24.6.2 Charges and Fields Inside a Conductor

FIGURE 24.31 shows a charged conductor with a hole inside. Can there be charge on this interior surface? To find out, we place a Gaussian surface around the hole, infinitesimally close but entirely within the conductor. The electric flux $Φ_{e}$ through this Gaussian surface is zero because the electric field is zero everywhere inside the conductor. Thus we must conclude that $Q_{in} = 0$ . There’s no net charge inside this Gaussian surface and thus no charge on the surface of the hole. Any excess charge resides on the exterior surface of the conductor, not on any interior surfaces.

The figure illustrates a positively charged conductor with a hole inside.

FIGURE 24.31 Full Alternative Text

Furthermore, because there’s no electric field inside the conductor and no charge inside the hole, the electric field inside the hole must also be zero. This conclusion has an important practical application. For example, suppose we need to exclude the electric field from the region in FIGURE 24.32a enclosed within dashed lines. We can do so by surrounding this region with the neutral conducting box of FIGURE 24.32b.

The figure consists of two parts labeled (“a”) and (b) that illustrate the electric field inside a parallel-plate capacitor.

FIGURE 24.32 Full Alternative Text

This region of space is now a hole inside a conductor, thus the interior electric field is zero. The use of a conducting box to exclude electric fields from a region of space is called screening. Solid metal walls are ideal, but in practice wire screen or wire mesh—sometimes called a Faraday cage—provides sufficient screening for all but the most sensitive applications. The price we pay is that the exterior field is now very complicated.

Finally, FIGURE 24.33 shows a charge q inside a hole within a neutral conductor. The electric field inside the conductor is still zero, hence the electric flux through the Gaussian surface is zero. But $Φ_{e} = 0$ requires $Q_{in} = 0$ . Consequently, the charge inside the hole attracts an equal charge of opposite sign, and charge $- q$ now lines the inner surface of the hole. There is a field inside the hole, but the charge on the inner surface keeps the field of the point charge from continuing into the conductor.

The figure illustrates a neutral conductor with a hole inside containing a positive point charge.

FIGURE 24.33 Full Alternative Text

The conductor as a whole is neutral, so moving $- q$ to the surface of the hole must leave $+ q$ behind somewhere else. Where is it? It can’t be in the interior of the conductor, as we’ve seen, and that leaves only the exterior surface. In essence, an internal charge polarizes the conductor just as an external charge would. Net charge $- q$ moves to the inner surface and net charge $+ q$ is left behind on the exterior surface.

NOTE When we say that the electric field is zero inside a conductor, we mean within the material that forms the conductor. Charges can create a field inside an enclosed cavity inside a conductor, but that field does not extend into the conducting medium.

TACTICS BOX 24.3 Finding the electric field of a conductor in electrostatic equilibrium

1 The electric field is zero at all points inside the conductor.

2 Any excess charge resides entirely on the exterior surface.

3 The external electric field at the surface of a charged conductor is perpendicular to the surface and of magnitude $η / ϵ_{0}$ , where $η$ is the surface charge density at that point.

4 The electric field is zero inside any hole within a conductor unless there is a charge in the hole.

Workbook Exercises 20–24

EXAMPLE 24.7 The electric field at the surface of a charged metal sphere

A 2.0-cm-diameter brass sphere has been given a charge of 2.0 nC. What is the electric field strength at the surface?

MODEL Brass is a conductor. The excess charge resides on the surface.

VISUALIZE The charge distribution has spherical symmetry. The electric field points radially outward from the surface.

SOLVE We can solve this problem in two ways. One uses the fact that a sphere, because of its complete symmetry, is the one shape for which any excess charge will spread out to a uniform surface charge density. Thus

η = \frac{q}{A_{sphere}} = \frac{q}{4 π R^{2}} = \frac{2.0 \times 10^{- 9} C}{4 π (0.010 {m)}^{2}} = 1.59 \times 10^{- 6} {C/m}^{2}

From Equation 24.20, we know the electric field at the surface has strength

E_{surface} = \frac{η}{ϵ_{0}} = \frac{1.59 \times 10^{- 6} {C/m}^{2}}{8.85 \times 10^{- 12} C^{2} / N m^{2}} = 1.8 \times 10^{5} N / C

Alternatively, we could have used the result, obtained earlier in the chapter, that the electric field strength outside a sphere of charge Q is $E_{outside} = Q_{in} / (4 π ϵ_{0} r^{2})$ . But $Q_{in} = q$ and, at the surface, $r = R$ . Thus

\begin{matrix} E_{surface} & = & \frac{1}{4 π ϵ_{0}} \frac{q}{R^{2}} = (9.0 \times 10^{9} N m^{2} / C^{2}) \frac{2.0 \times 10^{- 9} C}{(0.010 {m)}^{2}} \\ = & 1.8 \times 10^{5} N / C \end{matrix}

As we can see, both methods lead to the same result.

