6.3 Applying Gauss’s Law

Samuel J. Ling; William Moebs; Jeff Sanny

Chapter 6. Gauss’s Law

6.3 Applying Gauss’s Law

Learning Objectives

By the end of this section, you will be able to:

Explain what spherical, cylindrical, and planar symmetry are
Recognize whether or not a given system possesses one of these symmetries
Apply Gauss’s law to determine the electric field of a system with one of these symmetries

Gauss’s law is very helpful in determining expressions for the electric field, even though the law is not directly about the electric field; it is about the electric flux. It turns out that in situations that have certain symmetries (spherical, cylindrical, or planar) in the charge distribution, we can deduce the electric field based on knowledge of the electric flux. In these systems, we can find a Gaussian surface S over which the electric field has constant magnitude. Furthermore, if $\vec{E}$ is parallel to $\hat{n}$ everywhere on the surface, then $\vec{E} \cdot \hat{n} = E .$ (If $\vec{E}$ and $\hat{n}$ are antiparallel everywhere on the surface, then $\vec{E} \cdot \hat{n} = − E .$ ) Gauss’s law then simplifies to

Φ = \oint_{S} \vec{E} \cdot \hat{n} d A = E \oint_{S} d A = E A = \frac{q_{enc}}{ϵ_{0}},

where A is the area of the surface. Note that these symmetries lead to the transformation of the flux integral into a product of the magnitude of the electric field and an appropriate area. When you use this flux in the expression for Gauss’s law, you obtain an algebraic equation that you can solve for the magnitude of the electric field, which looks like

E \frac{q_{enc}}{ϵ_{0} area} .

The direction of the electric field at the field point P is obtained from the symmetry of the charge distribution and the type of charge in the distribution. Therefore, Gauss’s law can be used to determine $\vec{E} .$ Here is a summary of the steps we will follow:

Problem-Solving Strategy: Gauss’s Law

Identify the spatial symmetry of the charge distribution. This is an important first step that allows us to choose the appropriate Gaussian surface. As examples, an isolated point charge has spherical symmetry, and an infinite line of charge has cylindrical symmetry.
Choose a Gaussian surface with the same symmetry as the charge distribution and identify its consequences. With this choice, $\vec{E} \cdot \hat{n}$ is easily determined over the Gaussian surface.
Evaluate the integral $\oint_{S} \vec{E} \cdot \hat{n} d A$ over the Gaussian surface, that is, calculate the flux through the surface. The symmetry of the Gaussian surface allows us to factor $\vec{E} \cdot \hat{n}$ outside the integral.
Determine the amount of charge enclosed by the Gaussian surface. This is an evaluation of the right-hand side of the equation representing Gauss’s law. It is often necessary to perform an integration to obtain the net enclosed charge.
Evaluate the electric field of the charge distribution. The field may now be found using the results of steps 3 and 4.

Basically, there are only three types of symmetry that allow Gauss’s law to be used to deduce the electric field. They are

A charge distribution with spherical symmetry
A charge distribution with cylindrical symmetry
A charge distribution with planar symmetry

To exploit the symmetry, we perform the calculations in appropriate coordinate systems and use the right kind of Gaussian surface for that symmetry, applying the remaining four steps.

Charge Distribution with Spherical Symmetry

A charge distribution has spherical symmetry if the density of charge depends only on the distance from a point in space and not on the direction. In other words, if you rotate the system, it doesn’t look different. For instance, if a sphere of radius R is uniformly charged with charge density $ρ_{0}$ then the distribution has spherical symmetry (Figure 6.21(a)). On the other hand, if a sphere of radius R is charged so that the top half of the sphere has uniform charge density $ρ_{1}$ and the bottom half has a uniform charge density $ρ_{2} \neq ρ_{1},$ then the sphere does not have spherical symmetry because the charge density depends on the direction (Figure 6.21(b)). Thus, it is not the shape of the object but rather the shape of the charge distribution that determines whether or not a system has spherical symmetry.

Figure 6.21(c) shows a sphere with four different shells, each with its own uniform charge density. Although this is a situation where charge density in the full sphere is not uniform, the charge density function depends only on the distance from the center and not on the direction. Therefore, this charge distribution does have spherical symmetry.

