of Electricity in a uniform plane conducting Surface. 463 



or 



2 log r = n log fi. 



32. No part of the above reasoning will be disturbed if we 

 suppose that each of the equal poles hitherto discussed emits (or 

 absorbs) in unit of time a quantity of electricity equal to unity ; 

 and as this supposition will simplify the consideration of poles 

 of unequal strength, we will adopt it. Then, when the poles 

 are unequal, let them severally emit in unit of time quanti- 

 ties of electricity represented respectively by Q 1? Q 2 , Q 3 , .... 

 (the absorption of electricity by sinks being reckoned as nega- 

 tive emission), the effect will be the same as if Q L , Q 2 , .... poles 

 of unit strength coincided at the points occupied by the actual 

 poles. Hence, in such a case, the radii vectores from the several 

 poles must be taken Qj, Q 2 , .... times in the above formula, 

 and it thus becomes 



r 1 Q i.r 2 Q 2.r 3 Q 3.... = ^l 

 or V . . . . (16) 



2(Qlogr)=rcJog/z,, J 



which is a quite general expression for the equipotential lines 

 due to any distribution of poles of any strength. 



33. To find the actual potential at any point due to a given 

 set of poles, we must recur to equation (2), § 7. Let V lt 

 V 2 , .... be the potentials which the various poles would pro- 

 duce at the given point if each acted separately, and let <f> l9 

 (f>2, .... be the potentials produced by each separately at unit 

 distance from itself; then, taking the other symbols in the 

 senses already defined, we have 



and for the resultant potential at the given point, 



U=SV=S£-2^ & .2(Qlogr). . . . (17) 



Since, for a given set of poles, 2</> is constant, this gives 

 for a locus of constant potential 



£(Qlogr) = const. = 2t™;S(2<£-TJ), 



which is in accordance with (16) ; for we may write 



and by § 8, when Q = l, we have logyLt=27r/cS . Av. 



