129] 



LAW OF FORCE. 



249 



symmetry the charge is so distributed that equal areas possess 

 equal charges. Let the charge per unit area be <r, and let us find 

 the form of /(r), so that the resultant of all the forces <rdSf(r) 

 due to all the elements dS at distances r from a given point Q 

 within the sphere, shall be zero when resolved in any direction. 



On account of symmetry, the force acting 

 on Q must be in the direction of the radius 

 OQ. We shall accordingly consider the radial 

 component R. Let OQ=b, and let the radius 

 of the sphere be a. Let the distance PQ, 

 where P is any point on the surface of the 

 sphere, be r, and let the polar coordinates of 

 P be 0, (j>, the co-latitude being measured 

 from the radius OQ. Let the angle PQO be 8. Then the whole 

 force at Q resolved along the radius OQ is proportional to 



( I ) R = (ja-dS .f(r) cos 8 = cr f* ( 2 *f(r) cos 5 . a 2 sin Od6d<f>. 



FIG. 56. 



We may at once integrate with respect to 



(2) 



R = 



cos sin OdO. 



Now OQ is the sum of the projections of OP and PQ on the radius 

 r cos 8 4- a cos = 6, 



(3) 



~ 

 cos o = 



b a cos 6 



From the relation between the sides of the triangle POQ, 

 (4) r 2 = a 2 + 6 2 - 2a6 cos 0, 



we get on partial differentiation with respect to b, 



(5) 



r ~ b a cos 0, 



<7C> 



dr b a cos 



= cos o. 



Substituting this value of cos 8 in the integral, 



R = 27ra 2 <r [ " f(r) d ~ sin 6dO, 

 Jo ob 



and if we call f(r} = <E> 7 (r}. 



