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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jun 2007 00:24:18 IST
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Electric charges q,q and -2q are placed at the corners of an equilateral triangle ABC of side L.The magnitude of electric dipole moment of the system is? Please give detailed answers.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jun 2007 00:36:24 IST
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DIPOLE MOMENT= 3 *q *L Solution ther r 3 charges placed at A(q) B ( q)C(-2q) so thr r 2 dipoles AC and BC each of dipole moment p=qL inclined at 60* p(res)=2pcos30*= 3 *q *L
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jun 2007 12:45:57 IST
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why cos30?inclination is 60
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jun 2007 18:56:26 IST
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DIPOLE MOMENT= q *L
p(res)=2pcos60*= q *L
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Jun 2007 13:29:49 IST
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But the answer is given to be [ ] 3 q.l
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Jun 2007 16:53:30 IST
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make triangle assume that 2 charges of mag -q are lying at A
net p=2pcos30*= root3 *q *L The angle is 30 cos net p wud be along angle bisector
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devesh
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Jun 2007 22:07:00 IST
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There are two dipole moments of qL at an angle 60. Now if two vectors of same magnitude A are placed at angle , then the resultant has a magnitude of 2Acos( /2), along the angle bisector of the angle between the two vectors.
Applying the same here, the resultant is 2qLcos30 = 3 qL Along the angle bisector of the top angle.
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