| ||
2.3.3. Correct interpretation of inverse square law The specific feature of solutions presented in this book is to show that inverse square (distance) relation is independent of any requirements of limitations. The examples of such imposed requirements are for instance: field around charge in Gauss’ theorem should be uniform (identical in all directions) or that laws are applicable to whole area except the point where R=0. Such limitations were already pointed out in paragraph I.5 – “Incorrectly interpreted specific problems”. We are searching then for solutions which do not depend on limitations listed above. By changing the order of integration in equation (2.93), | ||
(2.95) | ||
we can change the integration by spherical angle Ω to integration
on whole surface of a sphere having specific radius = R. Elementary spherical angle dΩ can be treated as a product of two elementary angles in perpendicular directions. | ||
and | (2.96) | |
(where da and db
represent elementary sections located at vertical and horizontal circles
with radius R, ). Formula (2.95) then takes a form of: | ||
(2.97) | ||
Sections of circles da and db and radius R are depicted on the drawing (Fig. 4).
The arcs a and b shown on vertical and horizontal circles represent
the angles α = a/R and β = b/R respectively, showing the projection of the vector R onto the XY plane and the position of the vector R relative to the XY plane - and determine accordingly: w x = cos( b/R) · cos( a/R), w y = sin( b/R) · cos( a/R) and w z = sin( a/R). | ||
x
a
da
R
|