Simple Nature - Light and Matter

y considering its point charges individually, and the additivity-byregionproperty tells us that if we have a single point charge outsidea big region, we can break the region down into tiny cubes. If wecan prove that the flux through such a tiny cube really does cancelexactly, then the same must be true for any region, which we couldbuild out of such cubes, and any charge distribution, which we canbuild out of point charges.For simplicity, we will carry out this calculation only in the specialcase shown in figure f, where the charge lies along one axis ofthe cube. Let the sides of the cube have length 2b, so that the areaof each side is (2b) 2 = 4b 2 . The cube extends a distance b above,below, in front of, and behind the horizontal x axis. There is a distanced − b from the charge to the left side, and d + b to the rightside.There will be one negative flux, through the left side, and fivepositive ones. Of these positive ones, the one through the right sideis very nearly the same in magnitude as the negative flux throughthe left side, but just a little less because the field is weaker onthe right, due to the greater distance from the charge. The fluxesthrough the other four sides are very small, since the field is nearlyperpendicular to their area vectors, and the dot product E j · A j iszero if the two vectors are perpendicular. In the limit where b isvery small, we can approximate the flux by evaluating the field atthe center of each of the cube’s six sides, givingf / The flux through a tiny cubedue to a point charge.Φ = Φ left + 4Φ side + Φ right= |E left ||A left | cos 180 ◦ + 4|E side ||A side | cos θ side+ |E right ||A right | cos 0 ◦ ,and a little trig gives cos θ side ≈ b/d, soΦ = −|E left ||A left | + 4|E side ||A side | b d + |E right||A right |= ( 4b 2) ( −|E left | + 4|E side | b )d + |E right|= ( 4b 2) ( − kq(d − b) 2 + 4kq bd 2 d + kq )(d + b) 2( ) ( 4kqb21=d 2 −(1 − b/d) 2 + 4b)d + 1(1 + b/d) 2 .Using the approximation (1+ɛ) −2 ≈ 1−2ɛ for small ɛ, this becomesΦ =( ) ( 4kqb2d 2 −1 − 2bd + 4bd + 1 − 2b )d= 0 .Thus in the limit of a very small cube, b ≪ d, we have provedthat the flux due to this exterior charge is zero. The proof can be622 Chapter 10 Fields