Simple Nature - Light and Matter

|E| = kq totalr 2 ,where r is the radius of Flatcat. The flux is thenΦ = ∑ E j · A j ,and since the E j and A j vectors are parallel, the dot product equals|E j ||A j |, soΦ = ∑ kq totalr 2 |A j | .But the field strength is always the same, so we can take it outsidethe sum, givingΦ = kq totalr 2 ∑|Aj |= kq totalr 2 A total= kq totalr 2 4πr 2= 4πkq total .Not only have all the factors of r canceled out, but the result isthe same as for a disk!Everyone is pleasantly surprised by this apparent mathematicalcoincidence, but is it anything more than that? For instance, whatif the charge wasn’t concentrated at the center, but instead wasevenly distributed throughout Flatcat’s interior volume? Newton,however, is familiar with a result called the shell theorem (page 102),which states that the field of a uniformly charged sphere is the sameas if all the charge had been concentrated at its center. 10 We nowhave three different assumptions about the shape of Flatcat and thearrangement of the charges inside it, and all three lead to exactly thesame mathematical result, Φ = 4πkq total . This is starting to looklike more than a coincidence. In fact, there is a general mathematicaltheorem, called Gauss’ theorem, which states the following:For any region of space, the flux through the surface equals4πkq in , where q in is the total charge in that region.Don’t memorize the factor of 4π in front — you can rederive itany time you need to, by considering a spherical surface centered ona point charge.10 Newton’s human namesake actually proved this for gravity, not electricity,but they’re both 1/r 2 forces, so the proof works equally well in both cases.620 Chapter 10 Fields