Two point charges 100 μC and 5 μC are placed at points A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 μC from B to C, where BC = 30 cm, angle ABC =π2 and 14πε0=9×109Nm2/C2 ,is
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a
9 J
b
-8120 J
c
925 J
d
-94 J
answer is D.
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Detailed Solution
Work done in displacing charge of 5 μC from B to C isW=5×10-6VC-VB where VB=9×109×100×10-60.4=94×106 V and VC=9×109×100×10-60.5=95×106 V So W=5×10-6×95×106-94×106=-94 J