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

The field inside a conductor is zero

In a conductor (a metal, for instance) the outer electrons of the atoms are not bound to individual atoms. They can move freely through the whole body. When an external electric field is applied to a conductor, a force $- e E$ acts on each of these free electrons and they start to move. They pile up on one side of the body and leave a deficit of electrons (a net positive charge) on the other side. The charges so separated produce a field of their own, which inside the conductor opposes the applied field. The motion continues as long as there is any net field left to drive it, that is, until the net force (and with it the net electric field) inside the conductor has become zero (Figure 3.4.16).

A metal block placed in an applied field (orange arrows, pointing to the right). (a) At first the field penetrates the metal and pushes every free electron to the left. (b) Electrons pile up on the left face and leave positive charge on the right face. The field of these surface charges points from + to -, against the applied field, and the field inside weakens (pale arrows). (c) The motion stops when the field inside is exactly zero. Outside, the field lines end on the negative surface charge and start again on the positive charge, meeting the surface at right angles. — **Figure 3.4.16.** A metal block placed in an applied field (orange arrows, pointing to the right). (a) At first the field penetrates the metal and pushes every free electron to the left. (b) Electrons pile up on the left face and leave positive charge on the right face. The field of these surface charges points from $+$ to $-$ , against the applied field, and the field inside weakens (pale arrows). (c) The motion stops when the field inside is exactly zero. Outside, the field lines end on the negative surface charge and start again on the positive charge, meeting the surface at right angles.

The time needed to reach this equilibrium is called the relaxation time. For good conductors it is so short that, for all electrostatic purposes, equilibrium is reached instantaneously. (A simple model gives the relaxation time as $ε / σ$ , where $σ$ is the conductivity. It will be discussed with steady currents.) In electrostatics, where by definition nothing moves any more, we can therefore state:

The net electric field inside a conductor in electrostatic equilibrium is zero: $E = 0$ .

All charge resides on the surface

A conductor (region I) surrounded by a dielectric (region II). (a) Any closed surface S drawn inside the metal encloses no net charge, because \vec E=0 on it. (b) Any two points of the conductor are at the same potential, because \vec E=0 along every path joining them inside the metal. Here: (a) Gaussian surface S just inside the surface. — **Figure 3.4.17.** A conductor (region I) surrounded by a dielectric (region II). (a) Any closed surface S drawn inside the metal encloses no net charge, because $E = 0$ on it. (b) Any two points of the conductor are at the same potential, because $E = 0$ along every path joining them inside the metal. Here: (a) Gaussian surface S just inside the surface.

(b) Two points A and B of the conductor.

Since the field in a conductor is zero, Gauss's law lets us prove at once that all charges must reside on the surface of a conductor. Take any closed surface S lying entirely inside the metal, for example just beneath its surface (Figure 3.4.17a). On S, $E = 0$ and therefore $D = 0$ , so

\oint_{S} D \cdot d S = ε \oint_{S} E \cdot d S = total charge inside S = 0.

This holds for every surface inside the metal, however small, so the charge density in the interior is zero. Any net charge placed on a conductor, and any charge induced on it by an external field, must therefore sit on its surface, in practice in a layer only a few atoms thick, which we describe as a surface charge density $ρ_{S}$ .

Now apply the boundary conditions of Section 3.4.6 with region I the conductor and region II the surrounding dielectric of permittivity $ε_{2}$ (often air), $a_{n}$ pointing out of the conductor. In region I, $E_{1 τ} = 0$ and $D_{1 n} = 0$ , so

Boundary conditions at the surface of a conductor

E_{2 τ} = 0, D_{2 n} = ε_{2} E_{2 n} = ρ_{S} (a_{n} out of the conductor) . (3.4.19)

Just outside a conductor the field has no tangential component: it is perpendicular to the surface. Its normal component gives the surface charge density, and vice versa: we can read the charge from the field, or the field from the charge.

A conductor is an equipotential body

Since the net electric field in a conductor is zero, we can also prove that the potential in a conductor is constant. For any two points A and B of the conductor (Figure 3.4.17b) we choose a path from A to B lying inside the metal. Along it $E = 0$ , so

V_{B} - V_{A} = - \int_{A}^{B} E \cdot d l = 0.

