Mr maxwell, ON FARADAY'S LINES OF FORCE. 69 



Let a be the radius of the sphere, and p the given pressure, and let the given sources be 

 at distances b^ b^ 8ec. from the centre, and let their rates of production be 47rPi, 'iirPs &c. 



a^ a- 

 Then if at distances — , — &c. (measured in the same direction as 6j b^ Sec. from the 

 bi 62 



centre) we place negative sources whose rates are 



a a 



- 4nrPi - , - ^trPi - &c. 



the pressure at the surface r = a will be reduced to zero. Now placing a source 47r — at 

 the centre, the pressure at the surface will be uniform and equal to p. 



The whole amount of fluid emitted by the surface r = a may be found by adding the 

 rates of production of the sources within it. The result is 



-{^^^-}• 



To apply this result to the case of a conducting sphere, let us suppose the external sources 

 47rPi, ^wPi to be small electrified bodies, containing e^ 62 of positive electricity. Let us also sup- 

 pose that the whole charge of the conducting sphere is = E previous to the action of the external 

 points. Then all that is required for the complete solution of the problem is, that the surface 

 of the sphere shall be a surface of equal potential, and that the total charge of the surface shall 

 he E. 



If by any distribution of imaginary sources within the spherical surface we can effect this, 

 the value of the corresponding potential outside the sphere is the true and only one. The 

 potential inside the sphere must really be constant and equal to that at the surface. 



We must therefore find the images of the external electrified points, that is, for every 

 point at distance b from the centre we must find a point on the same radius at a distance 



— , and at that point we must place a quantity =—e — of imaginary electricity. 

 At the centre we must put a quantity E' such that 



E'=E+ei- +e.,-+kc.; 

 bi 62 



then if R be the distance from the centre, r^r^ &c. the distances from the electrified points, 

 and /j/g the distances from their images at any point outside the sphere, the potential at 

 that point will be 



jE' /I a \\ /I an„ 



R \ri birj ' Vi-2 i2rj 



E eifa 61 a\ e~>(a b^ a\ 



