Mr. 0. Heaviside on the Extra Current. 141 



ing a spark to pass. Then we have — = at P and Q. Let 



dQ at 



us first consider v and -=- resulting from the initial charge, 



supposing 47r 2 -5> 1. By Fourier's theorem, 



ttt /-. ®\ V ,2Vsl- cosot iirx 



where ^ is the final potential. Therefore, by (8), that part of 

 the potential due to the initial charge is 

 V 2 V _±® 1— coswr 



^ + _ e .2 - - . cos^ (cos + -jrimj — j (17) 



and by (10) that part of the current due to the initial charge 

 is 



2Y _A°° 1— cosi7r . wr« 2 . £m r s ..,„. 



To find the potential and current due to the initial current, 

 we have 



V __ 2 V si- cos iir . iirx ,. 

 kl trkl i i I 



therefore 



dQ 2V _l2l- cosi7r . farx , .. ^ . s tm'i 



Ht = M- e *? i «n T -(A,«». + B I « ft )- gri 



and 



TT/ , 2Vs 1- cos iir ittx _ 1 i A,- B* . 



v=V'H 2^ =2 cos-y-e 2a ( -— cos + -77 sin 



iTii z L \ 2 2 



, Bj-m',- A,m 



H 27— cos 



2 



- sin ) -r—i 



where V, A,-, and B* are constants. The conditions to deter- 

 mine them are that v = when £ = and when t = oo . Also 



-TJ7 =t-7 when £ = 0. Therefore 



V' = 0, A i= l, and Bi=--4* 

 Thus 

 dQ 2V _A^1— cosivr . iirx / 1 . \ tm'i 



_^ = — -.6 2a2 : SUl -y- • ( COS « gm ) « Q9) 



a — i _ 1 — cos iir iirx . tm'i 

 v 4V;^. e --S— jjr— oo.-j-.nn-g-. . . (20) 



