Question

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. $\frac{\rho q{R}^2}{4{\epsilon }_0}$
• B. $\frac{q\rho }{4{\epsilon }_0{R}^2}$
• C. $\frac{\rho q{R}^2}{2{\epsilon }_0}$
• D. $\frac{4{\epsilon }_0{R}^2}{q\rho }$

Answer 1

The correct option is A $\frac{\rho q{R}^2}{4{\epsilon }_0}$

The charge particle is revolving around the cylinder in the electric field $E$. So, we can write

$qE=\frac{m{v}^2}{r}$

Here, $r$ is the distance of the particle from the axis of the cylinder.

Now, $E=\frac{\rho r}{2{\epsilon }_0}$

$\Rightarrow \frac{m{v}^2}{r}=\frac{q\rho r}{2{\epsilon }_0}$

As, $r=R$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{\rho R^{2}q}{4{\epsilon }_0}$

Hence, $\left(A\right)$ is the correct answer.