Applications of Gauss's Law

One particularly interesting application of Gauss's law is Earnshaw's theorem, which states that it is impossible for a collection of electrically charged particles to remain in static equilibrium solely under the influence of (classical) electrostatic forces. For instance, consider the motion of the $i$th particle in the electric field, ${\bf E}({\bf r})$, generated by all of the other static particles. The equilibrium position of the $i$th particle corresponds to some point of displacement ${\bf r}_i$ at which ${\bf E}({\bf r}_i)={\bf0}$, because this implies that the particle is not subject to an electrical force. By implication, ${\bf r}_i$ does not correspond to the equilibrium displacement of any other particle in the system. However, in order for ${\bf r}_i$ to be the displacement of a stable equilibrium point, the $i$th particle must experience a restoring force when its displacement deviates slightly from ${\bf r}_i$ in any direction. Assuming that the $i$th particle is (say) positively charged, this implies that the electric field must be directed radially toward the point whose displacement is ${\bf r}_i$ at all neighboring points. Hence, if we consider a small sphere centered on displacement ${\bf r}_i$ then there must be a negative flux of ${\bf E}$ through the surface of this sphere. According to Gauss's law, this necessitates the presence of a negative charge at displacement ${\bf r}_i$. However, there is no such charge at displacement ${\bf r}_i$. Hence, we conclude that ${\bf E}$ cannot be directed radially toward the point whose displacement is ${\bf r}_i$ at all neighboring points. In other words, there must be some neighboring points at which ${\bf E}$ is directed away from the point whose displacement is ${\bf r}_i$. Hence, a positively charged particle placed at displacement ${\bf r}_i$ can always escape by moving to such neighboring points. One corollary of Earnshaw's theorem is that classical electrostatics cannot account for the stability of atoms and molecules.

As an example of the use of Gauss's law, let us calculate the electric field generated by a spherically symmetric charge annulus of inner radius $a$, and outer radius $b$, centered on the origin, and carrying a uniformly distributed electric charge $Q$. Now, by symmetry, we expect a spherically symmetric charge distribution to generate a spherically symmetric potential, $\phi(r)$, where $r$ is a spherical polar coordinate. (See Section A.23.) It therefore follows from Equation (2.17) that the electric field is both spherically symmetric and radial; that is, ${\bf E} = E_r(r)\,{\bf e}_r$. Let us apply Gauss's law to an imaginary spherical surface, of radius $r$, centered on the origin. See Figure 2.4. Such a surface is generally known as a Gaussian surface. According to Gauss's law, (2.58), the flux of the electric field out of the surface is equal to the enclosed charge, divided by $\epsilon_0$. The flux is easy to calculate because the electric field is everywhere perpendicular to the surface. We obtain

$$4\pi\,r^2\,E_r(r) = \frac{Q(r)}{\epsilon_0},$$ (2.59)

where $Q(r)$ is the charge enclosed by a Gaussian surface of radius $r$. However, simple arguments involving proportion reveal that

$$Q(r) = \left\{ \begin{array}{lcl} 0&\mbox{\hspace{1cm}}&r<a\\... ...3)\right] Q&&a\leq r\leq b\\ [0.5ex] Q&&b<r \end{array}\right..$$ (2.60)

Hence,

$$E_r(r) = \left\{ \begin{array}{lcl} 0&\mbox{\hspace{1cm}}&r<a... ... b\\ [0.5ex] Q/(4\pi\,\epsilon_0\,r^2)&&b<r \end{array}\right..$$ (2.61)

The previous electric field distribution illustrates two important points. First, the electric field generated outside a spherically symmetric charge distribution is the same as that which would be generated if all of the charge in the distribution was concentrated at its center. Second, zero electric field is generated inside an empty cavity surrounded by a spherically symmetric charge distribution.

Figure 2.4: An example use of Gauss's law.

We can easily determine the electric potential associated with the electric field (2.61) using

$$\frac{d\phi(r)}{d r} = -E_r(r).$$ (2.62)

[See Equation (2.17).] The boundary conditions are that $\phi(\infty)=0$, and that $\phi(r)$ is continuous at $r=a$ and $r=b$. (Of course, a discontinuous potential would lead to an infinite electric field, which is unphysical.) It follows that

$$\phi(r) = \left\{ \begin{array}{lcl} \left[Q/(4\pi\,\epsilon_... ...eq b\\ [0.5ex] Q/(4\pi\,\epsilon_0\,r)&&b<r \end{array}\right..$$ (2.63)

Hence, the work done in slowly moving a charge from infinity to the center of the distribution (which is minus the work done by the electric field) is

$$W = q\left[\phi(0)-\phi(\infty)\right] = \frac{q\,Q}{4\pi\,\epsilon_0}\,\frac{3}{2}\left(\frac{b^2-a^2}{b^3-a^3}\right).$$ (2.64)

[See Equation (2.23).]