Question 96 of 311

A long cylindrical volume contains a uniformly distributed charge of density $\rho$. The radius of cylindrical volume is $R$. A charge particle $\left(q\right)$ revolves around the cylinder in a circular path. The kinetic energy of the particle is :

A$\dfrac{\rho q{R}^{2}}{4{\epsilon }_{o}}$B$\dfrac{\rho q{R}^{2}}{2{\epsilon }_{o}}$C$\dfrac{q\rho }{4{\epsilon }_{0}{R}^{2}}$D$\dfrac{4{\epsilon }_{0}{R}^{2}}{q\rho }$

Show answer and solution▾

Answer: (A) $\dfrac{\rho q{R}^{2}}{4{\epsilon }_{o}}$

Electric field due to a long cylinder is,
$E=\dfrac{\lambda }{2\pi {\epsilon }_{0}r}=\dfrac{\rho A}{2\pi {\epsilon }_{0}r}$
If a charge $q$ is rotating in a circle of radius $r$
$\dfrac{m{v}^{2}}{r}=qE$ $\Rightarrow \dfrac{m{v}^{2}}{r}=q\dfrac{\rho A}{2\pi {\epsilon }_{0}r}\Rightarrow \dfrac{1}{2}m{v}^{2}=\dfrac{q\rho \pi {R}^{2}}{4\pi {\epsilon }_{0}}=\dfrac{q\rho {R}^{2}}{4{\epsilon }_{0}}$