6.8 The divergence theorem  (Page 5/12)

One of the most common applications of the divergence theorem is to electrostatic fields . An important result in this subject is Gauss’ law    . This law states that if S is a closed surface in electrostatic field E , then the flux of E across S is the total charge enclosed by S (divided by an electric constant). We now use the divergence theorem to justify the special case of this law in which the electrostatic field is generated by a stationary point charge at the origin.

If $(x,y,z)$ is a point in space, then the distance from the point to the origin is $r=\sqrt{x^2+y^2+z^2}.$ Let ${\text{F}}_r$ denote radial vector field ${\text{F}}_r=\frac{1}{r^2}⟨\frac{x}{r},\frac{y}{r},\frac{z}{r}⟩.$ The vector at a given position in space points in the direction of unit radial vector $⟨\frac{x}{r},\frac{y}{r},\frac{z}{r}⟩$ and is scaled by the quantity $1\text{/}{r}^2.$ Therefore, the magnitude of a vector at a given point is inversely proportional to the square of the vector’s distance from the origin. Suppose we have a stationary charge of q Coulombs at the origin, existing in a vacuum. The charge generates electrostatic field E given by

where the approximation ${\epsilon }_0=8.854\times {10}^{\mathrm{-12}}$ farad (F)/m is an electric constant. (The constant ${\epsilon }_0$ is a measure of the resistance encountered when forming an electric field in a vacuum.) Notice that E is a radial vector field similar to the gravitational field described in [link] . The difference is that this field points outward whereas the gravitational field points inward. Because

we say that electrostatic fields obey an inverse-square law    . That is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge (which in this case is at the origin). Given this vector field, we show that the flux across closed surface S is zero if the charge is outside of S , and that the flux is $q\text{/}{\epsilon }_0$ if the charge is inside of S . In other words, the flux across S is the charge inside the surface divided by constant ${\epsilon }_0.$ This is a special case of Gauss’ law, and here we use the divergence theorem to justify this special case.

To show that the flux across S is the charge inside the surface divided by constant ${\epsilon }_0,$ we need two intermediate steps. First we show that the divergence of ${\text{F}}_r$ is zero and then we show that the flux of ${\text{F}}_r$ across any smooth surface S is either zero or $4\pi .$ We can then justify this special case of Gauss’ law.

Notice that since the divergence of ${\text{F}}_r$ is zero and E is ${\text{F}}_r$ scaled by a constant, the divergence of electrostatic field E is also zero (except at the origin).

In other words, this theorem says that the flux of ${\text{F}}_r$ across any piecewise smooth closed surface S depends only on whether the origin is inside of S .