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Three particles have charges 20 mc each and they are fixed at the corners of an eq. triangle of side 0.5m The force on each of the particle has magnitude
Dear pallavi force between two charge F= 1/4∏ε (20 *10-6)2/(.5)2 = 9*109 * 4*10-10 /.25 =14.4 N this force act along the side of triangle . in one charge two force act along the direction of side and angel between then is 60 resultant force F'=[F2 +F2 +2FFcos60]1/2 =F√3 =24.94 N Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
Dear pallavi
force between two charge
F= 1/4∏ε (20 *10-6)2/(.5)2
= 9*109 * 4*10-10 /.25
=14.4 N
this force act along the side of triangle .
in one charge two force act along the direction of side and angel between then is 60
resultant force F'=[F2 +F2 +2FFcos60]1/2
=F√3
=24.94 N
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation.
All the best. Regards,Askiitians ExpertsBadiuddin
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