Question

A long cylindrical wire have linear density $+2\times {10}^{-8}\text{C/m}$. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron.

• A. $1\times {10}^{-17}\text{J}$
• B. $2\times {10}^{-17}\text{J}$
• C. $3\times {10}^{-17}\text{J}$
• D. $4\times {10}^{-17}\text{J}$

Answer 1

The correct option is C $3\times {10}^{-17}\text{J}$


Electric field at distance $r$ away from an infinite line charge is given as,

$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$

Now, force on charge $q$ is,

$F=qE=\frac{q\lambda }{2\pi {ϵ}_{0}r}$

This will provide the required centripetal force.

So, $m{\omega }^{2}r=\frac{q\lambda }{2\pi {ϵ}_{0}r}$

$\Rightarrow \frac{m{\omega }^2{r}^2}{2}=\frac{r}{2}\times \frac{q\lambda }{2\pi {ϵ}_{0}r}......\left(1\right)$

Now, kinetic energy, $E=\frac{1}{2}I{\omega }^2$

$\Rightarrow E=\frac{q\lambda }{4\pi {ϵ}_0}$ $\left[\text{from}\right(1\left)\right]$

$\Rightarrow E=kq\lambda [k=\frac{1}{4\pi {ϵ}_0}]$

$\Rightarrow E=9\times {10}^9\times 1.6\times {10}^{-19}\times 2\times {10}^{-8}$

$\Rightarrow E=3\times {10}^{-17}\text{J}$