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I don't see how using Gaussian surfaces can help me to calculate flux for example if there is a disk with radius R and center at (0,0,0) and a point charge at random (a,0,b). If I consider an hemisphere around the disk I still can't calculate the flux through it easily, since the electric field won't be perpendicular to the normal of the surface. Is there anyway to use a Gaussian surface to tackle these kind of problems? I mean something like this:

enter image description here

*I'm not asking for full calculations of course, just an abstract notion of how to treat these cases.

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enter image description here just for simplicity assume a symmetric closed surface like this above(imagine 3d) , you will get symmetrically equal field lines on area (flux) through both faces(shown in red) both faces are of equal area , hence flux through both faces are equal, but as we can see in figure flux through curved cylinder is also possible let us assume flux through both flat faces equal(by symmetry) as f1 and flux through curved part of cylinder is f2 then since it is closed surface from gauss law 2(f1) + f2= q/E0------1 this equation is valid for every case i.e wherever the charge q may be , draw a symmetric gaussian surface through surface (where you want flux) and find flux through equation 1. I hope now it helps if you take area as acos(theta) inclined on area a, the result would be same

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  • $\begingroup$ Why would the flux through the $acos\theta$ disk be the same as the one through the original one? could you elaborate? $\endgroup$
    – Darkenin
    Apr 1 '20 at 9:08
  • $\begingroup$ I have edited for more simplicity now, sorry for too much complexity earlier that was not needed $\endgroup$
    – maverick
    Apr 1 '20 at 10:26

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