217-220] Images 191 



Plane face with hemispherical boss. 



219. If we regard the whole equipotential V = C as a conductor, we 

 obtain the distribution of electricity on a plane conductor on which there 

 is a hemispherical boss of radius a. If we take the plane to be x = 0, we 

 have, by formula (126), 



At a point on the plane, 



- = _J^ 

 4-7T 



and on the hemisphere 



1 fdV\ F 



a- = -r ^ = . 3 cos 9. 

 4?r \orj r=a 4-7T 



The whole charge on the hemisphere is found on integration to be 



( 2 (-r- 3 cos 0] ZTTO? sin 6d6 = l Fa 2 , 

 Je=o \47r ) 



while, if the hemisphere were not present, the charge on the part of the 

 plane now covered by the base of the hemisphere would be 



Thus the presence of the boss results in there being three times as much 

 electricity on this part of the plane as there would otherwise be : this is 

 compensated by the diminution of surface-density on those parts of the plane 

 which immediately surround the boss. 



Capacity of a telegraph-wire. 



220. An important practical application of the method of images is the 

 determination of the capacity of a long straight wire placed parallel to an 

 infinite plane at potential zero, at a distance h from the plane. This may be 

 supposed to represent a telegraph-wire at height h above the surface of the 

 earth. 



Let us suppose that the wire has a charge e per unit length. To find 

 the field of force we imagine an image charged with a charge e per unit 

 length at a distance h below the earth's surface. The potential at a point at 

 distances r, r from the wire and image respectively is, by 75 and 100, 



<7-2elogr + 2elogr', 



and for this to vanish at the earth's surface we must take (7=0. Thus the 

 potential is 



2<? log - . 



