Theory

Conductors, capacitors and dielectrics

Why the interior field of a conductor in equilibrium is zero and the charge lies on the surface, how to compute the capacitance of parallel-plate, spherical and cylindrical capacitors and of their combinations, how much energy the electric field stores and how a dielectric changes capacitance, field and energy.

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Conductors in electrostatic equilibrium
Field at the surface and the effect of sharp points

Conductors in electrostatic equilibrium

Throughout this topic the charges are at rest, the conductors are homogeneous and the space around them is vacuum, except where dielectrics are introduced. Work is denoted by $W$ .

1. Conductor in electrostatic equilibrium

A conductor is a material that contains charges free to move inside it, such as the conduction electrons of a metal. A conductor is in electrostatic equilibrium when no charge inside it or on its surface is in macroscopic motion. When a conductor is charged or brought into an external field, the free charges move and equilibrium is reached in a very short time, which is neglected in electrostatics problems. All the properties that follow hold for conductors in equilibrium, and they serve to determine where the charge lies and what the field is without knowing the motion of the charges during the redistribution.

2. Electric field inside a conductor

At every point inside the material of a conductor in electrostatic equilibrium the electric field is zero:

E = 0 .

The interior field is zero even when the conductor is in an external field $E_{0}$ . In this case the free charges redistribute on the surface, with negative charge on the side the lines of $E_{0}$ come from and positive charge on the opposite side, and the field produced by these charges, called induced charges, cancels $E_{0}$ inside the material. The phenomenon is called electrostatic induction, and the total charge of an isolated neutral conductor remains zero.

Proof

Suppose that at an interior point $P$ the field is $E \neq = 0$ . A free charge $q$ at $P$ experiences the force $q E \neq = 0$ , therefore it starts moving. This contradicts the hypothesis that the conductor is in equilibrium, hence $E = 0$ at every interior point.

3. Charge on the surface of a conductor

In a conductor in electrostatic equilibrium the volume charge density is zero at every interior point, and any net charge of the conductor is distributed over its surface with surface density $σ$ .

The result says that to describe a charged conductor it is enough to know $σ$ on the surface. The distribution of $σ$ depends on the shape of the conductor and on nearby charges; on an isolated sphere, by symmetry, it is uniform, and a sphere of radius $R$ with charge $Q$ has $σ = \frac{Q}{4 π R ^{2}}$ .

Charged conducting sphere in equilibrium: the charge lies on the surface, the interior field is zero and the exterior field is perpendicular to the surface

Proof

Let $P$ be a point inside the material and let $S$ be a closed surface that encloses $P$ and lies entirely inside the material. By the theorem on the field inside a conductor $E = 0$ at every point of $S$ , therefore the flux through $S$ is zero. By Gauss’s law the charge enclosed by $S$ is then zero. Since $S$ can be chosen arbitrarily small around $P$ , the volume charge density at $P$ is zero. Hence the net charge, which is conserved, can only be on the surface.

4. Equipotential conductor

All points of a conductor in electrostatic equilibrium, interior and on the surface, are at the same potential. The surface of the conductor is therefore an equipotential surface, and the field at points just outside it is perpendicular to the surface.

A single value of the potential is therefore assigned to the conductor, the potential of the conductor, and two conductors connected by a conducting wire form a single conductor, so they reach the same potential. This property makes it possible to compute how the charge is shared among connected conductors.

Proof

Let $A$ and $B$ be two points of the conductor, joined by a path that runs inside the material. Along the path $E = 0$ , therefore $V_{A} - V_{B} = \int_{A}^{B} E \cdot d ℓ = 0$ . If the field just outside the surface had a component tangent to the surface, the free charges of the surface would move along it, contrary to the hypothesis of equilibrium; hence the exterior field is perpendicular to the surface, consistent with the fact that the field is perpendicular to equipotential surfaces.

Cavity in a conductor and electrostatic shielding

Consider a conductor with a cavity inside it.

If the cavity does not contain charges, the field in the cavity is zero and there is no charge on the wall of the cavity. Indeed, a field line in the cavity would go from a positive to a negative charge of the wall, and the integral of the field along that line would be positive, whereas the wall is equipotential. A closed conducting enclosure therefore isolates its interior from external fields, and it is called an electrostatic shield or Faraday cage.
If the cavity contains a charge $q$ , the charge $- q$ is induced on the wall of the cavity. Indeed, a closed surface contained in the material and surrounding the cavity has zero flux, so it encloses zero total charge. If the conductor has net charge $Q$ , the charge $Q + q$ is on the outer surface.

