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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jun 2007 15:04:33 IST
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A point charge q is placed at the center of a cube with edge a .Wat is the electric flux through each of the cube faces???????? plz ans wid explanations........
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5 Jun 2007 15:39:21 IST
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total flux=  = q /  now since thr r 6 faces in a cube.......... so flux through each face= /6 = q /6  |
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First of all understand that the point charge is placed symmetrically wrt to all the six faces of the cube. Now, flux is related to no of lines of electric field cutting a surface. As the point charge is placed at the center of the cube, the no of lines crossing each face must be same, which means that the fulx must be same.
Suppose the flux through each face is 
By Gauss's Law
6 = q/
= q/6
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7 Jun 2007 23:25:23 IST
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hey
see since the cube is a closed structure so gauss law is applicable here
further all the faces of the cube are symmetrical so the flux from them will be identical
now from gauss law the total flux coming out of cube=q/epsilon
hewnse due to symm. flux from each face= q/6epsilon
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everybody has a photographic memory, some lack film...........<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
         
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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11 Jun 2007 06:46:25 IST
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THE ANS IS q\epsilon
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11 Jun 2007 22:49:02 IST
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hey shubham u r wrong!
it has to be q/6epsilon
plz check out the explanation
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everybody has a photographic memory, some lack film...........<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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12 Jun 2007 12:20:48 IST
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q/6epsiolon naught
and its so obvious,u should know,because a cube has 6 surfaces
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