A particle of mass m and carrying charge -q_1 | vedclass

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Question

(A) The electrostatic force between the charges provides the necessary centripetal force for circular motion:$\frac{m v^2}{r} = \frac{1}{4 \pi \vareps

Vedclass · Accepted Answer

$\sqrt{\frac{16 \pi^3 \varepsilon_0 m r^3}{q_1 q_2}}$

Vedclass · Answer

(A) The electrostatic force between the charges provides the necessary centripetal force for circular motion:$\frac{m v^2}{r} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}$Solving for velocity $v$:$v^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{m r}$$v = \sqrt{\frac{q_1 q_2}{4 \pi \varepsilon_0 m r}}$The period of revolution $T$ is given by the circumference divided by the velocity:$T = \frac{2 \pi r}{v} = 2 \pi r \sqrt{\frac{4 \pi \varepsilon_0 m r}{q_1 q_2}}$$T = \sqrt{(2 \pi r)^2 \cdot \frac{4 \pi \varepsilon_0 m r}{q_1 q_2}}$$T = \sqrt{4 \pi^2 r^2 \cdot \frac{4 \pi \varepsilon_0 m r}{q_1 q_2}}$$T = \sqrt{\frac{16 \pi^3 \varepsilon_0 m r^3}{q_1 q_2}}$

$A$ particle of mass $m$ and carrying charge $-q_1$ is moving around a charge $+q_2$ along a circular path of radius $r$. Find the period of revolution of the charge $-q_1$.

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