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Electricity
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16 Sep 2009 17:02:05 IST
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the solid angle subtended by the circle at the point is............
2pi(1-cos(semi vertex angle))
semi vertex angle = 60 deg.............
solid angle=2pi(1/2)=pi..
flux k =solid angle * q/(4pi epsilon)
k=q/4epsilon...
q=4kepsilon.......................
u can also do this by integration
16 Sep 2009 17:20:41 IST
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i hv already mentioned ....here we hv to use solid angle which is outside JEE COURSE ...MOREOVER IN JEE COURSE WE DEAL WITH FLUX ONLY FOR SYMMETRIC gaussian surfaces...!!..for non symmetric surfaces ...we apply above method..!! ....thanx karna ...!! :)













well.vignesh ..!!
...i don`t think this question is under scope of iit...or aieee..
or any engineering exam..
as here there is no ...symmetric gausian surface possible...
so ..we can`t evaluate flux directly..
here we hv to take the concept of solid angle...
which is purely out of jee..
and moreover it will require intense integration..
frm ..where did u got this sum....
PLZZZZ....SOLVE THIS IF ANYBODY CAN..!!