# 4.4 Gauss’s Law

In this section, we use Maxwell’s equations to confirm the expressions for the electric field implied by Coulomb’s law, and propose alternative techniques for evaluating electric fields induced by electric charge. To that end, we restate Eq. (4.1a):

which is referred to as the differential form of Gauss’s law. The adjective “differential” refers to the fact that the divergence operation involves spatial derivatives. As we see shortly, Eq. (4.26) can be converted to an integral form. When solving electromagnetic problems, we often go back and forth between equations in differential and integral form, depending on which of the two happens to be the more applicable or convenient to use. To convert Eq. (4.26) into integral form, we multiply both sides by d υ and evaluate their integrals over an arbitrary volume υ:

∫υ∇⋅D d υ=∫υ ρ v d υ=Q.(4.27)

Here, Q is the total charge enclosed in υ. The divergence theorem, given by Eq. (3.98), states that the volume integral of the divergence of any vector over a volume υ equals the total outward flux of that vector through the surface S enclosing υ. Thus, for the vector D,

∫υ∇⋅D d υ=∮S D⋅d s.(4.28)

Comparison of Eq. (4.27) with Eq. (4.28) leads to

# Figure 4-8 The integral form of Gauss’s law states that the outward flux of D through a surface is proportional to the enclosed charge Q.

The integral form of Gauss’s law can be applied to determine D due to a single isolated point charge q by enclosing the latter with a closed, spherical, Gaussian surface S of arbitrary radius R centered at q (Fig. 4-9). From symmetry considerations and assuming that q is positive, the direction of D must be radially outward along the unit vector R ˆ, and D R, which is the magnitude of D, must be the same at all points on S. Thus, at any point on S,

D=R ˆ D R,(4.30)

and d s=R ˆ d s. Applying Gauss’s law gives

∮S D⋅d s=∮S R ˆ D R⋅R ˆ d s=∮S D R d s=D R(4 π R 2)=q.(4.31)

# Figure 4-9 Electric field D due to point charge q.

Figure 4 9 Full Alternative Text

Solving for D R and then inserting the result in Eq. (4.30) gives the expression for the electric field E induced by an isolated point charge in a medium with permittivity ε:

E=D ε=R ˆ q 4 π ε R 2(V/m).(4.32)

This is identical with Eq. (4.13) obtained from Coulomb’s law; after all, Maxwell’s equations incorporate Coulomb’s law. For this simple case of an isolated point charge, it does not matter whether Coulomb’s law or Gauss’s law is used to obtain the expression for E. However, it does matter which approach we follow when we deal with multiple point charges or continuous charge distributions. Even though Coulomb’s law can be used to find E for any specified distribution of charge, Gauss’s law is easier to apply than Coulomb’s law, but its utility is limited to symmetrical charge distributions.

Because at every point on the surface the direction of d s is along its outward normal, only the normal component of D at the surface contributes to the integral in Eq. (4.29). To successfully apply Gauss’s law, the surface S should be chosen from symmetry considerations so that, across each subsurface of S, D is constant in magnitude and its direction is either normal or purely tangential to the subsurface. These aspects are illustrated in Example 4-6.

# Example 4-6: Electric Field of an Infinite Line Charge

Use Gauss’s law to obtain an expression for E due to an infinitely long line with uniform charge density ρ ℓ that resides along the z axis in free space.

Solution: Since the charge density along the line is uniform, infinite in extent, and residing along the z axis, symmetry considerations dictate that D is in the radial r ˆ direction and cannot depend on ϕ or z. Thus, D=r ˆ D r. Therefore, we construct a finite cylindrical Gaussian surface of radius r and height h, which is concentric around the line of charge (Fig. 4-10). The total charge contained within the cylinder is Q=ρ ℓ h. Since D is along r ˆ, the top and bottom surfaces of the cylinder do not contribute to the surface integral on the left-hand side of Eq. (4.29); that is, only the curved surface contributes to the integral. Hence,

∫z=0 h∫ϕ=0 2 π r ˆ D r⋅r ˆ r d ϕ d z=ρ ℓ h

or

2 π h D r r=ρ ℓ h,

which yields

# Figure 4-10 Gaussian surface around an infinitely long line of charge (Example 4-6).

Figure 4 10 Full Alternative Text

Note that Eq. (4.33) is applicable for any infinite line of charge, regardless of its location and direction, as long as r ˆ is properly defined as the radial distance vector from the line charge to the observation point (i.e., r ˆ is perpendicular to the line of charge).

# Example 4-7: Two Infinite Lines of Charge

Figure 4-11 depicts the presence of two infinite lines of charge in free space: one residing in the x–y plane parallel to the x ˆ axis and carrying charge density ρ ℓ 1=1 (nC/m), and a second one residing in the y–z plane parallel to the y axis and carrying a charge density ρ ℓ 2=−2 (nC/m). Determine the electric field at the origin.

# Figure 4-11 Two infinite lines of charge (Example 4-7).

Figure 4 11 Full Alternative Text

Solution: The electric field E is the sum of two electric field components:

E=E 1+E 2,

where E 1 and E 2 are the electric fields due to line charges 1 and 2, respectively. According to Eq. (4.33), the direction of the electric field r ˆ is perpendicular to the direction of the line charge and points away from the line of charge (if ρ ℓ is positive). Hence, for the first line of charge, ρ ℓ 1=1 (nC/m), r ˆ 1=−y ˆ, r 1=2, and

E 1=r ˆ 1 ρ ℓ 1 2 π ε 0 r 1=−y ˆ 10−9 2 π×1 36 π×10−9×2=−y ˆ 9 V/m.

Similarly, for the second line of charge, ρ ℓ 2=−2 (nC/m), r ˆ 2=−z ˆ, r 2=6, and

E 2=r ˆ 2 ρ ℓ 2 2 π ε 0 r 2=−z ˆ(−2)×10−9 2 π×1 36 π×10−9×6=z ˆ 6 V/m.

Hence,

E=E 1+E 2=(−y ˆ 9+z ˆ 6)V/m.

Concept Question 4-7: Explain Gauss’s law. Under what circumstances is it useful?

Concept Question 4-8: How should one choose a Gaussian surface?