Two point charges 100μC and 5μC are placed at points A and B respectively with AB = 40cm. The work done by external force in displacing the charge 5μC from B to C, where BC = 30cm, angle $ABC = \frac{π}{2} and \frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2}$

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$\frac{81}{20} J$

$\frac{9}{25} J$

9 J

$- \frac{9}{4} J$

answer is D.

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Detailed Solution

Work done in displacing charge of 5 m C from B to C is
W = 5 x 10^-6(V_C - V_B) where
Question Image
$V_{B} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.4} = \frac{9}{4} \times 10^{6} V and V_{C} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.5} = \frac{9}{5} \times 10^{6} V So W = 5 \times 10^{- 6} \times (\frac{9}{5} \times 10^{6} - \frac{9}{4} \times 10^{6}) = - \frac{9}{4} J$