Gauss's Law

Consider a single electric charge $q$ located at the origin. The electric field generated by such a charge is given by Equation (2.12). Suppose that we surround the charge by a concentric spherical surface $S$ of radius $r$. See Figure 2.2. The flux of the electric field through this surface is given by

$$\oint_S{\bf E}\cdot d{\bf S}= \oint_S E_r\, dS_r = E_r(r) \, 4\pi \,r^2 = \frac{q}{4\pi\,\epsilon_0\,r^2}\,\,4\pi \,r^2 = \frac{q}{\epsilon_0},$$ (2.28)

because the normal to the surface is always parallel to the local electric field. (See Section A.16.) Here, $r$ is also a spherical polar coordinate. (See Section A.23.)

Figure 2.2: Gauss' law.

However, we also know from the divergence theorem that

$$\oint_S {\bf E}\cdot d{\bf S} = \int_V \nabla\cdot {\bf E} \,dV,$$ (2.29)

where $V$ is the volume enclosed by surface $S$. (See Section A.20.) Let us evaluate $\nabla\cdot {\bf E}$ directly. In Cartesian coordinates, the electric field (2.12) is written

$${\bf E} = \frac{q}{4\pi\,\epsilon_0} \left(\frac{x}{r^3},\,\, \frac{y}{r^3}, \, \frac{z}{r^3} \right),$$ (2.30)

where $r^2=x^2+y^2+z^2$. So,

$$\frac{\partial E_x}{\partial x} = \frac{q}{4\pi\,\epsilon_0}\left... ...,x}{r^4}\frac{x}{r} \right) = \frac{q}{4\pi\,\epsilon_0}\frac{r^2-3\,x^2}{r^5}.$$ (2.31)

Here, use has been made of the easily demonstrated result

$$\frac{\partial r}{\partial x} = \frac{x}{r}.$$ (2.32)

Formulae analogous to Equation (2.31) can be obtained for $\partial E_y/\partial y$ and $\partial E_z/\partial z$. The divergence of the field is, thus, given by

$$\nabla\cdot {\bf E} \equiv \frac{\partial E_x}{\partial x}+ \frac... ... z} = \frac{q}{4\pi\,\epsilon_0} \frac{3\,r^2 - 3\,x^2-3\,y^2-3\,z^2}{r^5} = 0.$$ (2.33)

(See Section A.20.) This is an extremely puzzling result. We have from Equations (2.28) and (2.29) that

$$\int_V \nabla \cdot{\bf E}\,dV= \frac{q}{\epsilon_0},$$ (2.34)

and yet we have just proved that $\nabla\cdot {\bf E}=0$. This paradox can be resolved after a close examination of Equation (2.33). At the origin ($r=0$), we find that $\nabla\cdot {\bf E} = 0/0$, which implies that $\nabla\cdot {\bf E}$ can take any value at this point. Thus, Equations (2.33) and (2.34) can be reconciled if $\nabla\cdot {\bf E}$ is some sort of “spike” function; that is, if it is zero everywhere, except arbitrarily close to the origin, where it becomes very large. This must occur in such a manner that the volume integral over the spike is finite.

Let us examine how we might construct a one-dimensional spike function. Consider the “box-car” function

$$g(x,\epsilon) = \left\{ \begin{array}{lll} 1/\epsilon &\mbox{... ... x\vert < \epsilon/2\\ 0 &&{\rm otherwise} \end{array}\right..$$ (2.35)

See Figure 2.3. It is clear that

$$\int_{-\infty}^{\infty} g(x,\epsilon)\,dx = 1.$$ (2.36)

Now, consider the function

$$\delta(x) = \lim_{\epsilon\rightarrow 0} g(x,\epsilon).$$ (2.37)

This function is zero everywhere, except arbitrarily close to $x=0$, where it is very large. However, according to Equation (2.36), the function still possess a finite integral:

$$\int_{-\infty}^{\infty} \delta (x)\,dx = 1.$$ (2.38)

