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

A long cylindrical volume of radius $$R$$ contains a uniformly distributed charge of density $$\rho$$. A charge particle $$q$$ revolves around the cylinder in a circular path.

Find the electric field using Gauss's Law: Consider a cylindrical Gaussian surface of radius $$r$$ (where $$r > R$$) and length $$L$$, coaxial with the charged cylinder.

The charge enclosed by the Gaussian surface:

$$Q_{\text{enc}} = \rho \cdot \pi R^2 \cdot L$$

By Gauss's Law:

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

$$E \cdot (2\pi r L) = \frac{\rho \pi R^2 L}{\varepsilon_0}$$

Solving for $$E$$:

$$E = \frac{\rho R^2}{2\varepsilon_0 r}$$

Apply the condition for circular motion: For the charged particle $$q$$ revolving in a circular orbit of radius $$r$$, the electrostatic force provides the centripetal force:

$$qE = \frac{mv^2}{r}$$

Find the kinetic energy: The kinetic energy is:

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}qEr$$

Substituting the expression for $$E$$:

$$KE = \frac{1}{2} \cdot q \cdot \frac{\rho R^2}{2\varepsilon_0 r} \cdot r$$

$$KE = \frac{1}{2} \cdot \frac{q\rho R^2}{2\varepsilon_0}$$

$$KE = \frac{q\rho R^2}{4\varepsilon_0}$$

The kinetic energy of the particle is $$\dfrac{q\rho R^2}{4\varepsilon_0}$$.

The correct answer is Option A.