Question

A long cylindrical wire carries a positive charge of linear density $2\times {10}^{-8}C{m}^{-1}$ . 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

From electric field due to linear charge density we can say,

$E=\frac{\lambda }{2\pi {\epsilon }_{O}r}$

Centripetal force is provided by force by electric field;

$E\times q=\frac{m{v}^2}{r}$ or, $\frac{\lambda }{2\pi {\epsilon }_{O}r}\times q=\frac{m{v}^2}{r}$

$\frac{\lambda }{2\pi {\epsilon }_{O}r}\times q=\frac{m{v}^2}{r}$

$\frac{\lambda }{2\pi {\epsilon }_O}\times q=m{v}^2$

As we know kinetic energy is given by: $\frac{1}{2}m{v}^2$

So, $\frac{1}{2}m{v}^2⟹\frac{1}{2}\times$ $\frac{\lambda }{2\pi {\epsilon }_O}\times q$

K.E.$=2\times {10}^9\times 2\times 2.8\times {10}^{-8}\times 1.6\times {10}^{-19}=2.88\times 10^{-17}J$