181-183] Green's Theorem 159 



Let us divide the sum on the right into I lt the integral over a single 

 closed surface enclosing any number of conductors, and / 2 the integrals over 

 the surfaces of the conductors. Thus 



-~\ 



where ~- denotes differentiation along the normal drawn into the surface. 



Thus ^ is equal to the component of intensity along this normal, and 



therefore to N t where N is the component along the outward normal. 

 Hence 



/|- 



At the surface of a conductor ^ = 4-Tro-, so that 



on 



7 2 = 4?rS 1 1 a-dS over conductors 



= 4-7T x total charge on conductors. 

 If there is any volume electrification, V 2 F= 4>7rp, so that 



rrr rrr 



I V 2 F dxdydz = 4?r III pdxdydz, 



JJJ JJJ 



and the integral on the right represents the total volume electrification. 

 Thus equation (102) becomes 



1 1 NdS = 4-7T x (total charge on conductors + total volume electrification), 

 so that the theorem reduces to Gauss' Theorem. 



183. Next put <E> and SP each equal to V. Then equation (100) becomes 



Take the surfaces now to be the surfaces of conductors, and a sphere of 

 radius r at infinity. At infinity V is of order - , so that ^ is of order 



1 dV 



, and hence ^o~ > integrated over the sphere at infinity, vanishes ( 178). 



The equation becomes 



- 4-7T If [p V dxdydz + fff^ 2 dxdydz - 4?r IjVa dS = 0. 



