490 Mr. J. Prescott on the 



is therefore true when r = a for all values of 6. Therefore 

 any combination of equations (25) and (26) will be true 

 all over the surface. 



Subtracting (25) from (26), we get 



- 2 ^S-™ }-/»^ = 0whenr=«.. (27) 

 Now equations (20) and (21) show that 



= - n dp SmZU l6\d? [r 



led 2 



B*=-4;-4H£ ( '' Se) -^4 



.-{Sw- 6 ™}]. ( 28 ) 



and this is the general expression for ^- at all points of 



the sphere. It can therefore be used in (27) . 

 Also, from (6) and (10), 



2 £S-™ } - {I &*>- 3 3i>>} sin 2 * (29 > 



Substituting from (28) and (29) in (27), and then dividing 

 by ^/o-j- sin 20, we find 



H£ (r3g) - 6 £ (re) }-M5 (A >- 6re } 



-YMl&W- 3 ^}- whenr=a. (30) 



For convenience, the other boundary condition is re- 

 written here : 



6e + r-j--l 2 ~r( T * e ) f — when r = a. . (30 a) 



We have to find e to satisfy equation (21) for all values 

 of r, and equations (30) and (30 a) when r=a. These 

 two boundary conditions are sufficient for the present 

 problem. If we had been dealing with a hollow sphere, 

 we should have had two more boundary conditions, and 

 these are replaced in the present case by the condition that 

 e must be finite at the centre. 



Solution of the Equations. 



We can get no further without assuming an expression 

 for p ; and to get a numerical result, we must assume a 



