432 Mr. R. Hargreaves on 



\=0 in the lower limit, so that <£ L ... are constants; then 

 the form of -^ above shows that first differential coefficients 

 are continuous at the boundary \ = 0. The constant is given 

 by &p = 22(pL +2p'Lo') = -4r(f> if (91) is true ; 



and <£> = — r = -C, i.e. C= -£; and 



V«Jo 4 



is the value of the external potential. To verify the observ- 

 ance of the condition (91) we have 



Since </> in (92) is — -r-ji, the condition is satisfied. 

 The field integral is given by 



S-j-2f+*,-^J i "^(l-2».i'-K.'P) | 



where 



- 2 If ,, A ,— 1 f 7 A' 



But by (86) 

 1 f B 4 VM n4/ /N f " d\ / Xtf,T\ 



aJo^ a " +2AV >=J, 7j( 3 -J^) j 



= Jo Vj +2 Jo Jal vj> x= Jo w' (93 ' 



since X/ a/ J vanishes at each limit. Hence 



s -jv*.=^f*,-£.j: 



• (94) 



VJ 



For the geometry of potentials it is often convenient to 

 write the constant of the internal potential separately, as 



and then ? ■ ( 95 ) 



X(AL + 2A'L ') = A a </> 0? and t( P L + 2p'L ') 



+ 2»'L ') = 2J 



