24.4 Gauss’s Law

The last section was long, but knowing how to calculate the electric flux through a closed surface is essential for the main topic of this chapter: Gauss’s law. Gauss’s law is equivalent to Coulomb’s law for static charges, although Gauss’s law will look very different.

The purpose of learning Gauss’s law is twofold:

Gauss’s law allows the electric fields of some continuous distributions of charge to be found much more easily than does Coulomb’s law.
Gauss’s law is valid for moving charges, but Coulomb’s law is not (although it’s a very good approximation for velocities that are much less than the speed of light). Thus Gauss’s law is ultimately a more fundamental statement about electric fields than is Coulomb’s law.

Let’s start with Coulomb’s law for the electric field of a point charge. FIGURE 24.18 shows a spherical Gaussian surface of radius r centered on a positive charge q. Keep in mind that this is an imaginary, mathematical surface, not a physical surface. There is a net flux through this surface because the electric field points outward at every point on the surface. To evaluate the flux, given formally by the surface integral of Equation 24.11, notice that the electric field is perpendicular to the surface at every point on the surface and, from Coulomb’s law, it has the same magnitude $E = q / 4 π ϵ_{0} r^{2}$ at every point on the surface. This simple situation arises because the Gaussian surface has the same symmetry as the electric field.

The figure illustrates the cross-sectional view of a spherical Gaussian surface with radius r centered on a positive point charge q.

FIGURE 24.18 Full Alternative Text

Thus we know, without having to do any hard work, that the flux integral is

Φ_{e} = \oint \vec{E} \cdot d \vec{A} = E A_{sphere}

(24.12)

The surface area of a sphere of radius r is $A_{sphere} = 4 π r^{2}$ . If we use $A_{sphere}$ and the Coulomb-law expression for E in Equation 24.12, we find that the electric flux through the spherical surface is

Φ_{e} = \frac{q}{4 π ϵ_{0} r^{2}} 4 π r^{2} = \frac{q}{ϵ_{0}}

(24.13)

You should examine the logic of this calculation closely. We really did evaluate the surface integral of Equation 24.11, although it may appear, at first, as if we didn’t do much. The integral was easily evaluated, we reiterate for emphasis, because the closed surface on which we performed the integration matched the symmetry of the charge distribution.

NOTE We found Equation 24.13 for a positive charge, but it applies equally to negative charges. According to Equation 24.13, $Φ_{e}$ is negative if q is negative. And that’s what we would expect from the basic definition of flux, $\vec{E} \cdot \vec{A}$ . The electric field of a negative charge points inward, while the area vector of a closed surface points outward, making the dot product negative.