Question

The time period of revolution of a charge $q_1$ and of mass $m$ moving in a circular path of radius $r$ due to Coulomb force of attraction with another charge $q_2$ at its centre is

• A. $\sqrt{\frac{16 \pi\varepsilon_{0} mr^{3}}{q_{1}q_{2}} }$
• B. $\sqrt{\frac{8\pi\varepsilon_{0} mr^{3}}{q_{1}q_{2}} }$
• C. $\sqrt{\frac{ \varepsilon_{0} mr^{3}}{ 16 q_{1}q_{2}} }$
• D. $\sqrt{\frac{16 \pi^3 \varepsilon_{0} mr^{3}}{q_{1}q_{2}} }$
• E. $\sqrt{\frac{\pi^2 \varepsilon_{0} mr^{3}}{ 8 q_{1}q_{2}} }$

Answer 1

Answer: (D)

Centripetal force $F=\frac{m v^{2}}{r}\,...(i)$
and electrostatic force
$F=K \cdot \frac{q_{1} \cdot q_{2}}{r^{2}}\,...(ii)$
Equating both equation
$\frac{m v^{2}}{r} =K \cdot \frac{q_{1} q_{2}}{r^{2}} $
$v =\sqrt{\frac{K}{m} \cdot \frac{q_{1} q_{2}}{r}}$
The time period of revolution
$T=\frac{2 \pi r}{V}$
$=2 \pi r \sqrt{\frac{m \cdot r}{K \cdot q_{1} q_{2}}} $
$ =\sqrt{\frac{4 \pi^{2} r^{2} m r}{K \cdot q_{1} q_{2}}} \,\left(\because K=\frac{1}{4 \pi \varepsilon_{0}}\right) $
$T=\sqrt{\frac{4 \pi^{2} r^{2} m r \cdot 4 \pi \varepsilon_{0}}{q_{1} q_{2}}} $
$=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m r^{3}}{q_{1} q_{2}}} $