Question
Fri July 01, 2011

# 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

Fri July 01, 2011
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
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# Two point charges q1 and q2, of magnitude 20 nC and –20 nC respectively, are placed 0.2 m apart. Calculate the electric field at points A, B and C.

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