Question

A long cylindrical wire carries a linear charge density of $3\times {10}^{-8}C{m}^{-1}$. An electron revolve around it in a circular path under the influence of the attractive force. $KE$ of the electron is :

• A. $1.44\times {10}^{-7}J$
• B. $2.88\times {10}^{-17}J$
• C. $4.32\times {10}^{-17}J$
• D. $8.64\times {10}^{-17}J$

Answer 1

The correct option is C $4.32\times {10}^{-17}J$

Let $r$ be the radius of the circular path.

For long wire the electric field at $r$ is $E=\frac{1}{4\pi {ϵ}_0}.\frac{2\lambda }{r}$

here, $\frac{m{v}^2}{r}=eE=\frac{e}{4\pi {ϵ}_0}.\frac{2\lambda }{r}$

$\frac{1}{2}m{v}^2=\frac{e}{4\pi {ϵ}_0}.\lambda$

$K.E=\frac{1}{2}m{v}^2=\frac{1.6\times {10}^{-19}}{4\pi \times 8.854\times {10}^{-12}}\times 3\times {10}^{-8}=4.32\times {10}^{-17}J$