CBSE Class 12-science Answered
ABC is a right angle triangle in which AB=3cm & BC=4cm . The three charges +15,+12,-20 e.s.u. are placed respectively on A,B,C.What is the force acting on B
Asked by Gaurav nagpal | 01 Jul, 2011, 12:12: AM
Expert Answer
We know that 1 esu = 3.34 x 10-10 coulomb
The force on charge placed at B by the charge on A, FBA = (1/4??0)(qAqB/rAB2)
The force on charge placed at B by the charge on C, FBc = (1/4??0)(qCqB/rBC2)
These two forces act on B mutually perpendicular to each other. then the net force on charge at B, F = sqrt (FAB2 + FBC2) = (1/4??0)sqrt {(qAqB/rAB2)2 + (qCqB/rBC2)2}
F = 9 x 109 x sqrt {(15 x 12)2/(0.03)4 + (12 x 20)2/(0.04)4 }x (3.34 x 10-10)2
This way by doing this simple airthmatic calculation the net force on the cahrge at B can be comuted. To find the direction of the net force on charge at B
tan ? = mag of FBC / mag of FAB
Answered by | 01 Jul, 2011, 10:10: AM
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