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Derivation of gauss's law for electrical fields please help

  • Thread starter Ragnar
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This is a very stupid question. extremely stupid. In fact I'm extremely embarassed.

I was reading a text on electromagnetism, and it said that since the flux due to a charge does not depend on the radius of the sphere then the formula, q/permitivitty applies to all closed surfaces. This is where i got confused. why does that make it apply to all closed surfaces?

please don't laugh at me.

Please I need an answer!
 
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  • #2
Hootenanny
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Try calculating the electric flux of spheres with various radii using [itex]\phi = \int\vec{E}\cdot d\vec{A}[/itex] and see where that gets you. Intuitively, one can consider the definition of electric flux (number of electric field lines passing through a unit area) and the definition of the electric field and how it relates to the density of electric field lines...
 
  • #3
robphy
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After calculating for spheres, try (say) a "northern" hemisphere at radius r1 and a "southern" hemisphere at radius r2 with an http://mathworld.wolfram.com/Annulus.html" [Broken] joining their "equators". Note the flux through this annulus.

Then, approximate ANY closed surface by portions of spheres and portions of annuli. ...take limits.
 
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