Figure a shows a uniformly colored sphere labeled rho 0. The figure is labeled spherically symmetric. Figure b shows a sphere whose top and bottom halves are differently colored. The top hemisphere is labeled rho 1 and the bottom one is labeled rho 2. The figure is labeled not spherically symmetric. Figure c shows a sphere, sectioned to show many concentric spheres of different colors within it. The figure is labeled spherically symmetric. — **Figure 6.21** Illustrations of spherically symmetrical and nonsymmetrical systems. Different shadings indicate different charge densities. Charges on spherically shaped objects do not necessarily mean the charges are distributed with spherical symmetry. The spherical symmetry occurs only when the charge density does not depend on the direction. In (a), charges are distributed uniformly in a sphere. In (b), the upper half of the sphere has a different charge density from the lower half; therefore, (b) does not have spherical symmetry. In (c), the charges are in spherical shells of different charge densities, which means that charge density is only a function of the radial distance from the center; therefore, the system has spherical symmetry.

One good way to determine whether or not your problem has spherical symmetry is to look at the charge density function in spherical coordinates, $ρ (r, θ, φ)$ . If the charge density is only a function of r, that is $ρ = ρ (r)$ , then you have spherical symmetry. If the density depends on $θ$ or $φ$ , you could change it by rotation; hence, you would not have spherical symmetry.

Consequences of symmetry

In all spherically symmetrical cases, the electric field at any point must be radially directed, because the charge and, hence, the field must be invariant under rotation. Therefore, using spherical coordinates with their origins at the center of the spherical charge distribution, we can write down the expected form of the electric field at a point P located at a distance r from the center:

Spherical symmetry: {\vec{E}}_{P} = E_{P} (r) \hat{r},

where $\hat{r}$ is the unit vector pointed in the direction from the origin to the field point P. The radial component $E_{P}$ of the electric field can be positive or negative. When $E_{P} > 0$ , the electric field at P points away from the origin, and when $E_{P} < 0$ , the electric field at P points toward the origin.

Gaussian surface and flux calculations

We can now use this form of the electric field to obtain the flux of the electric field through the Gaussian surface. For spherical symmetry, the Gaussian surface is a closed spherical surface that has the same center as the center of the charge distribution. Thus, the direction of the area vector of an area element on the Gaussian surface at any point is parallel to the direction of the electric field at that point, since they are both radially directed outward (Figure 6.22).

Figure shows a circle labeled charges with center O and radius R. A larger concentric circle shown with a dotted line is labeled Gaussian surface. An arrow labeled r marks the radius of the outer circle. An arrow labeled r hat is shown along r. A small patch where r touches the Gaussian surface is highlighted and labeled P. From here, another arrow points outward in the same direction as r. This is labeled vector E subscript P. Another arrow originates from the tip of vector E subscript P and points outward in the same direction. It is labeled delta vector A. — **Figure 6.22** The electric field at any point of the spherical Gaussian surface for a spherically symmetrical charge distribution is parallel to the area element vector at that point, giving flux as the product of the magnitude of electric field and the value of the area. Note that the radius R of the charge distribution and the radius r of the Gaussian surface are different quantities.

The magnitude of the electric field $\vec{E}$ must be the same everywhere on a spherical Gaussian surface concentric with the distribution. For a spherical surface of radius r,

Φ = \oint_{S} {\vec{E}}_{P} \cdot \hat{n} d A = E_{P} \oint_{S} d A = E_{P} 4 π r^{2} .

Using Gauss’s law

According to Gauss’s law, the flux through a closed surface is equal to the total charge enclosed within the closed surface divided by the permittivity of vacuum $ϵ_{0}$ . Let $q_{enc}$ be the total charge enclosed inside the distance r from the origin, which is the space inside the Gaussian spherical surface of radius r. This gives the following relation for Gauss’s law:

4 π r^{2} E = \frac{q_{enc}}{ϵ_{0}} .

Hence, the electric field at point P that is a distance r from the center of a spherically symmetrical charge distribution has the following magnitude and direction:

Magnitude: E (r) = \frac{1}{4 π ϵ_{0}} \frac{q_{enc}}{r^{2}}

Direction: radial from O to P or from P to O.