For example, a neutral conducting spherical shell containing a charge $q = 5.0 nC$ carries $- 5.0 nC$ on the inner surface and $+ 5.0 nC$ on the outer one, and outside the field is that of a charge $q$ placed at the center of the shell, whatever the position of $q$ in the cavity.

Field at the surface and the effect of sharp points

1. Coulomb's theorem

Let $σ$ be the surface charge density at a point of the surface of a conductor in equilibrium, and let $\overset{n}{^}$ be the unit normal to the surface pointing outward. At points just outside, the electric field is

E = \frac{σ}{ε _{0}} \overset{n}{^} .

The theorem gives the field near a conductor from the local charge density, and vice versa; the field points outward where $σ > 0$ and inward where $σ < 0$ . For example, where $σ = 1.0 μ C/ m^{2}$ the field is $E = \frac{1.0 \cdot 1 0 ^{- 6}}{8.85 \cdot 1 0 ^{- 12}} V/m \approx 1.1 \cdot 1 0^{5} V/m$ .

The small cylinder with bases of area

Δ A

straddling the surface encloses the charge

σ Δ A

; flux leaves only through the outer base

Proof

Consider a small cylinder with bases of area $Δ A$ parallel to the surface, one just outside and one just inside the conductor, and negligible height. The base area is so small that $σ$ and the exterior field can be considered constant on it. On the inner base the field is zero; on the lateral surface, of negligible area, the flux is negligible; on the outer base the field is perpendicular to the surface, by the theorem on the equipotential conductor, therefore the flux is $E_{n} Δ A$ , where $E_{n}$ is the component of $E$ along $\overset{n}{^}$ . The cylinder encloses the charge $σ Δ A$ , hence Gauss’s law gives

E_{n} Δ A = \frac{σ Δ A}{ε _{0}} ⟹ E_{n} = \frac{σ}{ε _{0}} .

Comparison with the charged plane

An isolated infinite plane with density $σ$ produces on each side the field $\frac{σ}{2 ε _{0}}$ , as shown in the topic on Gauss’s law. Near a conductor the field is twice as large, $\frac{σ}{ε _{0}}$ , because the field of the local charge adds to that of all the other charges of the conductor; these charges arrange themselves so that inside the conductor the two contributions cancel and outside they add. The factor $2$ reappears in the force between the plates of a capacitor.

2. Effect of sharp points

Consider two conducting spheres of radii $R_{1}$ and $R_{2}$ , at a distance much larger than the radii and connected by a thin conducting wire, on which the charge is negligible. At equilibrium the surface charge densities are inversely proportional to the radii:

\frac{σ _{1}}{σ _{2}} = \frac{R _{2}}{R _{1}} .

By Coulomb’s theorem the fields at the surfaces are in the same ratio. The result extends qualitatively to a conductor of any shape, on which the charge density and the field are more intense where the radius of curvature of the surface is smaller, that is, near points and edges. If the field near a point exceeds the dielectric strength of air, the air becomes conducting and the point discharges; the lightning rod works on this principle. With $R_{1} = 1 cm$ and $R_{2} = 10 cm$ we have $σ_{1} = 10 σ_{2}$ .

On a pointed conductor the charge density and the field are more intense where the radius of curvature is smaller, that is, near the point

Proof

The two spheres and the wire form a single conductor, therefore by the theorem on the equipotential conductor they are at the same potential. Since the spheres are far apart, the potential of each is that of an isolated sphere with its own charge, hence

\frac{k Q _{1}}{R _{1}} = \frac{k Q _{2}}{R _{2}} ⟹ \frac{Q _{1}}{Q _{2}} = \frac{R _{1}}{R _{2}} .

The densities are $σ_{i} = \frac{Q _{i}}{4 π R _{i}^{2}}$ , therefore $\frac{σ _{1}}{σ _{2}} = \frac{Q _{1}}{Q _{2}} \cdot \frac{R _{2}^{2}}{R _{1}^{2}} = \frac{R _{2}}{R _{1}}$ .

The theory continues with 6 more sections (capacitance of a conductor and capacitors, parallel plate spherical and cylindrical capacitors, capacitors in series and in parallel…), plus all worked examples, the solutions and the PDF.

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