Question

A long cylindrical wire carries a positive charge of linear density $2·0\times {10}^{-8}\mathrm{C}{\mathrm{m}}^{-1}$. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius.

Answer 1

Let the linear charge density of the wire be $\lambda$.

The electric field due to a charge distributed on a wire at a perpendicular distance r from the wire,

$\mathrm{E}=\frac{\lambda }{2\mathrm{\pi }{\in }_{0}r}$
The electrostatic force on the electron will provide the electron the necessary centripetal force required by it to move in a circular orbit. Thus,

$$\begin{gathered} qE=\frac{m{v}^2}{r} \\ \Rightarrow m{v}^2=qEr...(1) \end{gathered}$$
Kinetic energy of the electron, $K=\frac{1}{2}m{v}^2$
From (1),
$$\begin{gathered} K=\frac{qEr}{2} \\ K=\frac{qr}{2}\frac{\lambda }{2\mathrm{\pi }{\in }_{0}r}\left[\because E=\frac{\lambda }{2\mathrm{\pi }{\in }_{0}r}\right] \\ K=(1.6\times {10}^{-19})\times (2\times {10}^{-8})\times (9\times {10}^9)\mathrm{J} \\ K=2.88\times {10}^{-17}\mathrm{J} \end{gathered}$$