The direction of the field at point P depends on whether the charge in the sphere is positive or negative. For a net positive charge enclosed within the Gaussian surface, the direction is from O to P, and for a net negative charge, the direction is from P to O. This is all we need for a point charge, and you will notice that the result above is identical to that for a point charge. However, Gauss’s law becomes truly useful in cases where the charge occupies a finite volume.

Computing enclosed charge

The more interesting case is when a spherical charge distribution occupies a volume, and asking what the electric field inside the charge distribution is thus becomes relevant. In this case, the charge enclosed depends on the distance r of the field point relative to the radius of the charge distribution R, such as that shown in Figure 6.23.

Figure a shows a dotted circle S with center O and radius r, and a larger concentric circle with radius R. A small arrow points outward from S. This is labeled vector E subscript in. S is labeled Gaussian surface for vector E subscript in. Figure b shows a dotted circle S with center O and radius r, and a smaller concentric circle with radius R. A small arrow points outward from S. This is labeled vector E subscript out. S is labeled Gaussian surface for E vector subscript out. — **Figure 6.23** A spherically symmetrical charge distribution and the Gaussian surface used for finding the field (a) inside and (b) outside the distribution.

If point P is located outside the charge distribution—that is, if $r \geq R$ —then the Gaussian surface containing P encloses all charges in the sphere. In this case, $q_{enc}$ equals the total charge in the sphere. On the other hand, if point P is within the spherical charge distribution, that is, if $r < R,$ then the Gaussian surface encloses a smaller sphere than the sphere of charge distribution. In this case, $q_{enc}$ is less than the total charge present in the sphere. Referring to Figure 6.23, we can write $q_{enc}$ as

q_{enc} = {\begin{cases} \begin{array}{c} q_{tot} (total charge) if r \geq R \\ q_{within r < R} (only charge within r < R) if r < R \end{array} \end{cases}

.

The field at a point outside the charge distribution is also called ${\vec{E}}_{out}$ , and the field at a point inside the charge distribution is called ${\vec{E}}_{in} .$ Focusing on the two types of field points, either inside or outside the charge distribution, we can now write the magnitude of the electric field as

P outside sphere E_{out} = \frac{1}{4 π ϵ_{0}} \frac{q_{tot}}{r^{2}}

P inside sphere E_{in} = \frac{1}{4 π ϵ_{0}} \frac{q_{within r < R}}{r^{2}} .

Note that the electric field outside a spherically symmetrical charge distribution is identical to that of a point charge at the center that has a charge equal to the total charge of the spherical charge distribution. This is remarkable since the charges are not located at the center only. We now work out specific examples of spherical charge distributions, starting with the case of a uniformly charged sphere.

Example

Uniformly Charged Sphere

A sphere of radius R, such as that shown in Figure 6.23, has a uniform volume charge density $ρ_{0}$ . Find the electric field at a point outside the sphere and at a point inside the sphere.

Strategy

Apply the Gauss’s law problem-solving strategy, where we have already worked out the flux calculation.

Solution

Show Answer

The charge enclosed by the Gaussian surface is given by

q_{enc} = {\int ρ}_{0} d V = \int_{0}^{r} ρ_{0} 4 π {r^{'}}^{2} d r^{'} = ρ_{0} (\frac{4}{3} π r^{3}) .

The answer for electric field amplitude can then be written down immediately for a point outside the sphere, labeled $E_{out},$ and a point inside the sphere, labeled $E_{in} .$

\begin{array}{ccc} E_{out} & = & \frac{1}{4 π ϵ_{0}} \frac{q_{tot}}{r^{2}}, q_{tot} = \frac{4}{3} π R^{3} ρ_{0}, \\ E_{in} & = & \frac{q_{enc}}{4 π ϵ_{0} r^{2}} = \frac{ρ_{0} r}{3 ϵ_{0}}, since q_{enc} = \frac{4}{3} π r^{3} ρ_{0} . \end{array}

It is interesting to note that the magnitude of the electric field increases inside the material as you go out, since the amount of charge enclosed by the Gaussian surface increases with the volume. Specifically, the charge enclosed grows $\propto r^{3}$ , whereas the field from each infinitesimal element of charge drops off $\propto 1 / r^{2}$ with the net result that the electric field within the distribution increases in strength linearly with the radius. The magnitude of the electric field outside the sphere decreases as you go away from the charges, because the included charge remains the same but the distance increases. Figure 6.24 displays the variation of the magnitude of the electric field with distance from the center of a uniformly charged sphere.

