A particle of charge '$-q$' and mass '$m$' moves in a circle of radius '$r$' around an infinitely long line charge of linear charge density '$+\lambda$'. Then time period will be given as :
(Consider $k$ as Coulomb's constant)
Solution
<p>$$\begin{aligned}
& \frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{r}}=\mathrm{m} \omega^2 \mathrm{r} \\
& \omega^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\
& \left(\frac{2 \pi}{\mathrm{T}}\right)^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\
& \mathrm{~T}=2 \pi \mathrm{r} \sqrt{\frac{\mathrm{m}}{2 \mathrm{k} \lambda \mathrm{q}}}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
This question is part of PrepWiser's free JEE Main question bank. 132 more solved questions on Electrostatics are available — start with the harder ones if your accuracy is >70%.