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Next: Poisson's equation Up: Time-independent Maxwell equations Previous: The electric scalar potential

Gauss' law

Figure 26:
\epsfysize =2in
Consider a single charge located at the origin. The electric field generated by such a charge is given by Eq. (171). Suppose that we surround the charge by a concentric spherical surface $S$ of radius $r$ (see Fig. 26). 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)  ...
...q}{4\pi\epsilon_0 r^2}  4\pi  r^2 =
\end{displaymath} (187)

since the normal to the surface is always parallel to the local electric field. However, we also know from Gauss' theorem that
\oint_S {\bf E}\cdot d{\bf S} = \int_V \nabla\cdot {\bf E}   d^3{\bf r},
\end{displaymath} (188)

where $V$ is the volume enclosed by surface $S$. Let us evaluate $\nabla\cdot {\bf E}$ directly. In Cartesian coordinates, the field is written
{\bf E} = \frac{q}{4\pi \epsilon_0} \left(\frac{x}{r^3},   \frac{y}{r^3},
\end{displaymath} (189)

where $r^2=x^2+y^2+z^2$. So,
\frac{\partial E_x}{\partial x} = \frac{q}{4\pi \epsilon_0}...
...r} \right) =
\frac{q}{4\pi \epsilon_0}\frac{r^2-3 x^2}{r^5}.
\end{displaymath} (190)

Here, use has been made of
\frac{\partial r}{\partial x} = \frac{x}{r}.
\end{displaymath} (191)

Formulae analogous to Eq. (190) 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} = \frac{\partial E_x}{\partial x}+
...pi \epsilon_0} \frac{3 r^2 - 3 x^2-3 y^2-3 z^2}{r^5} = 0.
\end{displaymath} (192)

This is a puzzling result! We have from Eqs. (187) and (188) that
\int_V \nabla \cdot{\bf E}  d^3{\bf r}= \frac{q}{\epsilon_0},
\end{displaymath} (193)

and yet we have just proved that $\nabla\cdot {\bf E}=0$. This paradox can be resolved after a close examination of Eq. (192). At the origin ($r=0$) we find that $\nabla\cdot
{\bf E} = 0/0$, which means that $\nabla\cdot {\bf E}$ can take any value at this point. Thus, Eqs. (192) and (193) can be reconciled if $\nabla\cdot {\bf E}$ is some sort of ``spike'' function: i.e., 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.

Figure 27:
\epsfysize =2in
Let us examine how we might construct a one-dimensional spike function. Consider the ``box-car'' function
g(x,\epsilon) = \left\{
1/\epsilon &\mbox...
... x\vert < \epsilon/2\\
0 &&{\rm otherwise}
\end{displaymath} (194)

(see Fig. 27). It is clear that that
\int_{-\infty}^{\infty} g(x,\epsilon) dx = 1.
\end{displaymath} (195)

Now consider the function
\delta(x) = \lim_{\epsilon\rightarrow 0} g(x,\epsilon).
\end{displaymath} (196)

This is zero everywhere except arbitrarily close to $x=0$. According to Eq. (195), it also possess a finite integral;
\int_{-\infty}^{\infty} \delta (x) dx = 1.
\end{displaymath} (197)

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 the Cambridge physicist 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,
\end{displaymath} (198)

where $f(x)$ is a function which is well-behaved in the vicinity of $x=0$. Since the delta-function is zero everywhere apart from very 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),
\end{displaymath} (199)

where use has been made of Eq. (197). The above equation, which is valid for any well-behaved function, $f(x)$, is effectively the definition of a delta-function. A simple change of variables allows us to define $\delta(x-x_0)$, which is a spike function centred on $x=x_0$. Equation (199) gives
\int_{-\infty}^{\infty} f(x) \delta(x-x_0) dx = f(x_0).
\end{displaymath} (200)

We actually want a three-dimensional spike function: i.e., a function which is zero everywhere apart from arbitrarily close to the origin, 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).
\end{displaymath} (201)

This function is clearly zero everywhere except the origin. But is its volume integral unity? Let us integrate over a cube of dimensions $2 a$ which is centred on the origin, and aligned along the Cartesian axes. This volume integral is obviously separable, so that
\int \delta({\bf r}) d^3{\bf r} = \int_{-a}^{a} \delta(x) dx
\int_{-a}^{a} \delta(y) dy \int_{-a}^{a} \delta(z) dz.
\end{displaymath} (202)

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 $\int_{-\infty}^{\infty} \delta(x) dx = 1$, so it follows from the above equation that
\int \delta({\bf r}) d^3{\bf r} =1,
\end{displaymath} (203)

which is the desired result. A simple generalization of previous arguments yields
\int f({\bf r})  \delta({\bf r}) d^3{\bf r} =f({\bf0}),
\end{displaymath} (204)

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'),
\end{displaymath} (205)

which is a three-dimensional spike function centred on ${\bf r} = {\bf r}'$. It is easily demonstrated that
\int f({\bf r}) \delta({\bf r}- {\bf r}') d^3{\bf r}= f({\bf r}').
\end{displaymath} (206)

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

Let us now return to the problem in hand. The electric field generated by a 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}  d^3{\bf r} = \frac{q}{\epsilon_0}
\end{displaymath} (207)

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}),
\end{displaymath} (208)

where use has been made of Eq. (203).

