Answer to Question #97432 in Field Theory for Grey Elseyn

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:

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

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

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