54-58] 



Tubes of Force 



45 



57. COULOMB'S LAW. If R is the outward intensity at a point just 

 outside a conductor, then R = 47rcr, where a- is the surface density of electri- 

 fication on the conductor. 



We have already seen that the whole electrification of a conductor must 

 reside on the surface. Therefore we no longer deal with a volume density 

 of electrification p, such that the charge in the element of volume dxdydz is 

 pdxdydz, but with a surface-density of electrification <r such that the charge 

 on an element dS of the surface of the conductor is &dS. 



The surface of the conductor, as we have seen, is an equipotential, so that 

 by the theorem of p. 29, the intensity is in a direction normal to the 

 surface. Let us draw perpendiculars to the surface at every 

 point on the boundary of a small element of area dS, these per- 

 pendiculars each extending a small distance into the conductor 

 in one direction and a small distance away from the conductor 

 in the other direction. We can close the cylindrical surface so 

 formed, by two small plane areas, each equal and parallel to the 

 original element of area dS. Let us now apply Gauss' Theorem 

 to this closed surface. The normal intensity is zero over every 

 part of this surface except over the cap of area dS which is 

 outside the conductor. Over this cap the outward normal in- 

 tensity is R, so that the value of the surface integral of normal 

 intensity taken over the closed surface, consists of the single term RdS. 

 The total charge inside the surface is <rdS, so that by Gauss' Theorem, 



FIG. 13. 



.(21), 



and Coulomb's Law follows on dividing by dS. 



58. Let us draw the complete tube of force which is formed by the 

 lines of force starting from points on the boundary of the element dS of the 

 surface of the conductor. Let us suppose that the surface density on this- 

 element is positive, so that the area dS forms the normal cross-section at 



FIG. 14. 



the positive end, or beginning, of the tube of force. Let us suppose that at 

 the negative end of the tube of force, the normal cross- section is dS', that 



