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 :

\rho qR^2 \over 4\varepsilon _0

\rho qR^2 \over 2\varepsilon _0

Electrostatics PYQ — JEE Main 2022 · 24 Jun · Shift 2 · Q60 with Solution

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Correct answer: A

Electric field due to a uniformly charged long cylinder

For an infinitely long cylinder of radius $R$ with uniform volume charge density $\rho$ , the electric field at a distance $r$ from the axis is found using Gauss's law.

Since the particle is revolving around the cylinder, we take it to move in a circular path outside the cylinder, i.e. at $r=R$ for the required expression in options.

For $r \ge R$ ,

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

At the surface $r=R$ ,

E=\frac{\rho R}{2\varepsilon_0}

Force on the charge particle

The electrostatic force on charge $q$ is

F=qE=q\cdot \frac{\rho R}{2\varepsilon_0}

This provides the centripetal force for circular motion:

\frac{mv^2}{R}=\frac{q\rho R}{2\varepsilon_0}

Find kinetic energy

Multiply both sides by $\dfrac{R}{2}$ :

\frac{1}{2}mv^2=\frac{q\rho R^2}{4\varepsilon_0}

But

K=\frac{1}{2}mv^2

Hence,

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

Match with options

This matches Option A:

\boxed{\frac{\rho q R^2}{4\varepsilon_0}}

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