Question

an electron is sitting 10 cm away from a small object of charge -5nc it is let go and move away from the charge object when it reaches a distance of 15 cm away. find its speed

Answer 1

Using the energy conservation principle:

$$\frac{kq_1q_2}{r_0}=\frac{kq_1q_2}{r_1}+\frac{mv^2}{2}$$

So:

$$v=\sqrt{\frac{2}{m}(\frac{kq_1q_2}{r_0}-\frac{kq_1q_2}{r_1})}$$

$$v=\sqrt{\frac{2}{9.1*10^{-31}}(9*10^9*1.6*10^{-19}*5*10^{-9}(\frac{1}{0.1}-\frac{1}{0.15}))}$$

$$v=7.26*10^6 m/s$$