252 Prof. P. J. Daniell on the 



But 



Q 2r (^) =*-<*■+!> f (£ + V? 2 + 1 cosh u)-< 2r +^du 

 or 



Q, r (0) = £-(2r+l) f C osh U-^+^ll. 



Also 



^Qar(tr) = t- (2r+1) J (f+ V?+lcoshu)-^+ 2 )(-2r-l) 



or when £=0 this 



Then 



X (lH ==eoshwWw, 



j-(2r+i)(__2r-l) I cosh w-^+a)^. 

 Jo 



(-iIL=?? Wf)C - 



where 0*= (2r + 1) f " (cosh u)-V+»dul f " (cosh u)-<»-+i><fo 



_ r nr+i) y 



X7 2 

 t>2r ^2r 



4r+l" 



Between the regions I. and II. the boundary of non- 

 conducting material is continuous, that is to say there is no 

 chance of any current escaping between the regions I. and 

 II. except through the open end itself. The total current 

 passing into the region I must then be the same as ttZ, the 

 total current in the region II. This total current is 



and =7rZ on the other hand. 

 Then& =J*. 



