Method of finding Scalar and Vector Potentials. 229 

 Thus 3A 1 



(cf+Xr), 



S< (r-f) 3 

 and 



d« 2 " (r-f) 6 + (r-f)'^ + 'BJ 



(r-f) 5 C'-f) 1 ' 



where 



n = [<t>' , '{.i—4>)+f"(y-i') + x"(t-x) 



Also B>' ,r-<£ ~dr _y — yjr B?' _ z — % 



Be " 7="f cty ~ 7h?' B~e ~ ^|' 



B.*' c c" >' — f By c c'r— £' B- c c'r — f 

 ~d\ _ 4>' f M' % — <b ~d\ _^\r" I* y — yfr BA, __ %'' ^ z—% 

 B.r c c' r — f By o c' r — f' B~ a c ' r — % 



Hence 



and 



c2 P"= (^ [+'(«-0-(«+>.)(*-#)]'» 



+ ^[-- , f'(--^)-^+M- ( ^f 2 -o(c+x)], 



with similar expressions for c 2 ^^- and t* 2 — „. Addin<>- 



Br B- 2 ° 



these three expressions, Ave easily find that 



- (r _ f) » ( ,._^ (r _g)i (r _ f) « + (r-B« 



_ 3c -f + 3W + c\r 2 + c\g 3 + 4cXrg ^ 



(»—*)* ~ + -0—?) 4 



_ X(3\ + c) r 2 + 4\f|r - i- <■( 3c + X)g 2 . yx?- 2 



and, therefore, 



V 2 A-4 |£ = 0. 

 c- Ot- 



