An electron revolves around an infinite cylindrical wire having uniform linear charge density $2\times10^{-8}\ \text{C m}^{-1}$ in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is ______ $\times10^6\ \text{m s}^{-1}$. Given mass of electron $=9\times10^{-31}\ \text{kg}$.
Numerical value type. Enter your answer.
Answer: 8
$E=\dfrac{\lambda}{2\pi\epsilon_0r}$ provides the centripetal force: $\dfrac{e\lambda}{2\pi\epsilon_0r}=\dfrac{mv^2}{r}$, so $v$ is independent of $r$:
$$v^2=\frac{2k e\lambda}{m}=\frac{2\times9\times10^9\times1.6\times10^{-19}\times2\times10^{-8}}{9\times10^{-31}}=6.4\times10^{13}$$
$v=8\times10^6\ \text{m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics