384 Mr Searle, On the Calculation of Capacities 



I shall, however, first solve the problem by the aid of the general 

 equations obtained in § 6. 



If C 1+2 denote the capacity of [1] and [2] " in parallel," and 

 v be the common potential of the two conductors, then, by 

 Definition A, 



since e x + e 2 is the charge upon the single conductor formed 

 by joining [1] and [2]. Putting v 1 = v 2 = v in (12), we find 

 at once that, when [3], ... [m] are insulated and [m + 1], ... [n] are 

 kept at zero potential, 



A u + A 12 + A 21 + A 22 

 1+2 = A (15). 



'-»ll'-»22 ^^12 *-*21 



§9. I now pass on to calculate, in terms of the original n 2 

 coefficients, the (n — l) 2 coefficients which occur when [1] and [2] 

 are joined. If we put v x = v 2 = v in Maxwell's equations (3), 

 and then add together the first two equations, we obtain the 

 n — 1 equations 



(?n+ ?u + q*. + #22) v + (?is + #23) v 3 + (q u + # 24 ) v 4 + . . . = e!+e 2 

 (g 3 i + 2s2> +$33 v s +q 2i v 4 + ... = e 3 



(qa + q*)v +q 4z v 3 +qu v 4 + ... = e A 



.(16). 



If, for brevity, the single conductor formed by joining [1] and [2] 

 is denoted by the symbol [0], and if the coefficients for the new 

 system are denoted by accented letters, we see, on comparing the 

 above equations with Maxwell's scheme, that 



q\o = ffn + q u + g 2 i + q* (17), 



q'o S = q u + 9W> q'm = qsi + qsz (18), 



q rs == qrs } q sr = qsr \ -*■ «v> 



w 7 here s and r range from 3 to n. 



§ 10. If A' denote the determinant of m — 1 rows, formed on 

 the plan of (6) from the coefficients for the m — 1 conductors 

 [0], [3], ... [m], and if A' 00 denote the minor of A' complementary 

 to q' 0Q , it follows from (6) that 



A' = 



?33 ^34 



q i3 qu 



All. 22- 



