6.8 The divergence theorem  (Page 6/12)

Proof

The logic of this proof follows the logic of [link] , only we use the divergence theorem rather than Green’s theorem.

First, suppose that S does not encompass the origin. In this case, the solid enclosed by S is in the domain of ${\text{F}}_r,$ and since the divergence of ${\text{F}}_r$ is zero, we can immediately apply the divergence theorem and find that ${\displaystyle {\iint }_S\text{F}·d\text{S}}$ is zero.

Now suppose that S does encompass the origin. We cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. Let $S_a$ be a sphere of radius a inside of S centered at the origin. The outward normal vector field on the sphere, in spherical coordinates, is

(see [link] ). Therefore, on the surface of the sphere, the dot product ${\text{F}}_r·\text{N}$ (in spherical coordinates) is

The flux of ${\text{F}}_r$ across $S_a$ is

Now, remember that we are interested in the flux across S , not necessarily the flux across $S_a.$ To calculate the flux across S , let E be the solid between surfaces $S_a$ and S . Then, the boundary of E consists of $S_a$ and S . Denote this boundary by $S-S_a$ to indicate that S is oriented outward but now $S_a$ is oriented inward. We would like to apply the divergence theorem to solid E. Notice that the divergence theorem, as stated, can’t handle a solid such as E because E has a hole. However, the divergence theorem can be extended to handle solids with holes, just as Green’s theorem can be extended to handle regions with holes. This allows us to use the divergence theorem in the following way. By the divergence theorem,

Therefore,

and we have our desired result.

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Now we return to calculating the flux across a smooth surface in the context of electrostatic field $\text{E}=\frac{q}{4\pi {\epsilon }_0}{\text{F}}_r$ of a point charge at the origin. Let S be a piecewise smooth closed surface that encompasses the origin. Then

If S does not encompass the origin, then

Therefore, we have justified the claim that we set out to justify: the flux across closed surface S is zero if the charge is outside of S , and the flux is $q\text{/}{\epsilon }_0$ if the charge is inside of S .

This analysis works only if there is a single point charge at the origin. In this case, Gauss’ law says that the flux of E across S is the total charge enclosed by S . Gauss’ law can be extended to handle multiple charged solids in space, not just a single point charge at the origin. The logic is similar to the previous analysis, but beyond the scope of this text. In full generality, Gauss’ law states that if S is a piecewise smooth closed surface and Q is the total amount of charge inside of S , then the flux of E across S is $Q\text{/}{\epsilon }_0.$