CHAPTER 24 CHALLENGE EXAMPLE The electric field of a slab of charge

An infinite slab of charge of thickness 2a is centered in the xy-plane. The volume charge density is $ρ = ρ_{0} (1 - | z | / a)$ . Find the electric field strengths inside and outside this slab of charge.

MODEL The charge density is not uniform. Starting at $ρ_{0}$ in the xy-plane, it decreases linearly with distance above and below the xy-plane until reaching zero at $z = \pm a$ , the edges of the slab.

VISUALIZE FIGURE 24.34 shows an edge view of the slab of charge and, as Gaussian surfaces, side views of two cylinders with cross-section area A. By symmetry, the electric field must point away from the xy-plane; the field cannot have an x- or y-component.

The figure illustrates a closed slab centered in the x y plane with side views of two Gaussian cylinders.

FIGURE 24.34 Full Alternative Text

SOLVE Gauss’s law is

Φ_{e} = \oint \vec{E} \cdot d \vec{A} = \frac{Q_{in}}{ϵ_{0}}

With symmetry, finding the net flux is straightforward. The electric field is perpendicular to the faces of the cylinders and pointing outward, so the total flux through the faces is $Φ_{faces} = 2 E A$ , where E may depend on distance z. The field is parallel to the walls of the cylinders, so $Φ_{wall} = 0$ . Thus the net flux is simply

Φ_{e} = 2 E A

Because the charge density is not uniform, we need to integrate to find $Q_{in}$ , the charge inside the cylinder. We can slice the cylinder into small slabs of infinitesimal thickness dz and volume $d V = A d z$ . Figure 24.34 shows one such little slab at distance z from the xy-plane. The charge in this little slab is

d q = ρ d V = ρ_{0} (1 - \frac{z}{a}) A d z

where we assumed that z is positive. Because the charge is symmetric about $z = 0$ , we can avoid difficulties with the absolute value sign in the charge density by integrating from 0 and multiplying by 2. For the Gaussian cylinder that ends inside the slab of charge, at distance z, the total charge inside is

\begin{matrix} Q_{in} & = & \int_{}^{} d q = 2 \int_{0}^{z} ρ_{0} (1 - \frac{z}{a}) A d z \\ = & 2 ρ_{0} A [z |_{0}^{z} - \frac{1}{2 a} z^{2} |_{0}^{z}] \\ = & 2 ρ_{0} A z (1 - \frac{z}{2 a}) \end{matrix}

Gauss’s law inside the slab is then

Φ_{e} = 2 E_{inside} A = \frac{Q_{in}}{ϵ_{0}} = \frac{2 ρ_{0} A z}{ϵ_{0}} (1 - \frac{z}{2 a})

The area A cancels, as it must because it was an arbitrary choice, leaving

E_{inside} = \frac{ρ_{0} z}{ϵ_{0}} (1 - \frac{z}{2 a})

The field strength is zero at $z = 0$ , then increases as z increases. This expression is valid only above the xy-plane, for $z > 0$ , but the field strength is symmetric on the other side.

For the Gaussian cylinder that extends outside the slab of charge, the integral for Q has to end at $z = a$ . Thus

Q_{in} = 2 ρ_{0} A a (1 - \frac{a}{2 a}) = ρ_{0} A a

independent of distance z. With this, Gauss’s law gives

E_{outside} = \frac{Q_{in}}{2 ϵ_{0} A} = \frac{ρ_{0} a}{2 ϵ_{0}}

This matches $E_{inside}$ at the surface, $z = a$ , so the field is continuous as it crosses the boundary.

REVIEW Outside a sphere of charge, the field is the same as that of a point charge at the center. Similarly, the field outside an infinite slab of charge should be the same as that of an infinite charged plane. We found, by integration, that the total charge in an area A of the slab is $Q = ρ_{0} A a$ . If we squished this charge into a plane, the surface charge density would be $η = Q / A = ρ_{0} a$ . Thus our expression for $E_{outside}$ could be written $η / 2 ϵ_{0}$ , which matches the field we found in Example 24.6 for a plane of charge.

24.6.2 Charges and Fields Inside a Conductor

FIGURE 24.31 A Gaussian surface surrounding a hole inside a conductor.

FIGURE 24.32 The electric field can be excluded from a region of space by surrounding it with a conducting box.

FIGURE 24.33 A charge in the hole causes a net charge on the interior and exterior surfaces.

TACTICS BOX 24.3 Finding the electric field of a conductor in electrostatic equilibrium

EXAMPLE 24.7 The electric field at the surface of a charged metal sphere

CHAPTER 24 CHALLENGE EXAMPLE The electric field of a slab of charge

FIGURE 24.34 Two cylindrical Gaussian surfaces for an infinite slab of charge.