46, 47] Cavendish's Proof of Law of Force 39 



point at which the potential is evaluated, to the element a 2 sin 0ddd(f) of 

 spherical surface. 



If we agree to evaluate the potential at a point situated on the axis 6 = 

 at a distance c from the centre, we may write 



r 2 = a? + c 2 - 2ac cos 6. 



Since c is a constant, we obtain as the relation between dr and dd, by 

 differentiation of this last equation, 



rdr acsin 6dQ (15)- 



If we integrate expression (14) with respect to <, the limits being of 

 course < = and <f> = 2?r, we obtain 



V=\E I*'" (\ $ (r) dr} sin Odd, 



J 0=0 \J r ' 



or, on changing the variable from 6 to r, by the help of relation (15) 



rr=a c / / \ r ,~, 



F*|*{ ( f<r).*)*S: 



J r=a c V r / dC 



If we introduce a new function / (r), defined by 



we obtain as the value of V, 



If the inner and outer spheres are in electrical contact, their potentials 

 are the same ; and if, as experiment shews to be the case, there is no charge 

 on the inner sphere, then the whole potential must be that just found. This 

 expression must, accordingly, have the same value whether c represents the 

 radius of the outer sphere or that of the inner. Since this is true whatever 

 the radius of the inner sphere may be, the expression must be the same for 

 all values of c. We must accordingly have 



20^1^ 



-g- =f(a + c)-f(a-c), 



where V is the same for all values of c. Differentiating this equation twice 

 with respect to c, we obtain 



Since by definition, f(r) depends only on the law of force, and not on a or c. 

 it follows from the relation 



/" (a + o) -/"(a-c), 

 that f " (r) must be a constant, say C. 



