Q. Charge $q_{2}$ of mass $m$ revolves around a stationary charge $q_{1}$ in a circular orbit of radius$\ r$. The orbital periodic time of $q_{2}$ would be
A) $\left\lbrack \frac{4\pi^{3}mr^{2}}{kq_{1}q_{2}} \right\rbrack^{1/2}$
B) $\left\lbrack \frac{kq_{1}q_{2}}{4\pi^{2}mr^{2}} \right\rbrack^{1/2}$
C) $\left\lbrack \frac{4\pi^{2}mr^{4}}{kq_{1}q_{2}} \right\rbrack^{1/2}$
D) $\left\lbrack \frac{4\pi^{2}mr^{2}}{kq_{1}q_{2}} \right\rbrack^{1/2}$
Explanation
(a) $\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}} = mr\omega^{2} = \frac{4\pi^{2}mr^{2}}{T^{2}}$
$T^{2} = \frac{\left( 4\pi\varepsilon_{0} \right)r^{2}(4\pi^{2}mr)}{q_{1}q_{2}}$
$T = \left\lbrack \frac{4\pi^{3}mr^{2}}{kq_{1}q_{2}} \right\rbrack^{1/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.