A particle of mass 100 gm and charge 2 µC is released from a distance of 50 cm from a fixed charge of 5µC. Find the speed of the particle when its distance from the fixed charge becomes 3 m. Neglect any other force.

1.73 m/s

Loss of potential energy = gain in kinetic energy

$U_1-U_2=\Delta K$

$KQq \left[\frac{1}{r_1}-\frac{1}{r_2}\right]=1 / 2 mv^2-0$

$v=\sqrt{\frac{2 kQq}{m}\left[\frac{r_2-r_1}{r_1 r_2}\right]}$

$=\sqrt{\frac{2 \times 9 \times 10^9 \times 5 \times 10^{-6} \times 2 \times 10^{-6} \times 2.5}{0.1 \times 3 \times 0.5}}$ = 1.73 m/s