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Q.
A charged particle moves with a speed $v$ in a circular path of radius $R$ around a long uniformly charged conductor.
Electric Charges and Fields
Solution:
$ E_{\text {rod }}=\frac{\lambda}{2 \pi \varepsilon_{0} R}$
If charge goes around rod, it means the electric force provides the centripetal force, then
$\frac{m v^{2}}{R}=q E$
$\frac{m v^{2}}{R}=q\left(\frac{\lambda}{2 \pi \varepsilon_{0} R}\right) \Rightarrow v=\left(\frac{q \lambda}{2 \pi \varepsilon_{0} m}\right)^{\frac{1}{2}}$
$\Rightarrow v$ is independent of $R$