Figure shows a graph of E versus r. The curve rises in a straight line labeled E proportional to r, peaks and falls in a curved line labeled E proportional to 1 by r squared. The peak has an x value of R and a y value of E subscript R. — **Figure 6.24** Electric field of a uniformly charged, non-conducting sphere increases inside the sphere to a maximum at the surface and then decreases as $1 / r^{2}$ . Here, $E_{R} = \frac{ρ_{0} R}{3 ϵ_{0}}$ . The electric field is due to a spherical charge distribution of uniform charge density and total charge Q as a function of distance from the center of the distribution.

The direction of the electric field at any point P is radially outward from the origin if $ρ_{0}$ is positive, and inward (i.e., toward the center) if $ρ_{0}$ is negative. The electric field at some representative space points are displayed in Figure 6.25 whose radial coordinates r are $r = R / 2$ , $r = R$ , and $r = 2 R$ .

Figure shows three concentric circles. The smallest one is dotted and labeled r equal to R by 2. The middle one is labeled r equal to R and the largest one, also dotted, is labeled r equal to 2R. Arrows labeled vector E originate from each circle and point outward, perpendicular to the circle. The ones on the outer circle are smallest and the ones on the middle circle are the longest. — **Figure 6.25** Electric field vectors inside and outside a uniformly charged sphere.

Significance

Notice that $E_{out}$ has the same form as the equation of the electric field of an isolated point charge. In determining the electric field of a uniform spherical charge distribution, we can therefore assume that all of the charge inside the appropriate spherical Gaussian surface is located at the center of the distribution.

Example

Non-Uniformly Charged Sphere

A non-conducting sphere of radius R has a non-uniform charge density that varies with the distance from its center as given by

ρ (r) = a r^{n} (r \leq R; n \geq 0),

where a is a constant. We require $n \geq 0$ so that the charge density is not undefined at $r = 0$ . Find the electric field at a point outside the sphere and at a point inside the sphere.

Strategy

Apply the Gauss’s law strategy given above, where we work out the enclosed charge integrals separately for cases inside and outside the sphere.

Solution

Show Answer

Since the given charge density function has only a radial dependence and no dependence on direction, we have a spherically symmetrical situation. Therefore, the magnitude of the electric field at any point is given above and the direction is radial. We just need to find the enclosed charge $q_{enc},$ which depends on the location of the field point.

A note about symbols: We use $r^{'}$ for locating charges in the charge distribution and r for locating the field point(s) at the Gaussian surface(s). The letter R is used for the radius of the charge distribution.

As charge density is not constant here, we need to integrate the charge density function over the volume enclosed by the Gaussian surface. Therefore, we set up the problem for charges in one spherical shell, say between $r^{'}$ and $r^{'} + d r^{'},$ as shown in Figure 6.26. The volume of charges in the shell of infinitesimal width is equal to the product of the area of surface $4 π {r^{'}}^{2}$ and the thickness $d r^{'}$ . Multiplying the volume with the density at this location, which is $a {r^{'}}^{n}$ , gives the charge in the shell:

d q = a {r^{'}}^{n} 4 π {r^{'}}^{2} d r^{'} .

Figure shows four concentric circles. Starting from the smallest, their radii are labeled: r prime, r prime plus d r prime, R and r. The outermost circle is dotted and labeled Gaussian surface. — **Figure 6.26** Spherical symmetry with non-uniform charge distribution. In this type of problem, we need four radii: R is the radius of the charge distribution, r is the radius of the Gaussian surface, $r^{'}$ is the inner radius of the spherical shell, and $r^{'} + d r^{'}$ is the outer radius of the spherical shell. The spherical shell is used to calculate the charge enclosed within the Gaussian surface. The range for $r^{'}$ is from 0 to r for the field at a point inside the charge distribution and from 0 to R for the field at a point outside the charge distribution. If $r > R$ , then the Gaussian surface encloses more volume than the charge distribution, but the additional volume does not contribute to $q_{enc}$ .

