Q140 / 180 · KCET 2016 · Physics
A particle of mass \( 1 \mathrm{gm} \) and charge \( 1 \mu \mathrm{C} \) is held at rest on a frictionless horizontal surface at
distance \( 1 \mathrm{~m} \) from the fixed charge \( 2 \mathrm{mC} \). If the particle is released, it will be repelled. The
speed of the particle when it is at a distance of \( 10 \mathrm{~m} \) from the fixed charge
distance \( 1 \mathrm{~m} \) from the fixed charge \( 2 \mathrm{mC} \). If the particle is released, it will be repelled. The
speed of the particle when it is at a distance of \( 10 \mathrm{~m} \) from the fixed charge
- A\( 60 m s^{-1} \)
- B\( 100 \mathrm{~ms}^{-1} \)
- C\( 90 m s^{-1} \)
- Correct answerD\( 180 \mathrm{~ms}^{-1} \)
Solution
Force between two particles is given as
\( F=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \) Now, work done by force \( =\int_{1}^{10} F d r=\int_{1}^{10} \frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} d r \) \( =\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2} \int_{1}^{10} \frac{1}{r^{2}} d r=\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2}\left(\left.\cdot \frac{-1}{4}\right|_{1} ^{10}\right) \) \( =\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2}\left(1-\frac{1}{10}\right)=\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2} \times \frac{9}{10} \rightarrow(1) \)
This work is stored in form of kinetic energy. Therefore,
\[ \begin{array}{l} W=\frac{1}{2} m v^{2} \\ \Rightarrow v^{2}=\frac{2 W}{m}=\frac{2}{1 \times 10^{-3}} \frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2} \times \frac{9}{10}[\text { [Using Eq. (1)] } \\ \Rightarrow v^{2}=\frac{2}{1 \times 10^{-3}} \times 9 \times 10^{9} \times\left(1 \times 10^{-6} \times 2 \times 10^{-3}\right) \times \frac{9}{10} \\ =2 \times 2 \times 9 \times 9 \times 100 \\ \Rightarrow v=2 \times 9 \times 10=180 \mathrm{~ms}^{-1} \end{array} \]
\( F=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \) Now, work done by force \( =\int_{1}^{10} F d r=\int_{1}^{10} \frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} d r \) \( =\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2} \int_{1}^{10} \frac{1}{r^{2}} d r=\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2}\left(\left.\cdot \frac{-1}{4}\right|_{1} ^{10}\right) \) \( =\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2}\left(1-\frac{1}{10}\right)=\frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2} \times \frac{9}{10} \rightarrow(1) \)
This work is stored in form of kinetic energy. Therefore,
\[ \begin{array}{l} W=\frac{1}{2} m v^{2} \\ \Rightarrow v^{2}=\frac{2 W}{m}=\frac{2}{1 \times 10^{-3}} \frac{1}{4 \Pi \varepsilon_{0}} q_{1} q_{2} \times \frac{9}{10}[\text { [Using Eq. (1)] } \\ \Rightarrow v^{2}=\frac{2}{1 \times 10^{-3}} \times 9 \times 10^{9} \times\left(1 \times 10^{-6} \times 2 \times 10^{-3}\right) \times \frac{9}{10} \\ =2 \times 2 \times 9 \times 9 \times 100 \\ \Rightarrow v=2 \times 9 \times 10=180 \mathrm{~ms}^{-1} \end{array} \]
Hence, the correct answer is option: 'D'.