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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Jul 2007 11:20:27 IST
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Along the x-axis, three charges q/2 , -q and q/2 are placed at x=0 , x=a and x=2a respectively.The resultant electric potential at a point 'p' located at a distance r from the charge -q(a<<r) is ______________
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Bad news is that time always flies,
Good news is that u r the pilot.
yesterday is history,
tomorrow is a mystery,
today is a gift and that is why it's called "the present". |
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23 Jul 2007 11:23:43 IST
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plz c ur prev.reply on "electric field" topic
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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plz check if this question is complete..
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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23 Jul 2007 11:50:41 IST
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There is a circle surrounding charge -q of radius r in which each point is at a distance of r from the charge. You must be specific about that point at which the potential is to be calculated.
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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25 Jul 2007 17:49:06 IST
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hey divalli ans pasted by you is wrong .Check its dimensions...
so write down complete question and right amswer
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29 Jul 2007 19:56:11 IST
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hey!!!!! anyone pls answer.......
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Bad news is that time always flies,
Good news is that u r the pilot.
yesterday is history,
tomorrow is a mystery,
today is a gift and that is why it's called "the present". |
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29 Jul 2007 21:39:28 IST
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sorry u have to wait for the next months edition on this book i too buy this monthly edition book from which u have asked the question
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29 Jul 2007 21:48:15 IST
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i think ans is
-1/Eo*(aQ/r(a+r)) if wrong plz correct
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31 Jul 2007 21:27:16 IST
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Potential at P due to q/2 at ( 0,0 ) : Kq / 2(r + a)
Potential at P due to q at ( a,0 ) : K( - q / r )
Potential at P due to q/2 at ( 2a,0 ) : Kq / 2(r - a)
SO TOTAL POTENTIAL AT P IS : K [ (q / 2(r + a)) - (q / r) + (q / 2(r - a)) ] -------------(1)
BUT : a << r ; SO : eliminating 'a' in (1) to get the potential at P as zero.
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