A charge q_2 of mass m revolves around a stat | vedclass

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(A) For a charge $q_2$ to revolve in a circular orbit,the electrostatic force between $q_1$ and $q_2$ provides the necessary centripetal force.The cen

Vedclass · Accepted Answer

$\left|\frac{4 \pi^2 m r^3}{k q_1 q_2}\right|^{\frac{1}{2}}$

Vedclass · Answer

(A) For a charge $q_2$ to revolve in a circular orbit,the electrostatic force between $q_1$ and $q_2$ provides the necessary centripetal force.The centripetal force is given by $F_c = \frac{m v^2}{r}$,where $v$ is the orbital velocity.The electrostatic force is given by Coulomb's law as $F_e = \frac{k q_1 q_2}{r^2}$.Equating the two forces: $\frac{m v^2}{r} = \frac{k q_1 q_2}{r^2}$.Since $v = r \omega$,where $\omega$ is the angular velocity,we have $\frac{m (r \omega)^2}{r} = \frac{k q_1 q_2}{r^2}$.Simplifying,$m r \omega^2 = \frac{k q_1 q_2}{r^2}$,which gives $\omega^2 = \frac{k q_1 q_2}{m r^3}$.Since $\omega = \frac{2 \pi}{T}$,where $T$ is the periodic time,we have $\left(\frac{2 \pi}{T}\right)^2 = \frac{k q_1 q_2}{m r^3}$.$\frac{4 \pi^2}{T^2} = \frac{k q_1 q_2}{m r^3}$.Therefore,$T^2 = \frac{4 \pi^2 m r^3}{k q_1 q_2}$.Taking the square root,$T = \left[\frac{4 \pi^2 m r^3}{k q_1 q_2}\right]^{\frac{1}{2}}$.

$A$ 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 . . . . . . .

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