Motion of Charges in an Electric Field question from JEE Main 2022 (24 Jun, Shift 2)

Answer: (a)

1. Gauss's law outside the cylinder ($r > R$): $E \cdot 2\pi r l = \frac{\rho\pi R^2 l}{\varepsilon_0} \Rightarrow E = \frac{\rho R^2}{2\varepsilon_0 r}$
2. Centripetal force: $\frac{mv^2}{r} = qE = \frac{q\rho R^2}{2\varepsilon_0 r}$
3. $KE = \frac{1}{2}mv^2 = \frac{\rho q R^2}{4\varepsilon_0}$
4. So the answer is (a).