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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Jun 2008 18:38:48 IST
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A point charge Q is located just above the centrre of the face of a hemisphere of radius R ...
find the flux
(a) through the curved face
(b) through flat face
some explain in detail plzzzzzzzzzzzz
the concept is more imp than the problem
so detailed solution rates for sure
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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 18:49:15 IST
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naa its wrong balganesh
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17 Jun 2008 18:49:51 IST
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Is the answer zero....
pls temme if m correct ill post the solution....
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17 Jun 2008 18:49:55 IST
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2] base area * E
= pier^2E
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17 Jun 2008 18:51:02 IST
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according to me the whole hemisphere does not enclose any charge( the charge is juyst outside the hemisphere)
so i think the answer shud b zero...
still u post me the ans...
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17 Jun 2008 18:53:16 IST
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yeah read wrong its just above centre , won;t be 0
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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
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17 Jun 2008 20:51:50 IST
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in my opinion ur explaination is correct,,,
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
      
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
if i helped u plzzzzz rate me,,,,,,, |
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17 Jun 2008 23:32:09 IST
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on the curved surface =q/2eo
zero on flat surface
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17 Jun 2008 23:38:37 IST
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well accordin to me ganesh1991 is right
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FAILURE, THE FIRST STEP TO SUCCESS |
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17 Jun 2008 23:40:59 IST
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YUPP HES RITE
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[
[url=http://sig.graphicsfactory.com/]
    
[/url]
Glitter Graphics
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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
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18 Jun 2008 00:23:10 IST
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Are the answers:
A)
b)0
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Sleep to study....study to sleep |
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19 Jun 2008 23:12:55 IST
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yup ganesha is rite
the net flux shud b zero as i said
however when v find it thru the flat surface it is q/2piepsilon
therefore for the curved surface it will be -q/2piepsilon
gr8 job ganesha
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