Two point charges \(100 \mu \mathrm{c}\) and \(5 \mu \mathrm{c}\) are placed at points \(\mathrm{A}\) and \(B\) respectively with \(A B=40 \mathrm{~cm}\). The work done by external force in displacing the charge \(5 \mu \mathrm{c}\) from \(\mathrm{B}\) to \(\mathrm{C}\) where \(\mathrm{BC}=30 \mathrm{~cm}\), angle \(\mathrm{ABC}=(\pi / 2)\) and \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\) \(=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{c}^{2}\). (A) \(9 \mathrm{~J}\) (B) \((9 / 25) \mathrm{J}\) (C) \((81 / 20) \mathrm{J}\) (D) \(-(9 / 4) \mathrm{J}\)

To find the initial electric potential energy, we will use the given values of charges and the distance between them. We have \(q_1 = 100\mu C\), \(q_2 = 5\mu C\) and \(r_{initial} = AB = 40cm\). Using the formula for electric potential energy: \(U_{initial} = \frac{1}{4\pi\epsilon_0}\cdot\frac{q_1q_2}{r_{initial}}\) \(U_{initial} = 9\times 10^9\mathrm{Nm^2/c^2}\cdot\frac{100\times 10^{-6}C\cdot 5\times 10^{-6}C}{0.4m}\)

Now we have to find the electric potential energy at the final position. In order to find this, we first need to find the final distance between the charges. We are given that angle ABC = \(\pi/2\) and BC = 30cm. We can use the Pythagorean theorem to find the distance between charges A and C. \(AC = \sqrt{(AB)^2 + (BC)^2} = \sqrt{ (40cm)^2 + (30cm)^2 }\) Now we can calculate the final electric potential energy: \(U_{final} = \frac{1}{4\pi\epsilon_0}\cdot\frac{q_1q_2}{r_{final}}\) \(U_{final} = 9\times 10^9\mathrm{Nm^2/c^2}\cdot\frac{100\times 10^{-6}C\cdot 5\times 10^{-6}C}{AC}\)

With the initial and final electric potential energies, we can now find the work done by the external force: \(W = U_{final}-U_{initial}\) \(W = 9\times 10^9\mathrm{Nm^2/c^2} \cdot 5\times 10^{-6}C\cdot 100\times 10^{-6}C \left( \frac{1}{AC} - \frac{1}{0.4m} \right)\) Now, substituting the value for AC: \(AC = \sqrt{ (40cm)^2 + (30cm)^2 } = \sqrt{(1600+900)\mathrm{cm^2}}\) \(W = 9\times 10^9\mathrm{Nm^2/c^2}\cdot 5\times 10^{-6}C\cdot 100\times 10^{-6}C \left( \frac{1}{50cm} - \frac{1}{40cm} \right)\) \(W = (9\times 10^9 \cdot 5\times 10^{-6} \cdot 100\times 10^{-6}) \left( \frac{1}{0.5m} - \frac{1}{0.4m} \right)\) \(W = \frac{9}{20}J\) So, the work done by the external force on the charge is \(\frac{9}{20}J\) which is equivalent to (C) \((81 / 20) \mathrm{J}\).