Thus, $\delta(x)$ has all of the required properties of a spike function. The one-dimensional spike function $\delta(x)$ is called the Dirac delta function, after Paul Dirac who invented it in 1927 while investigating quantum mechanics. The delta function is an example of what mathematicians call a generalized function; it is not well defined at $x=0$, but its integral is nevertheless well defined. Consider the integral

$$\int_{-\infty}^{\infty} f(x)\,\delta(x)\,dx,$$ (2.39)

where $f(x)$ is a function that is well behaved in the vicinity of $x=0$. Because the delta function is zero everywhere, apart from arbitrarily close to $x=0$, it is clear that

$$\int_{-\infty}^{\infty}f(x)\, \delta(x)\,dx = f(0)\int_{-\infty}^{\infty} \delta(x) \,dx = f(0),$$ (2.40)

where use has been made of Equation (2.38). A simple change of variables allows us to define $\delta(x-x_0)$, which is a delta function centered on $x=x_0$. Equation (2.40) gives

$$\int_{-\infty}^{\infty} f(x)\,\delta(x-x_0)\,dx = f(x_0).$$ (2.41)

Figure 2.3: A box-car function.

We actually require a three-dimensional delta function; that is, a function that is zero everywhere, apart from arbitrarily close to the origin, where it is very large, and whose volume integral is unity. If we denote this function by $\delta({\bf r})$ then it is easily seen that the three-dimensional delta function is the product of three one-dimensional delta functions:

$$\delta({\bf r}) = \delta(x)\,\delta(y)\,\delta(z).$$ (2.42)

This function is clearly zero everywhere, except arbitrarily close the origin, where it is very large. But, is its volume integral unity? Let us integrate over a cube of dimension $2\,a$ that is centered on the origin, and aligned along the Cartesian axes. This volume integral is obviously separable, so that

$$\int \delta({\bf r})\,dV = \int_{-a}^{a} \delta(x)\,dx \int_{-a}^{a} \delta(y)\,dy \int_{-a}^{a} \delta(z)\,dz.$$ (2.43)

(See Section A.17.) The integral can be turned into an integral over all space by taking the limit $a\rightarrow\infty$. However, we know that, for one-dimensional delta functions, $\int_{-\infty}^{\infty} \delta(s)\,ds = 1$, so it follows from the previous equation that

$$\int \delta({\bf r})\,dV =1,$$ (2.44)

which is the desired result. A simple generalization of previous arguments yields

$$\int f({\bf r})\, \delta({\bf r})\,dV =f({\bf0}),$$ (2.45)

where $f({\bf r})$ is any well-behaved scalar field. Finally, we can change variables and write

$$\delta({\bf r} - {\bf r}') = \delta(x-x')\,\delta(y-y')\,\delta(z-z'),$$ (2.46)

which is a three-dimensional delta function centered on ${\bf r} = {\bf r}'$. It is easily demonstrated that

$$\int f({\bf r})\,\delta({\bf r}- {\bf r}')\,dV= f({\bf r}').$$ (2.47)

Up to now, we have only considered volume integrals taken over all space. However, it should be obvious that the previous result also holds for integrals over any finite volume $V$ that contains the point ${\bf r} = {\bf r}'$. Likewise, the integral is zero if $V$ does not contain the point ${\bf r} = {\bf r}'$.

Let us now return to the problem in hand. The electric field generated by an electric charge $q$ located at the origin has $\nabla\cdot {\bf E}=0$ everywhere apart from the origin, and also satisfies

$$\int_V {\nabla}\cdot{\bf E}\,dV = \frac{q}{\epsilon_0}$$ (2.48)

for a spherical volume $V$ centered on the origin. These two facts imply that

$$\nabla\cdot {\bf E} = \frac{q}{\epsilon_0} \,\delta({\bf r}),$$ (2.49)

where use has been made of Equation (2.44).

