A particle of mass 100 g and charge \( 2\,\mu\mathrm{C} \) is released from a distance of 50 cm from a fixed charge of \( 5\,\mu\mathrm{C} \). Find the speed of the particle when its distance from the fixed charge becomes 3 m. Neglect any other force. (Take \( \frac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \).)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.73 m/s
Only the electric force acts, so the sum of the kinetic and potential energies is conserved. The like charges repel, so the particle moves away from rest. The loss in electrostatic potential energy \( U=kqQ/r \) appears as its kinetic energy.
In SI units the data are:
\( q=2\,\mu\mathrm{C}=2\times10^{-6}\,\mathrm{C} \)
\( Q=5\,\mu\mathrm{C}=5\times10^{-6}\,\mathrm{C} \)
\( m=100\,\mathrm{g}=0.1\,\mathrm{kg} \)
\( r_1=50\,\mathrm{cm}=0.5\,\mathrm{m} \)
\( r_2=3\,\mathrm{m} \)
The potential energies at the two distances are:
\( \begin{aligned} U_1 &= \frac{kqQ}{r_1} \\ &= \frac{9\times10^{9}\times2\times10^{-6}\times5\times10^{-6}}{0.5}\,\mathrm{J} \\ &= 0.18\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} U_2 &= \frac{kqQ}{r_2} \\ &= \frac{9\times10^{9}\times2\times10^{-6}\times5\times10^{-6}}{3}\,\mathrm{J} \\ &= 0.03\,\mathrm{J} \end{aligned} \)
The kinetic energy gained is:
\( \begin{aligned} \frac{1}{2}mv^{2} &= U_1-U_2 \\ &= 0.18\,\mathrm{J}-0.03\,\mathrm{J} \\ &= 0.15\,\mathrm{J} \end{aligned} \)
The speed is then:
\( \begin{aligned} v &= \sqrt{\frac{2\times0.15\,\mathrm{J}}{0.1\,\mathrm{kg}}} \\ &= \sqrt{3}\,\mathrm{m/s} \\ &\approx 1.73\,\mathrm{m/s} \end{aligned} \)