306 ELECTROSTATICS. [PT. II. CH. VII. 



If we call 7 the angular half-opening of the bowl, a/R = sin 7, 

 a/c tan 7 and the charge of the bowl is 



(14) e = R (IT 7) + R sin 7, 



giving as its capacity 



To complete the problem we have to find the surface density. 

 We find 



3F = 1 1 8X R I _JL_SX / 



dnt ~ X ' 2a \/X fa* I -, V 

 a 2 a 2 



Rdl fir l Vx7\ 



__ _ _ ton * _ 1 







aF 7 = 1 1 8X jR 1 1 d\' 



dn e ~ l \ ' 2a VX 9w e ^ x V 

 a 2 a 2 



R dl /TT 



- i; o~ ^ 



t 2 9w e \2 



Now on the surface S 



v/ _ av_ ax 7-7? ^_ _8[_ 



:A " 97i""8^' ^ 8r <e ~ 9^" 

 and therefore 



dV_ 1 /TT -^ 



-" 



. V\\ a 9X 



" - 



Now the direction cosines of the normal n e being a/R,yJR,zlR, 

 we have 



8X _ ^ 8X i/ 8X 8X 



~ 



The quadratic for X is, cleared of fractions, 

 (21} X (a 2 + X) - O 2 + 7/ 2 ) X - - c) 2 (a 2 + X) = 0, 



