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Two point charges 100μC 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=9×109 Nm2/C2

a
9 J
b
8120 J
c
925 J
d
-94 J

detailed solution

Correct option is D

Work done in displacing charge of 5 μ C from B to C is W=5×10−6VC−VB whereVB=9×109×100×10−60.4=94×106 Vand VC=9×109×100×10−60.5=95×106 VSo W=5×10−6×95×106−94×106=−94 J

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