206 Prof. G. M. Minchin on the Magnetic Field 



Let PN be the perpendicular on the plate from P ; let 

 AB be the diameter in which it is cut by the plane through 

 the axis of the plate and P ; let Q, S be two very close 

 points on the circumference,, NQ making the angle 6 with 

 NA. Then we shall suppose the plate broken up into tri- 

 angular strips such as QNS, and calculate the potential of 

 each strip at P. 



Let L be any point on 'NQ, let NL = f ; then the poten- 

 tial of the element mPd£d6 at P is — T ~L , or md6 , . 



LP ' v^ + f 2 



where PN = £. If NQ = r we find by integration that the 

 potential produced by the strip QNS is 



rod0(v7* + ?-s), ...... (1) 



and the potential, V, of the whole plate is 



2m J ( */?+?- z)dd (2) 



Now if, as in the figure, the point N falls within the plate, 

 the limits of 6 are and ir ; if N falls on the edge of the 



rrr 



plate, at B, the limits are and -a ; and if it falls outside the 



z 



plate, the limits are and 0. Taking 6 as the independent 

 variable would, then, give us three different expressions for V, 

 according to the position of N ; and hence we must choose a 

 more convenient variable than 6. Let (f> be the angle QOA, 

 and change the expression (2) into one in which </> is the 

 independent variable. We shall then have 



2 2i2,n j j a a sin 6 



t z == a + or + lax cos 9 ; tan v — 



tf? + acos <f) 

 so that 



tt o C n */a 2 + z? + z 2 + 2aa:cos6—z, . ,>.,, 



\=2ma\ r y — K r^ (a + w cos <p)d(b ; 



J a 2 + x 2 + 2a%cos<j> v YJ r 



or if we put D = a 2 + x 2 + 2ax cos $, 



)W 