The whole conductor, including its surface, is an equipotential volume. This is consistent with $E_{2 τ} = 0$ : a field perpendicular to a surface does no work along it, so the surface is an equipotential surface, and field lines meet equipotentials at right angles (Section 3.3).

What the field looks like

An uncharged hollow metal sphere in an originally uniform field \vec E_0 (pointing to the right), computed field lines. The free electrons move until the field in the metal vanishes: negative charge gathers on the upstream side, positive charge on the downstream side (the size of the signs indicates |\rho_S|\propto|\cos\theta|, \theta measured from the direction of \vec E_0). Every field line outside meets the surface at right angles (E_\tau=0) and ends on the surface charge (\varepsilon_0E_n=\rho_S), and none enters the metal or the cavity. — **Figure 3.4.18.** An uncharged hollow metal sphere in an originally uniform field $E_{0}$ (pointing to the right), computed field lines. The free electrons move until the field in the metal vanishes: negative charge gathers on the upstream side, positive charge on the downstream side (the size of the signs indicates $∣ ρ_{S} ∣ \propto ∣ cos θ ∣$ , $θ$ measured from the direction of $E_{0}$ ). Every field line outside meets the surface at right angles ( $E_{τ} = 0$ ) and ends on the surface charge ( $ε_{0} E_{n} = ρ_{S}$ ), and none enters the metal or the cavity.

Figure 3.4.18 shows these properties at work for an uncharged hollow metal sphere placed in a uniform field (the curved version of Figure 3.4.16). Solving Laplace's equation in spherical coordinates (a technique beyond this section) gives the induced surface charge density $ρ_{S} = 3 ε_{0} E_{0} cos θ$ , so that the field just outside the surface, $E_{n} = ρ_{S} / ε_{0}$ , reaches three times the applied field at the two poles. All the properties of conductors stem from one cause:

Three facts, one cause

$E = 0$ inside every conductor in equilibrium implies, through Gauss's law and the two boundary conditions:

Charge on the surface only: any Gaussian surface inside the metal encloses zero net charge.
The conductor is an equipotential: $V_{B} - V_{A} = 0$ for any two of its points, and $E$ just outside is normal to the surface.
Field just outside: $D_{n} = ε_{2} E_{2 n} = ρ_{S}$ .

A fourth consequence is electrostatic shielding. Suppose the conductor encloses a cavity containing no charge. Could there be a field in the cavity? A field line in the cavity would have to start on the wall and end on the wall (there is no charge in the cavity for it to start or end on). Integrating $E \cdot d l$ along this line would give a non-zero result, because $E$ and $d l$ are parallel all the way. But the integral equals the potential difference between its two end points, which both lie on the same equipotential wall, and is therefore zero. The contradiction shows that no field line can exist in the cavity: $E = 0$ and $V$ is constant inside, whatever the field outside. This is the Faraday cage of Section 3.2.

Inside: $E = 0$ , $V =$ const. Surface: $E_{τ} = 0$ , $ρ_{S} = ε E_{n}$

Quick check

A closed, uncharged metal box has an empty cavity. (a) A charge is brought near the outside of the box. Is there a field in the cavity? (b) Instead, a charge is placed inside the cavity. Is there a field in the cavity now? In the metal?

(a) No: the cavity is charge-free and bounded by an equipotential, so $E = 0$ there. (b) Yes, in the cavity: the shielding argument assumed an empty cavity, and the field lines of the charge now end on induced charge on the cavity wall. In the metal itself $E = 0$ , as always in equilibrium.

Common mistake · Zero inside does not mean zero just outside

$E = 0$ holds inside the metal. Just outside, the field is normal to the surface and $E_{n} = ρ_{S} / ε$ , which is in general not zero. Nor is $ρ_{S}$ uniform on an arbitrary conductor: it is largest where the surface is most sharply curved or nearest to other charges (Figure 3.4.18).

3.4.7 Conductors

The field inside a conductor is zero

All charge resides on the surface

A conductor is an equipotential body

What the field looks like