" INVERSE SQUARE " SYSTEMS 35 



side.^ of the tube the intensity is parallel to the surface of the 

 tube and has no component perpendicular to it, and therefore 

 J NWS = 0. But J NWS = 47T</, since q is the charge within. Then 

 irq = and q = 0. 



Then the supposition that the tube, starting from any real 

 charge g, can end at a point in the field where there is no charge, 

 is inadmissible. 



Now suppose that the tube begins at S x , Fig. 27, on a charge q 1 

 and ends at S 2 on a charge </ 2 . By ending we mean that within S 2 



Flo. -27. 



I lu- inlcsiMtv IN /ero. Prolong the tube a> dotted within S x and S, 

 and thu> make- U<> ciuk Apply (iati^V theorem to the closed 

 surface thus formed. Then J NWS = 0, since N = at every point, 

 and l-('/i+'7 2 ) = 0. 



Hence q^= q r 



The product intensity x cross-section is constant 

 along a tube of force which contains no charge. Let SjSg, 



ed ion- of >uch a tube, and let I^ be the inten- 

 at those section.-. Let us apply Gauss's theorem to the closed 



Fio. 28. 



MII face formed by the tube between Sjand S 2 and by the ends SjSg. 

 N ha> no value over the tides of the tube since the intensity is along 



the tube, 



Over the end S 2 , N(/S = 



o . rtheendS^ NWS a I 1 S 1 

 IjS r ne^ali\. I inuard-. Then J Nf/S = 1^ I 2 S 2 . 



But >iii( e there i, no charge within the tube between S x and 

 N 3 ' 'iid thi-refon- 1^! I 2 S 2 . 



