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Electricity
17 Jun 2008 18:38:48 IST
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BALGANESH
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17 Jun 2008 18:48:00 IST
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i think for the first 1 it would be 0 , logic i think is same as that we use for a cube with charge at one of the corner ,and those prob we take the flux to be 0 for the three sides at which the charge is located
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17 Jun 2008 19:00:17 IST
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the answers are = Q/2eo
and -Q/2e0
my working is draw a shell around it gaussian surface
for this the flux = Q/eo
so for the hemisphere it is Q/2eo
but am i correct
and the flux is not zero bcos Q makes some chrges appear on the surface
but not sure
explain in detail someone
and -Q/2e0
my working is draw a shell around it gaussian surface
for this the flux = Q/eo
so for the hemisphere it is Q/2eo
but am i correct
and the flux is not zero bcos Q makes some chrges appear on the surface
but not sure
explain in detail someone
18 Jun 2008 00:09:07 IST
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If it's a point charge located just above the centre of the surface, imagine a gaussian surface in the form of a sphere of radius R.. This sphere would almost coincide with your given hemisphere. Net flux through this surface is Q/e0
Since Half the flux is above and half is below, flux through top and bottom hemisphere's is each Q/2e0
Now replace the bottom hemisphere of the gaussian surface by a flat plate.. Since No charge is located in the volume you just excluded, flux through this flat surface remains the same..
The gaussian surface you now get is coinciding with the surface given to you.. So flux through bottom face is also q/2e0
Here e0 can be assumed to be permittivity of free space everywhere
Since Half the flux is above and half is below, flux through top and bottom hemisphere's is each Q/2e0
Now replace the bottom hemisphere of the gaussian surface by a flat plate.. Since No charge is located in the volume you just excluded, flux through this flat surface remains the same..
The gaussian surface you now get is coinciding with the surface given to you.. So flux through bottom face is also q/2e0
Here e0 can be assumed to be permittivity of free space everywhere
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