(a) Field at a point outside the charge distribution. In this case, the Gaussian surface, which contains the field point P, has a radius r that is greater than the radius R of the charge distribution, $r > R$ . Therefore, all charges of the charge distribution are enclosed within the Gaussian surface. Note that the space between $r^{'} = R$ and $r^{'} = r$ is empty of charges and therefore does not contribute to the integral over the volume enclosed by the Gaussian surface:

q_{enc} = \int d q = \int_{0}^{R} a {r^{'}}^{n} 4 π {r^{'}}^{2} d r^{'} = \frac{4 π a}{n + 3} R^{n + 3} .

This is used in the general result for ${\vec{E}}_{out}$ above to obtain the electric field at a point outside the charge distribution as

{\vec{E}}_{out} = [\frac{a R^{n + 3}}{ϵ_{0} (n + 3)}] \frac{1}{r^{2}} \hat{r},

where $\hat{r}$ is a unit vector in the direction from the origin to the field point at the Gaussian surface.

(b) Field at a point inside the charge distribution. The Gaussian surface is now buried inside the charge distribution, with $r < R$ . Therefore, only those charges in the distribution that are within a distance r of the center of the spherical charge distribution count in $r_{enc}$ :

q_{enc} = \int_{0}^{r} a {r^{'}}^{n} 4 π {r^{'}}^{2} d r^{'} = \frac{4 π a}{n + 3} r^{n + 3} .

Now, using the general result above for ${\vec{E}}_{in},$ we find the electric field at a point that is a distance r from the center and lies within the charge distribution as

{\vec{E}}_{in} = [\frac{a}{ϵ_{0} (n + 3)}] r^{n + 1} \hat{r},

where the direction information is included by using the unit radial vector.

Check Your Understanding

Check that the electric fields for the sphere reduce to the correct values for a point charge.

Show Solution

In this case, there is only ${\vec{E}}_{out} . So, yes .$

Charge Distribution with Cylindrical Symmetry

A charge distribution has cylindrical symmetry if the charge density depends only upon the distance r from the axis of a cylinder and must not vary along the axis or with direction about the axis. In other words, if your system varies if you rotate it around the axis, or shift it along the axis, you do not have cylindrical symmetry.

Figure 6.27 shows four situations in which charges are distributed in a cylinder. A uniform charge density $ρ_{0} .$ in an infinite straight wire has a cylindrical symmetry, and so does an infinitely long cylinder with constant charge density $ρ_{0} .$ An infinitely long cylinder that has different charge densities along its length, such as a charge density $ρ_{1}$ for $z > 0$ and $ρ_{2} \neq ρ_{1}$ for $z < 0$ , does not have a usable cylindrical symmetry for this course. Neither does a cylinder in which charge density varies with the direction, such as a charge density $ρ_{1}$ for $0 \leq θ < π$ and $ρ_{2} \neq ρ_{1}$ for $π \leq θ < 2 π$ . A system with concentric cylindrical shells, each with uniform charge densities, albeit different in different shells, as in Figure 6.27(d), does have cylindrical symmetry if they are infinitely long. The infinite length requirement is due to the charge density changing along the axis of a finite cylinder. In real systems, we don’t have infinite cylinders; however, if the cylindrical object is considerably longer than the radius from it that we are interested in, then the approximation of an infinite cylinder becomes useful.

Figures a through d show a cylinder. In figure a, labeled cylindrically symmetrical, the cylinder is uniformly colored and labeled rho zero. In figure b, labeled not cylindrically symmetrical, the top and bottom halves of the cylinder are different in color. The top is labeled rho 1 and the bottom is labeled rho 2. In figure c, labeled not cylindrically symmetrical, the left and right halves of the cylinder are different in color. The left is labeled rho 1 and the right is labeled rho 2. In figure d, many concentric sections are seen within the cylinder. The figure is labeled cylindrically symmetrical. — **Figure 6.27** To determine whether a given charge distribution has cylindrical symmetry, look at the cross-section of an “infinitely long” cylinder. If the charge density does not depend on the polar angle of the cross-section or along the axis, then you have cylindrical symmetry. (a) Charge density is constant in the cylinder; (b) upper half of the cylinder has a different charge density from the lower half; (c) left half of the cylinder has a different charge density from the right half; (d) charges are constant in different cylindrical rings, but the density does not depend on the polar angle. Cases (a) and (d) have cylindrical symmetry, whereas (b) and (c) do not.