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GUJCET · Physics · Electric Charges and Fields

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 __________.

  1. A \(\left|\frac{4 \pi^2 m r^3}{k q_1 q_2}\right|^{\frac{1}{2}}\)
  2. B \(\left[\frac{k q_1 q_2}{4 \pi^2 m r^3}\right]^{\frac{1}{2}}\)
  3. C \(\left[\frac{4 \pi^2 m r^4}{k q_1 q_2}\right]^{\frac{1}{2}}\)
  4. D \(\left[\frac{4 \pi^2 m r^2}{k q_1 q_2}\right]^{\frac{1}{2}}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

(A)
image
\(\frac{m v^2}{r}=\frac{k q_1 q_2}{r^2}\)
\(\begin{aligned} & \therefore & \frac{m(r \omega)^2}{r} & =\frac{k q_1 q_2}{r^2} \\ & \therefore & \omega^2 & =\frac{k q_1 q_2}{m r^3} \\ & \therefore & \frac{4 \pi^2}{T^2} & =\frac{k q_1 q_2}{m r^3} \\ & \therefore & T^2 & =\frac{4 \pi^2 m r^3}{k q_1 q_2} \\ & \therefore & T & =\left[\frac{4 \pi^2 m r^3}{k q_1 q_2}\right]^{\frac{1}{2}}\end{aligned}\)