580 Problems in Electrostatics. 



This potential, however, does not satisfy (C) as JJ(p) must 

 do. Bat we can determine another, ih{p), having at ®, on 

 the side of the conductor, a normal derivative equal and 

 opposite to that of v(p), and continuous elsewhere. This- 

 function v^p) is given by 



*fj0^f<WP*) £(*)«* 



Put U(j>) equal to the sum of these two potentials, i e. 



= 4 a +1 (po-)/i(<r)^cr. 

 «/ S 



This sum satisfies both (A) and (C). It will be the required 

 potential to supplement n(p) if we can find yu,(<x) so that (23) 

 holds. Now at X the potential v x {p) has a continuous normal 

 derivative, while that of v(/>) is discontinuous. It follows 

 then that 



^<•->+!^•*Hi G( *' , * , ''" 



s and a denoting points on the surface 2. On substituting 



in (23) the values of j— («") and ^ ( 5+ ) derived from 

 these, we find for /jl(s) the integral equation 



This is a Fredholm equation from which /*(«) is uniquely 

 determined. Though the kernel becomes infinite at s = a 



like - the second iterated kernel is finite and FredhoWs 

 r 



method is applicable. 



That (24) admits a unique finite and continuous solution 

 will follow if we can show that the corresponding homo- 

 geneous equation does not. That such is the case is easily 



