Q.

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=9×109 Nm2/C2

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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 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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