346 Steady Currents in continuous Media [CH. x 



the circular mouth at B is 7/7ra 2 . In order that the potentials in the 

 electrostatic problem may be the same, we must have a uniform surface 

 density of electricity 



/ \ T/ 



0- = J or 



m>/yS 



on the surface of the disc. 



The heat generated is PR, where 72 is the resistance of the conductor C. 

 It is also 



taken through the conductor C. Now if W is the electrostatic energy of 

 a disc of 

 we have 



a disc of radius a, having a uniform surface density cr = 2 on each side, 



^ * ^ 



o- +1-5-] + o- \dxdydz, 

 dxJ \dyj \dzJ) 



where the integral is taken through all space, or again, 



where the integral is taken through the semi-infinite space on one side of 

 the disc, i.e. through the space C, if the disc is made to coincide with the 

 mouth B. On substituting for the volume integral in expression (323), we 

 find that 



Following Maxwell, we shall find it convenient to calculate W directly 

 from the potential. If a disc of radius r has a uniform surface density cr 

 on each side, the potential at a point P on its edge will be 



where the integral is taken over one side of the disc, and r is the distance 

 from P to the element dxdy. Taking polar coordinates, with P as origin, 

 the equation of the circle will be r = 26 cos 0, we may replace dxdy by 

 rdrdd, and obtain 



/r=2bcos9 ["0=- 

 r= J=- 



drd0 



On increasing the radius of the disc to b + db, we bring up a charge 

 4i7rbcrdb from infinity to potential 8bcr, so that the work done is 



