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Question

Physics Question on Electric charges and fields

The time period of revolution of a charge q1q_1 and of mass mm moving in a circular path of radius rr due to Coulomb force of attraction with another charge q2q_2 at its centre is

A

16πε0mr3q1q2\sqrt{\frac{16 \pi\varepsilon_{0} mr^{3}}{q_{1}q_{2}} }

B

8πε0mr3q1q2\sqrt{\frac{8\pi\varepsilon_{0} mr^{3}}{q_{1}q_{2}} }

C

ε0mr316q1q2\sqrt{\frac{ \varepsilon_{0} mr^{3}}{ 16 q_{1}q_{2}} }

D

16π3ε0mr3q1q2\sqrt{\frac{16 \pi^3 \varepsilon_{0} mr^{3}}{q_{1}q_{2}} }

Answer

16π3ε0mr3q1q2\sqrt{\frac{16 \pi^3 \varepsilon_{0} mr^{3}}{q_{1}q_{2}} }

Explanation

Solution

Centripetal force F=mv2r...(i)F=\frac{m v^{2}}{r}\,...(i) and electrostatic force F=Kq1q2r2...(ii)F=K \cdot \frac{q_{1} \cdot q_{2}}{r^{2}}\,...(ii) Equating both equation mv2r=Kq1q2r2\frac{m v^{2}}{r} =K \cdot \frac{q_{1} q_{2}}{r^{2}} v=Kmq1q2rv =\sqrt{\frac{K}{m} \cdot \frac{q_{1} q_{2}}{r}} The time period of revolution T=2πrVT=\frac{2 \pi r}{V} =2πrmrKq1q2=2 \pi r \sqrt{\frac{m \cdot r}{K \cdot q_{1} q_{2}}} =4π2r2mrKq1q2(K=14πε0)=\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=4π2r2mr4πε0q1q2T=\sqrt{\frac{4 \pi^{2} r^{2} m r \cdot 4 \pi \varepsilon_{0}}{q_{1} q_{2}}} =16π3ε0mr3q1q2=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m r^{3}}{q_{1} q_{2}}}