Certain Problems of Two- Dimensional Physics. 11Z 



the proper multiple of the term (6) to V;. This term is 

 equal to 



-{2E/(K + l)}log{2sini[^ + ^ + t (2X ^|)]sini[T + ^ -.(2\ + f )]} 



= — {E/ (K + 1) } log {4 sin ((/> + cu) sin ($ — ia) sin (\/r — ice) sin ty + iu) \ 



E , 2 sin ($ + lu) sin (ty — to) 

 ~"K+1 g 2sin(^>-ta)sin(^ + ^) ' 



where 4> = \(fi + V Q \ ^ = i( T + Vo), a =^ + i? - 



The term to be added to V is found by multiplying the 

 coefficient of the second logarithm by K. Hence 



V = -E log {cosh (|r_f )_cos (v - Vo)} 



+ {(K-1)/(K + 1)}E log {cosh (£ + f )-cos (rj- % )\ 

 -Elog{cosh(2\+? + f)-cos(^ + ^ )} 



+ {(K-l)/(K + l)}Elog{cosh(2\ + ? -?)-cos(7 ? + ^ )}. 



And we have V { = - {4E/(K + 1)} log n. 



This, however, is not the solution of the problem we set 

 out to solve, for there is a line charge of strength 



-{(K-1)/(K + 1)}E at ?=2X. + f , V = 2tt- Vo . 



In the case of the circle X is infinite (c->0 in such a way 

 that ^ce^ = a, the radius of the circle) and the terms given 

 are sufficient for the complete solution. In the more general 

 case we superimpose the solution for a charge 



{(K-1)/(K + 1)}E at 2\ + f 0) Z*- % , 

 thus obtaining the solution of a problem involving a charge 



-{(K-1)/(K + 1)} 2 E at 4X+? ,%- 

 And proceeding in this way we have, finally, 



Vi— {4fl/(K + l)f 2 {(K-1)/(K+1)}- 



n=0 



X |log [cosh (2;a+f -£)-cos {v-(~) n v] 



+ log [oosh{(2n+2)X + fc+f}-ooB{iy + (->^,}]}, 



V = - E log {cosh (fo-f)- cos (y -v)\ 



+ {(K-1)/(K+1)}E log {cosh (&+0-«»(? o -i,)} 



-{4KE (K+l) 2 } 



% {(K-l)/(K + l)}«-ilog [cosh(2,a+f + f) 



n=l 



-cos{«7-(-) w ^ }], 



