MODEL The charge distribution within the sphere need not be uniform (i.e., the charge density might increase or decrease with r), but it must have spherical symmetry in order for us to use Gauss’s law. We will assume that it does.

VISUALIZE FIGURE 24.23 shows a sphere of charge Q and radius R. We want to find $\vec{E}$ outside this sphere, for distances $r>R$. The spherical symmetry of the charge distribution tells us that the electric field must point radially outward from the sphere. Although Gauss’s law is true for any surface surrounding the charged sphere, it is useful only if we choose a Gaussian surface to match the spherical symmetry of the charge distribution and the field. Thus a spherical surface of radius $r>R$ concentric with the charged sphere will be our Gaussian surface. Because this surface surrounds the entire sphere of charge, the enclosed charge is simply $Q_{\text{in}}=Q$.

FIGURE 24.23 A spherical Gaussian surface surrounding a sphere of charge.

SOLVE Gauss’s law is

To calculate the flux, notice that the electric field is everywhere perpendicular to the spherical surface. And although we don’t know the electric field magnitude E, spherical symmetry dictates that E must have the same value at all points equally distant from the center of the sphere. Thus we have the simple result that the net flux through the Gaussian surface is

where we used the fact that the surface area of a sphere is $A_{\text{sphere}}=4\pi r^2$. With this result for the flux, Gauss’s law is

Thus the electric field at distance r outside a sphere of charge is

Or in vector form, making use of the fact that $\vec{E}$ is radially outward,

where $\widehat{r}$ is a radial unit vector.

REVIEW The field is exactly that of a point charge Q, which is what we wanted to show.

MODEL We haven’t considered a situation like this before. To begin, we don’t know if the field strength is increasing or decreasing as we move outward from the center of the sphere. But the field inside must have spherical symmetry. That is, the field must point radially inward or outward, and the field strength can depend only on r. This is sufficient information to solve the problem because it allows us to choose a Gaussian surface.

VISUALIZE FIGURE 24.24 shows a spherical Gaussian surface with radius $r\le R$ inside, and concentric with, the sphere of charge. This surface matches the symmetry of the charge distribution, hence $\vec{E}$ is perpendicular to this surface and the field strength E has the same value at all points on the surface.

FIGURE 24.24 A spherical Gaussian surface inside a uniform sphere of charge.

SOLVE The flux integral is identical to that of Example 24.3:

Consequently, Gauss’s law is

The difference between this example and Example 24.3 is that $Q_{\text{in}}$ is no longer the total charge of the sphere. Instead, $Q_{\text{in}}$ is the amount of charge inside the Gaussian sphere of radius r. Because the charge distribution is uniform, the volume charge density is

The charge enclosed in a sphere of radius r is thus

The amount of enclosed charge increases with the cube of the distance r from the center and, as expected, $Q_{\text{in}}=Q$ if $r=R$. With this expression for $Q_{\text{in}}$, Gauss’s law is

Thus the electric field at radius r inside a uniformly charged sphere is

The electric field strength inside the sphere increases linearly with the distance r from the center.

REVIEW The field inside and the field outside a sphere of charge match at the boundary of the sphere, $r=R$, where both give $E=Q\text{/}4\pi {ϵ}_0{R}^2$. In other words, the field strength is continuous as we cross the boundary of the sphere. These results are shown graphically in FIGURE 24.25.

FIGURE 24.25 The electric field strength of a uniform sphere of charge of radius $R$.

MODEL A long, charged wire can be modeled as an infinitely long line of charge.

VISUALIZE FIGURE 24.26 shows an infinitely long line of charge. We can use the symmetry of the situation to see that the only possible shape of the electric field is to point straight into or out from the wire, rather like the bristles on a bottle brush. The shape of the field suggests that we choose our Gaussian surface to be a cylinder of radius r and length L, centered on the wire. Because Gauss’s law refers to closed surfaces, we must include the ends of the cylinder as part of the surface.

FIGURE 24.26 A Gaussian surface around a charged wire.

SOLVE Gauss’s law is

where $Q_{\text{in}}$ is the charge inside the closed cylinder. We have two tasks: to evaluate the flux integral, and to determine how much charge is inside the closed surface. The wire has linear charge density $\text{λ}$, so the amount of charge inside a cylinder of length L is simply

Finding the net flux is just as straightforward. We can divide the flux through the entire closed surface into the flux through each end plus the flux through the cylindrical wall. The electric field $\vec{E}$, pointing straight out from the wire, is tangent to the end surfaces at every point. Thus the flux through these two surfaces is zero. On the wall, $\vec{E}$ is perpendicular to the surface and has the same strength E at every point. Thus

where we used $A_{\text{cyl}}=2\pi rL$ as the surface area of a cylindrical wall of radius r and length L. Once again, the proper choice of the Gaussian surface reduces the flux integral merely to finding a surface area. With these expressions for $Q_{\text{in}}$ and ${\text{Φ}}_{\text{e}}$, Gauss’s law is

Thus the electric field at distance r from a long, charged wire is

REVIEW This agrees exactly with the result of the more complex derivation in Chapter 23. Notice that the result does not depend on our choice of L. A Gaussian surface is an imaginary device, not a physical object. We needed a finite-length cylinder to do the flux calculation, but the electric field of an infinitely long wire can’t depend on the length of an imaginary cylinder.

MODEL A uniformly charged flat electrode can be modeled as an infinite plane of charge.

VISUALIZE FIGURE 24.27 shows a uniformly charged plane with surface charge density $\eta$. We will assume that the plane extends infinitely far in all directions, although we obviously have to show “edges” in our drawing. The planar symmetry allows the electric field to point only straight toward or away from the plane. With this in mind, choose as a Gaussian surface a cylinder with length L and faces of area A centered on the plane of charge. Although we’ve drawn them as circular, the shape of the faces is not relevant.

FIGURE 24.27 The Gaussian surface extends to both sides of a plane of charge.

SOLVE The electric field is perpendicular to both faces of the cylinder, so the total flux through both faces is ${\text{Φ}}_{\text{faces}}=2EA$. (The fluxes add rather than cancel because the area vector $\vec{A}$ points outward on each face.) There’s no flux through the wall of the cylinder because the field vectors are tangent to the wall. Thus the net flux is simply

The charge inside the cylinder is the charge contained in area A of the plane. This is

With these expressions for $Q_{\text{in}}$ and ${\text{Φ}}_{\text{e}}$, Gauss’s law is

Thus the electric field of an infinite charged plane is

This agrees with the result in Chapter 23.

REVIEW This is another example of a Gaussian surface enclosing only some of the charge. Most of the plane’s charge is outside the Gaussian surface and does not contribute to the flux, but it does affect the shape of the field. We wouldn’t have planar symmetry, with the electric field exactly perpendicular to the plane, without all the rest of the charge on the plane.