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

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$\frac{81}{20} J$

$\frac{9}{25} J$

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

answer is D.

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Detailed Solution

$\begin{array}{l} W = {PE}_{2} - {PE}_{1} \\ w = \frac{1}{4 {πε}_{0}} [\frac{500 \times 10^{- 12}}{50 \times 10^{2}} - \frac{500 \times 10^{- 12}}{40 \times 10^{2}}] \\ w = - \frac{9}{4} J \end{array}$

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