774 Dr. J. R. Wilton on the Solution of 



provided that these series converge, — as they evidently do, 

 for all finite values of f, except that at £=f , V~Va-> Vo 

 becomes infinite like — 2E log?*!. And when (■ is infinite we 

 have 



V =-{2E/(K + 1)}£-{4KE/(K-h1)'} I {(K-1)/(K + 1)}- 1 £ 



2E log r v 



So that all the necessary conditions are satisfied. 



There is little interest in the case of the circular cylinder, 

 as the solution can immediately be derived by the method of 

 images. But there is some interest in the case of the thin 

 plate of dielectric, X = 0, for the series for V and V; can 

 then be summed. For all points outside the plate (which 

 cuts the xy plane in the line y = 0, —c<x~< c) we have 



V=— E log {cosh (£o— £) — cos Ofo—T?)} 

 -E log {cosh (f + £)-cos (rjo + v)}, 

 where x-\-i-y = c cosh (f + irj). 



Gravitational Potential. 



10. If Y and Yi be respectively the external and internal 

 potentials of a gravitating cylinder bounded by the curve (1), 

 we have 



V 2 Vi=-±7i7>, V 2 V =0, 

 and when d = r, 



V != V„, dV,-/3f=9V /9f. 



Also Y -> —A log r as r -*> go , 



where A is a constant, which in the case of constant density 

 is 2p X the area of cross-section of the cylinder. 

 Let ijr be determined so that 



-4tt / o = V 2 ^. 

 Then 



V*=*0e, y) + i{f(6) f/(r)}+ lj{F(e)-Y(r)}, 



And we have immediately 



g=f+f (X, Y), 



G'=F-(X'^-Y'^)*(X,Y). 



