the Electrical Capacities of two Condensers. 375 



channel. We must therefore make 



+ R + R' 



4pZ 2 



I 



a minimum. We find 



^-^=0; (23) 



and we get finally 



G=R+R'. 



Now for a given value of a, R and W must be as high as 

 possible ; therefore we must make the resistance of our galva- 

 nometer as high as possible. 



Returning to the general case in which there is a leakage 

 in the condensers ; and putting t = r, t being so large that the 

 terms involving e~ HT may be neglected, we get 



* = G\WTp'-K+pS (24) 



Thus there is a steady current through the galvanometer, and 

 the needle is permanently deflected. 



Again, if t = 0, x—0. But let us suppose t very small, so 

 that, on expanding e~ nt , powers higher than the first may be 

 neglected, and find the initial current. We find 



_E/ R'C'-RC [1/1 ly p p' \\ ( , 



x ~~ at rcr'C' + gIc + c / Ar+ /5 B,'+ P r Jy> ^ D) 



which reduces, if we neglect powers of - above the first, to 



E fB'C-RO 1/1 I\/^_?U/ f9a\ 



x ~ at rcr'C' + av.o + 07V P JS ' { } 



By altering R the sign of this may be made to change, and 

 thus the initial current may be in the same or in the opposite 

 direction to the final. In practice this is indicated by a short 

 kick of the needle in one direction, followed by a deflection in 

 the other. 



On the same assumptions as to p, //, let us find the quantity 

 of electricity which passes through the galvanometer in time t, 

 t being very short, but yet so long that e~ nr maybe neglected. 

 Integrating (16) and calling the quantity P, we have 



. { « S] ( Bo +IW i] + M-!>. . m 



