Simple Nature - Light and Matter

Note that although region and its surface had a definite physicalexistence in our story — they are the planet Flatcat and the surfaceof planet Flatcat — Gauss’ law is true for any region and surface wechoose, and in general, the Gaussian surface has no direct physicalsignificance. It’s simply a computational tool.Rather than proving Gauss’ theorem and then presenting someexamples and applications, it turns out to be easier to show some examplesthat demonstrate its salient properties. Having understoodthese properties, the proof becomes quite simple.self-check KSuppose we have a negative point charge, whose field points inward,and we pick a Gaussian surface which is a sphere centered on thatcharge. How does Gauss’ theorem apply here? ⊲ Answer, p. 92610.6.2 Additivity of fluxFigure e shows two two different ways in which flux is additive.Figure e/1, additivity by charge, shows that we can break down acharge distribution into two or more parts, and the flux equals thesum of the fluxes due to the individual charges. This follows directlyfrom the fact that the flux is defined in terms of a dot product, E·A,and the dot product has the additive property (a+b)·c = a·c+b·c.To understand additivity of flux by region, e/2, we have to considerthe parts of the two surfaces that were eliminated when theywere joined together, like knocking out a wall to make two smallapartments into one big one. Although the two regions shared thiswall before it was removed, the area vectors were opposite: the directionthat is outward from one region is inward with respect tothe other. Thus if the field on the wall contributes positive flux toone region, it contributes an equal amount of negative flux to theother region, and we can therefore eliminate the wall to join the tworegions, without changing the total flux.10.6.3 Zero flux from outside chargesA third important property of Gauss’ theorem is that it onlyrefers to the charge inside the region we choose to discuss. In otherwords, it asserts that any charge outside the region contributes zeroto the flux. This makes at least some sense, because a charge outsidethe region will have field vectors pointing into the surface on oneside, and out of the surface on the other. Certainly there shouldbe at least partial cancellation between the negative (inward) fluxon one side and the positive (outward) flux on the other. But whyshould this cancellation be exact?To see the reason for this perfect cancellation, we can imaginespace as being built out of tiny cubes, and we can think of any chargedistribution as being composed of point charges. The additivity-bychargeproperty tells us that any charge distribution can be handlede / 1. The flux due to twocharges equals the sum of thefluxes from each one. 2. Whentwo regions are joined together,the flux through the new regionequals the sum of the fluxesthrough the two parts.Section 10.6 Fields by Gauss’ Law 621