﻿390 
  Prof. 
  A. 
  Anderson 
  on 
  the 
  Problem 
  of 
  Two 
  and 
  

   But 
  if 
  E 
  is 
  the 
  charge 
  on 
  A, 
  A 
  = 
  B 
  -f 
  — 
  , 
  and 
  therefore 
  

  

  E 
  = 
  aA 
  — 
  aB 
  . 
  

  

  Thus 
  

  

  v 
  _ 
  a 
  f 
  ,a 
  2 
  b 
  /_1 
  , 
  ab 
  

  

  J!i 
  - 
  A 
  l 
  rt+ 
  T\OJ 
  1 
  + 
  OA 
  . 
  OA 
  . 
  2 
  I 
  2 
  

  

  a 
  2 
  5 
  2 
  

  

  o,i! 
  . 
  ojs 
  • 
  oa 
  . 
  o 
  2 
  i 
  2 
  . 
  o 
  2 
  i 
  4 
  

  

  a 
  n 
  h 
  r 
  ' 
  

  

  + 
  -" 
  + 
  o 
  1 
  i 
  1 
  ....o 
  1 
  w.o 
  2 
  i 
  s 
  ....oj 
  2 
  , 
  + 
  -;/' 
  

  

  j>ab 
  T 
  1 
  afr 
  a 
  2 
  /r 
  

  

  * 
  i 
  i+ 
  o^toj; 
  + 
  oji 
  . 
  oa 
  . 
  o 
  2 
  i 
  2 
  . 
  o 
  2 
  i 
  4 
  + 
  ••• 
  

  

  + 
  J 
  2 
  . 
  OA 
  . 
  . 
  . 
  . 
  J 
  2 
  „_! 
  . 
  2 
  I 
  2 
  T77707l 
  2? 
  ; 
  + 
  '•• 
  J 
  * 
  

  

  Thus 
  (7 
  n 
  and 
  q 
  l2 
  have 
  been 
  found, 
  and, 
  of 
  course, 
  also 
  ^ 
  22 
  , 
  

   by 
  a 
  simple 
  application 
  of 
  the 
  above 
  equation. 
  

  

  The 
  same 
  method 
  is 
  applicable 
  to 
  the 
  case 
  of 
  several 
  

   conducting 
  spheres. 
  For 
  three 
  spheres 
  whose 
  centres 
  are 
  

   at 
  the 
  corners 
  of 
  a 
  triangle 
  the 
  work 
  is 
  necessarily 
  much 
  

   longer 
  than 
  that 
  for 
  two, 
  but 
  it 
  is 
  possible 
  to 
  find 
  the 
  values 
  

   of 
  the 
  coefficients 
  of 
  induction 
  and 
  capacity 
  to 
  any 
  degree 
  of 
  

   approximation. 
  

  

  We 
  require 
  for 
  the 
  solution 
  of 
  the 
  problem 
  a 
  set 
  of 
  points 
  

   I 
  l9 
  I 
  3 
  , 
  I 
  5 
  ,... 
  inside 
  B, 
  and 
  a 
  set 
  I 
  2 
  , 
  I 
  4 
  , 
  I 
  6 
  ,... 
  inside 
  A, 
  as 
  in 
  

   the 
  problem 
  for 
  two 
  spheres, 
  and 
  corresponding 
  to 
  these, 
  

   for 
  A 
  and 
  C, 
  a 
  set 
  K 
  1? 
  K 
  3 
  , 
  K 
  5 
  ,... 
  inside 
  0, 
  and 
  a 
  set 
  K 
  2 
  , 
  K 
  4 
  , 
  

   K 
  G 
  ,... 
  inside 
  A. 
  But 
  other 
  points 
  besides 
  these 
  are 
  needed. 
  

   Take 
  one 
  of 
  the 
  points, 
  say 
  I 
  3 
  , 
  inside 
  B. 
  We 
  take 
  its 
  image 
  

   in 
  C 
  and 
  denote 
  it 
  by 
  L 
  31 
  , 
  the 
  image 
  of 
  this 
  in 
  A 
  by 
  L 
  S2 
  , 
  

   the 
  image 
  of 
  L 
  32 
  in 
  B 
  by 
  L 
  33 
  , 
  and 
  so 
  on, 
  going 
  round 
  again 
  

   and 
  again 
  in 
  the 
  positive 
  direction. 
  

  

  For 
  the 
  I 
  points 
  inside 
  A 
  we 
  do 
  a 
  similar 
  thing, 
  the 
  image 
  

   of 
  I 
  4 
  in 
  being 
  L 
  41 
  , 
  and 
  that 
  of 
  this 
  in 
  B 
  L 
  42 
  , 
  and 
  so 
  on, 
  

   going 
  round 
  in 
  the 
  negative 
  direction. 
  We 
  do 
  the 
  same 
  

   thing 
  for 
  the 
  K 
  points. 
  Thus 
  the 
  image 
  of 
  K 
  5 
  in 
  B 
  is 
  M 
  51 
  , 
  

   and 
  the 
  image 
  of 
  this 
  in 
  A 
  is 
  M 
  52 
  , 
  and 
  so 
  on. 
  But 
  this 
  does 
  

   not 
  exhaust 
  all 
  the 
  points 
  required. 
  Starting 
  from 
  any 
  

   L 
  or 
  M 
  point, 
  we 
  reverse 
  the 
  direction 
  and 
  find 
  an 
  infinite 
  

   series 
  of 
  points 
  for 
  it. 
  Thus, 
  taking 
  the 
  point 
  M 
  52 
  , 
  its 
  

   image 
  in 
  is 
  M 
  53 
  , 
  but 
  we 
  also 
  take 
  its 
  image 
  in 
  B 
  and 
  call 
  

   it 
  M 
  52 
  i, 
  the 
  image 
  of 
  this 
  in 
  C, 
  M 
  522 
  , 
  and 
  so 
  on. 
  A 
  few 
  of 
  

   the 
  points 
  are 
  shown 
  in 
  fig. 
  2. 
  

  

  