Q 12-01-095JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
A long cylindrical volume contains a uniformly distributed charge of density $\rho$. The radius of the cylindrical volume is $R$. A charged particle $(q)$ revolves around the cylinder in a circular path. The kinetic energy of the particle is:
Answer: (A) $\dfrac{\rho qR^2}{4\varepsilon_0}$
By Gauss's law, outside the cylinder at distance $r$: $E\cdot2\pi rl = \dfrac{\rho\pi R^2l}{\varepsilon_0}\Rightarrow E = \dfrac{\rho R^2}{2\varepsilon_0 r}$.
The electric force supplies the centripetal force:
$$\frac{mv^2}{r} = qE = \frac{q\rho R^2}{2\varepsilon_0 r}\ \Rightarrow\ \frac12mv^2 = \frac{q\rho R^2}{4\varepsilon_0}$$
Solution by Sreeraj P, M.Sc Physics