518 



Prof. John Perry on a 



distance of any point from a diametral plane, being 6 per square 

 centimetre where greatest : find the potential A inside and B 

 outside. Taking the diametral plane as the equator and 6 as 

 the co-latitude, it is obvious that 



cr = 6yu, 2 . 

 The expansion of /j? in spherical harmonics is already given as 



/* — q r 0T g -t 2 » 



So that we have a in spherical harmonics, 

 <7=2P + 4P 2 . 



Hence, as A and B are derivable from the same surface 

 harmonics, 



A=A P +A 2 rT 2 , 



B 



B 0p , B 2 -p 



~ r 0"r TJ r 23 



where A , A 2 , B , B 2 are constants to be found. 



Now at the surface, that is where r=l, A = B, and we can 

 apply this to every harmonic separately. Hence 



A = i3 , A 2 = B 2 . 



Again, we know from the theory of attraction that the 

 resultant force just outside and just inside the shell differs by 

 the amount 47rcr, or 



dA_dB =4 



dr dr ' 



and this is to be applied to every harmonic separately. Thus, 

 taking terms involving P , we have 



or putting r=l, 



+ B ^- 2 =47rx2 



B n =87T. 



Again, taking the second terms, 



2A 2 r + 3B 2 ^- 4 = 47rx4; 

 or, as A 3 = B 2 , and putting r = l, 



5B 2 = 167T, .*. B 2 =-=-7T. 



