So I was reading Electricity and magnetism by Purcell and I came across the Gauss law. Now in here its said I quote "The area of the patch $A$ is larger than that of the patch $a$ by two factors: first, by the ratio of the distance squared $(\frac{R}{r})^2$ and second, owing to its inclination, by the factor $\frac{1}{\cos \theta}$ "

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Now I understood the "distance squared" part. But I cant wrap my head around the inclination part.

What I tried to understand: The vector $A$ can be resolved into two components, one along the radial and the other perpendicular to it.

Likewise I may even resolve the radial vector into components along the vector $A$ and perpendicular to it.

Any help is appreciated. :)


Forget the vector $\overrightarrow{A}$ and imagine the surface $A$ itself.

It is inclined at an angle $\theta $ to the surface that is normal to the radius. The projected surface, orthogonal to the radius is ${{A}_{\bot }}=A\cos (\theta )$ (If you incline the surface, it is greater !)

It is this surface ${{A}_{\bot }}$ that you have to compare to $a$ a by using the factor ${{\left( R/r \right)}^{2}}$ : $\frac{{{A}_{\bot }}}{{{R}^{2}}}=\frac{a}{{{r}^{2}}}$

  • $\begingroup$ That does make sense. But why cant I project the radial surface onto the surface A? Yes It may be smaller than it but why cant I say that $A' \cos \theta = A$? I'm sorry but I'm not really getting on board with that idea. (btw I loved how you referred to the surface as a "she") $\endgroup$ – The Jade Emperor Feb 10 at 9:16
  • $\begingroup$ Sorry for my english. English is not my native langage. In french, surface is a feminine word ! I have changed.You can project because you reason on elementary surfaces. We can assimilate the surface to its tangent plane and reason with flat surfaces. $\endgroup$ – Vincent Fraticelli Feb 10 at 9:24

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