At this stage, vector field theory has yet to show its worth.. After all, we have just spent an inordinately long time proving something using vector field theory which we previously proved in one line [see Eq. (187)] using conventional analysis. It is time to demonstrate the power of vector field theory. Consider, again, a charge $q$ at the origin surrounded by a spherical surface $S$ which is centered on the origin. Suppose that we now displace the surface $S$, so that it is no longer centered on the origin. What is the flux of the electric field out of S? This is not a simple problem for conventional analysis, because the normal to the surface is no longer parallel to the local electric field. However, using vector field theory this problem is no more difficult than the previous one. We have

\oint_S {\bf E} \cdot d{\bf S} = \int_V \nabla\cdot {\bf E}  d^3{\bf r}
\end{displaymath} (209)

from Gauss' theorem, plus Eq. (208). From these equations, it is clear that the flux of ${\bf E}$ out of $S$ is $q/\epsilon_0$ for a spherical surface displaced from the origin. However, the flux becomes zero when the displacement is sufficiently large that the origin is no longer enclosed by the sphere. It is possible to prove this via conventional analysis, but it is certainly not easy. Suppose that the surface $S$ is not spherical but is instead highly distorted. What now is the flux of ${\bf E}$ out of $S$? This is a virtually impossible problem in conventional analysis, but it is still easy using vector field theory. Gauss' theorem and Eq. (208) tell us that the flux is $q/\epsilon_0$ provided that the surface contains the origin, and that the flux is zero otherwise. This result is completely independent of the shape of $S$.

Consider $N$ charges $q_i$ located at ${\bf r}_i$. A simple generalization of Eq. (208) gives

\nabla\cdot {\bf E} = \sum_{i=1}^N
\frac{q_i}{\epsilon_0}  \delta({\bf r} - {\bf r}_i).
\end{displaymath} (210)

Thus, Gauss' theorem (209) implies that
\int_S {\bf E}\cdot d{\bf S} = \int_V \nabla\cdot{\bf E}  
d^3{\bf r}= \frac{Q}{\epsilon_0},
\end{displaymath} (211)

where $Q$ is the total charge enclosed by the surface $S$. This result is called Gauss' law, and does not depend on the shape of the surface.

Suppose, finally, that instead of having a set of discrete 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 position ${\bf r}$ is $Q=\rho({\bf r}) dx dy dz$. However, if we integrate $\nabla\cdot {\bf E}$ over this volume element we obtain

\nabla \cdot {\bf E}   dx dy dz = \frac{Q}{\epsilon_0}=
\frac{\rho dx dy dz}{\epsilon_0},
\end{displaymath} (212)

where use has been made of Eq. (211). Here, the volume element is assumed to be sufficiently small that $\nabla\cdot {\bf E}$ does not vary significantly across it. Thus, we obtain
\nabla \cdot {\bf E} = \frac{\rho}{\epsilon_0}.
\end{displaymath} (213)

This is the first of four field equations, called Maxwell's equations, which together form a complete description of electromagnetism. Of course, our derivation of Eq. (213) is only valid for electric fields generated by stationary charge distributions. In principle, additional terms might be required to describe fields generated by moving charge distributions. However, it turns out that this is not the case, and that Eq. (213) is universally valid.

Equation (213) is a differential equation describing 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 Eq. (172),

{\bf E}({\bf r}) = \frac{1}{4\pi  \epsilon_0} \int \rho({\b...
...f r} - {\bf r}'}{\vert{\bf r} - {\bf r}'\vert^3} d^3{\bf r}'.
\end{displaymath} (214)

Equations (213) and (214) can be reconciled provided
\nabla\cdot\left(\frac{{\bf r} - {\bf r}'}{\vert{\bf r} - {\...
...f r} - {\bf r}'\vert}\right)= 4\pi \delta({\bf r}
-{\bf r}'),
\end{displaymath} (215)

where use has been made of Eq. (175). It follows that
$\displaystyle \nabla\cdot{\bf E}({\bf r})$ $\textstyle =$ $\displaystyle \frac{1}{4\pi \epsilon_0}\int
\rho({\bf r}') \nabla\cdot\left(\frac{{\bf r} - {\bf r}'}
{\vert{\bf r} - {\bf r}'\vert^3}\right) d^3{\bf r}'$  
  $\textstyle =$ $\displaystyle \int \frac{\rho({\bf r}')}{\epsilon_0}  \delta({\bf r} - {\bf r}') 
d^3{\bf r}' = \frac{\rho({\bf r})}{\epsilon_0},$ (216)

which is the desired result. The most general form of Gauss' law, Eq. (211), is obtained by integrating Eq. (213) over a volume $V$ surrounded by a surface $S$, and making use of Gauss' theorem:
\oint_S {\bf E}\cdot d{\bf S} = \frac{1}{\epsilon_0} \int_V \rho({\bf r}) d^3{\bf r}.
\end{displaymath} (217)

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

next up previous
Next: Poisson's equation Up: Time-independent Maxwell equations Previous: The electric scalar potential
Richard Fitzpatrick 2006-02-02