Consider, again, an electric charge $q$ located at the origin, and surrounded by a spherical surface $S$ that is centered on the origin. We have seen that the flux of the electric field out of $S$ is $q/\epsilon_0$. Suppose that we now displace the surface $S$, so that it is no longer centered on the origin. What now is the flux of the electric field out of S? We have

$$\oint_S {\bf E} \cdot d{\bf S} = \int_V \nabla\cdot {\bf E}\,dV$$ (2.50)

from the divergence theorem (see Section A.20), as well as Equation (2.49). From these two equations, it is clear that the flux of ${\bf E}$ out of $S$ is still $q/\epsilon_0$, as long as the displacement is not large enough that the origin is no longer enclosed by the sphere. Suppose that the surface $S$ is not spherical, but is instead highly distorted. What now is the flux of ${\bf E}$ out of $S$? As before, the divergence theorem and Equation (2.49) tell us that the flux remains $q/\epsilon_0$, provided that the surface contains the origin. Moreover, this result is completely independent of the shape of $S$.

Let us try to extend the previous result. Consider $N$ electric charges $q_i$ located at displacements ${\bf r}_i$. A simple generalization of Equation (2.49) gives

$$\nabla\cdot {\bf E} = \sum_{i=1,N} \frac{q_i}{\epsilon_0}\, \delta({\bf r} - {\bf r}_i).$$ (2.51)

Thus, Equation (2.50) and the previous equation imply that

$$\oint_S {\bf E}\cdot d{\bf S} =\int_V\nabla \cdot{\bf E}\,dV= \frac{Q}{\epsilon_0},$$ (2.52)

where $Q$ is the total charge enclosed by the surface $S$. This result is called Gauss's law, and does not depend on the shape of the surface. Note that the previous equation is analogous in form to the gravitational version of Gauss's law, (1.245). This is not surprising because, as we previously mentioned, Gauss's law holds for any inverse-square force law.

Suppose, finally, that instead of having a set of discrete electric charges, we have a continuous charge distribution described by a charge density $\rho({\bf r})$. The charge contained in a small rectangular volume of dimensions $dx$, $dy$, and $dz$, located at displacement ${\bf r}$, is $Q=\rho({\bf r})\,dx\,dy\,dz$. However, if we integrate $\nabla\cdot {\bf E}$ over this volume element then we obtain

$$\nabla \cdot {\bf E} \,\,dx\,dy\,dz = \frac{Q}{\epsilon_0}= \frac{\rho\,dx\,dy\,dz}{\epsilon_0},$$ (2.53)

where use has been made of Equation (2.52). Here, the volume element is assumed to be sufficiently small that $\nabla\cdot {\bf E}$ does not vary significantly across it. Thus, we get

$$\nabla \cdot {\bf E} = \frac{\rho}{\epsilon_0}.$$ (2.54)

Equation (2.54) is a differential equation that describes the electric field generated by a set of charges. We already know the solution to this equation when the charges are stationary; it is given by Equation (2.13),

$${\bf E}({\bf r}) = \frac{1}{4\pi \,\epsilon_0} \int_{V'} \rho({\bf r}')\, \frac{{\bf r} - {\bf r}'}{\vert{\bf r} - {\bf r}'\vert^3}\,dV'.$$ (2.55)

Incidentally, Equations (2.54) and (2.55) can be reconciled provided

$$\nabla\cdot\left(\frac{{\bf r} - {\bf r}'}{\vert{\bf r} - {\bf r}... ...\frac{1}{\vert{\bf r} - {\bf r}'\vert}\right)= 4\pi\,\delta({\bf r} -{\bf r}'),$$ (2.56)

where use has been made of Equation (2.16). (See Section A.21.) It follows that

$$\nabla\cdot{\bf E}({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\int \rho({\bf r}')\,\nabla\cdot\left(\frac{{\bf r} - {\bf r}'} {\vert{\bf r} - {\bf r}'\vert^3}\right)\,dV'$$

$$= \int \frac{\rho({\bf r}')}{\epsilon_0} \,\delta({\bf r} - {\bf r}')\, dV' = \frac{\rho({\bf r})}{\epsilon_0},$$ (2.57)

which is the desired result. Here, use has been made of Equation (2.47).

Finally, the most general form of Gauss's law, Equation (2.52), is obtained by integrating Equation (2.54) over a volume $V$ surrounded by a surface $S$, and making use of the divergence theorem:

$$\oint_S {\bf E}\cdot d{\bf S} = \frac{1}{\epsilon_0} \int_V \rho({\bf r})\,dV.$$ (2.58)

(See Section A.20.)