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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Feb 2008 23:36:36 IST
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In region der exists an electric field E(bar)=(4x^3i+3y^2j+8zk)\(4x^4+3y^3+8z^2).
the charge enclosed within a sphere of radius R cntered at origin(x,y,z=0) is
a)2pie(epsilon not) R
b)4pie(epsilon not) R
c)(4x^3+3y^2+8z)(epsilon not)
d)zero..
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IIT is more toxic beauty,
it intoxicates both the holder n the beholder!.. |
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24 Feb 2008 14:05:38 IST
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pls reply...
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IIT is more toxic beauty,
it intoxicates both the holder n the beholder!.. |
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24 Feb 2008 15:13:42 IST
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yaar plz make the Q. clear, the E field clearer, icant understand wat it is actually! and u can use symbols instead of typing epsilon or pi
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24 Feb 2008 15:14:39 IST
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and superscriptions for powers
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24 Feb 2008 17:16:33 IST
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Is the answer (b) ?
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______________________________________________
Siddhant Shah
The trouble with doing something right the first time is that nobody appreciates how difficult it is.
You can't control the wind, but you can adjust your sails.
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24 Feb 2008 17:21:35 IST
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see the normal vector of the sphere at point ( x, y, z ) is n =( xi + yj + zk)/ ( R ) ( since x^2 + y^2 + z^2 = R^2 ) so E. dS = E. n dS = 1/R dS ( see, here E.n = 1 ) So charge =  o E .dS =1/R 4 pioR^2 = 4 pi epsilon naught R
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24 Feb 2008 21:28:06 IST
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thnk u feynman!...
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