $E_{0}$ in the metal. A Gaussian surface $S$ inside the metal encloses zero charge, and the cavity is free of field.

Field lines around a long conducting tube (cross-section, radius a) placed in a uniform field \vec E_0 pointing to the right, computed from the exact potential V=-E_0(\rho-a^2/\rho)\cos\phi (the potential is introduced in Section 3.3, and this V satisfies Laplace's equation and is zero on the tube). The lines meet the metal at right angles and end on the induced surface charge \rho_S=2\varepsilon_0E_0\cos\phi (negative on the side facing the oncoming field, positive on the far side), and no line reaches the cavity. Far away the lines are equally spaced, so each carries the same flux, and their crowding near \phi=0 and \phi=\pi shows the doubled field 2E_0 there. — **Figure 3.2.14.** Field lines around a long conducting tube (cross-section, radius $a$ ) placed in a uniform field $E_{0}$ pointing to the right, computed from the exact potential $V = - E_{0} (ρ - a^{2} / ρ) cos ϕ$ (the potential is introduced in Section 3.3, and this $V$ satisfies Laplace's equation and is zero on the tube). The lines meet the metal at right angles and end on the induced surface charge $ρ_{S} = 2 ε_{0} E_{0} cos ϕ$ (negative on the side facing the oncoming field, positive on the far side), and no line reaches the cavity. Far away the lines are equally spaced, so each carries the same flux, and their crowding near $ϕ = 0$ and $ϕ = π$ shows the doubled field $2 E_{0}$ there.

The shield need not be a solid sheet. For time-varying fields, which go beyond electrostatics, a mesh also shields provided its holes are much smaller than the wavelength. The door of a microwave oven is the standard example: at $2.45 GHz$ the free-space wavelength is about $12 cm$ , while the holes of the metal mesh are a few millimetres across. Screened rooms for electromagnetic-compatibility (EMC) testing and for magnetic resonance imaging (MRI), and the braided outer conductor of a coaxial cable, work on the same principle.

Everyday shielding

A car struck by lightning largely protects its occupants because the current and charge stay on the outer metal body, the cage, and do not enter the interior. A mobile phone often loses its signal in a metal lift for the same reason: the metal box screens the incoming radio waves.

Common mistake · “The field in a cavity is always zero”

The cage shields the cavity from charges outside. If a charge $q$ is placed inside the cavity, there is a field in the cavity: the lines of $q$ run to the induced charge $- q$ on the cavity wall (Worked example 3.2.2). It is the metal in which $E = 0$ , always.

Application 3.2.2 · Coaxial cable

The coaxial line, patented by O. Heaviside in 1880, is Problems 3.2.1 and 3.2.2 made into a product. It consists of an inner conductor of radius $a$ and a coaxial outer conductor, the screen, whose inner surface has radius $b$ . The space between them is filled with an insulator (air in what follows). In normal operation the inner conductor carries $+ ρ_{L}$ per metre and the screen $- ρ_{L}$ per metre (Figure 3.2.15). Three Gaussian cylinders tell us everything:

$S_{1}$ ( $a < ρ < b$ ) encloses only the charge of the inner conductor, $Q_{enc} = ρ_{L} L$ , so exactly as in Problem 3.2.1

$D = \frac{ρ _{L}}{2 π ρ} a_{ρ}, E = \frac{ρ _{L}}{2 π ε _{0} ρ} a_{ρ} .$

The field is largest at the inner conductor, $ρ = a$ , which is where the insulation is most stressed and where breakdown starts.
A Gaussian cylinder drawn inside the metal of the screen has $D = 0$ on it, so it encloses zero charge: the inner face of the screen carries exactly $- ρ_{L}$ per metre. The surface charge densities are $ρ_{S} = ρ_{L} / (2 π a)$ on the inner conductor and $- ρ_{L} / (2 π b)$ on the inner face of the screen. Both agree with $D_{n} = ρ_{S}$ .
$S_{2}$ ( $ρ > b$ ) encloses $ρ_{L} - ρ_{L} = 0$ : the outer face of the screen carries no charge and $E = 0$ outside. The field is confined to the insulator between the conductors: nothing leaks out, and (by Application 3.2.1) external static fields do not reach the signal.

If the screen were isolated and neutral, Gauss's law in its metal would still put $- ρ_{L}$ on its inner face, but neutrality would then put $+ ρ_{L}$ on its outer face and the cable would have the external field of a line charge. This is one reason why the screen is earthed: the outer charge flows away to earth (Worked example 3.2.1).

(a) Cross-section with two Gaussian cylinders.

(b) |\vec E| versus \rho (here b=3.5a). A coaxial cable with +\rho_L per metre on the inner conductor and -\rho_L on the screen. (a) The Gaussian cylinder S_1 gives the 1/\rho field between the conductors, while S_2 encloses no net charge, so there is no field outside. The inner conductor carries +\rho_L, the inner face of the screen -\rho_L, its outer face nothing. (b) The field is zero in the metal, jumps to its maximum at the surface of the inner conductor, falls as 1/\rho across the insulator to a/b of that value at the screen, and is zero in the screen and outside. — **Figure 3.2.15.** (b) $∣ E ∣$ versus $ρ$ (here $b = 3.5 a$ ). A coaxial cable with $+ ρ_{L}$ per metre on the inner conductor and $- ρ_{L}$ on the screen. (a) The Gaussian cylinder $S_{1}$ gives the $1/ ρ$ field between the conductors, while $S_{2}$ encloses no net charge, so there is no field outside. The inner conductor carries $+ ρ_{L}$ , the inner face of the screen $- ρ_{L}$ , its outer face nothing. (b) The field is zero in the metal, jumps to its maximum at the surface of the inner conductor, falls as $1/ ρ$ across the insulator to $a / b$ of that value at the screen, and is zero in the screen and outside.

Worked example 3.2.1 puts numbers in: for $a = 1 mm$ , $b = 3.5 mm$ and $ρ_{L} = 20 nC/m$ the field is $360 kV/m$ at the inner conductor and $103 kV/m$ at the screen, and since air breaks down at about $3 MV/m$ an air-spaced line of these dimensions can carry at most about $167 nC/m$ .

Solve the cross-section of a coaxial line numerically below: the potential and the field lines show the $1/ ρ$ field confined between the conductors and crowding at the inner one.

InteractiveLoading simulation…

Where you meet it

Antenna feeders, oscilloscope probes, cable television and the old 10BASE2 Ethernet on RG-58 cable. Filled with gas, the same geometry is the proportional counter and the Geiger–Müller tube of radiation detectors: the electron avalanche develops in the strong $1/ ρ$ field close to the thin central wire.

Quick check

The screen of a coaxial cable is earthed and the inner conductor carries $+ ρ_{L}$ . What is $E$ outside the cable?

Zero. A Gaussian cylinder in the screen metal forces $- ρ_{L}$ onto the inner face of the screen, and earthing removes any charge from the outer face. A cylinder outside the cable then encloses $ρ_{L} - ρ_{L} = 0$ .

Table 3.2.2. The applications of Gauss's law in this section.

Application	Physics used	Worked out in
Faraday cage	$E = 0$ in the conductor, $Q_{enc} = 0$ for the cavity	Worked example 3.2.2
Coaxial cable	cylindrical symmetry, $E \propto 1/ ρ$	Problems 3.2.1–3.2.2, Worked example 3.2.1

3.2.5 Applications of Gauss's law

Three basic results about conductors