520 



Prof. Jolm Perry on a 

 For 6 = 4:5°. 



r ... 







•1 



•2 



•3 



■4 



•5 



•6 



•7 



•8 



•9 



10 



a ... 



1 



1-001 



1-004 



1-009 



1-016 



1-025 



1-036 



1-049 



1-064 



1-081 



1-100 



T ... 



1-2 



1-4 



1-6 



1-8 



2-0 



2-2 



2-6 



3-0 



3-5 



4 



5 



p... 



•8912 



•7507 



■65 



•572 



•512 













•2 



Now on a sheet of squared paper I plotted the values of r 

 and u or /3, and so found the values of v for such particular 

 values of a or /3 as seemed suitable for curve-drawing. In 

 fact I found the values of r for = 45° for various equi- 

 potential surfaces. 



Kepeating this for other values of and drawing radial 

 lines on a sheet of paper, it was easy to draw the equipoten- 

 tial surfaces. 



The figure (PI. I.) was obtained in this way by Mr. Joselin. 

 It shows the equipotential surfaces from 0*5 to 1*4. These 

 surfaces are surfaces of revolution. Of course the resultant 

 force anywhere is inversely as the normal distance apart of 

 the equipotential surfaces, and the direction of the force is 

 everywhere normal to the equipotential surfaces. 



Example II. 



One circular spire of wire of radius a has an electric cur- 

 tric current C flowing in it. Find the electromagnetic potential 

 everywhere. 



At any point on the axis, z centimetres from the centre, 

 the potential is 



V s/z' + a*' 



(1) 



It is always worth while at first to get an idea of the range of 

 values of V. 



Putting 



0=0, 



£=CO 



V = 2tt0. 



V = -293x27rC. 



v=o. 



