Question

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 \(\left(k=\frac{1}{4\pi \epsilon _{0}}\right)\)

• A. \(\left[\frac{4\pi ^{2}mr^{3}}{kq_{1}q_{2}}\right]^{\frac{1}{2}}\)
• B. \(\left[\frac{kq_{1}q_{2}}{4\pi ^{2}mr^{3}}\right]^{\frac{1}{2}}\)
• C. \(\left[\frac{4\pi ^{2}mr^{4}}{kq_{1}q_{2}}\right]^{\frac{1}{2}}\)
• D. \(\left[\frac{4\pi ^{2}mr^{2}}{kq_{1}q_{2}}\right]^{\frac{1}{2}}\)

Hint:

The Coulomb force supplies the centripetal force, with a = 4 pi squared r over T squared.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The electrostatic force provides the centripetal force for the circular orbit.

Step 2: Force Balance
\(\dfrac{kq_1q_2}{r^{2}}=m\omega^{2}r=m\dfrac{4\pi^{2}}{T^{2}}r\).

Step 3: Solve for T Squared
\(T^{2}=\dfrac{4\pi^{2}mr^{3}}{kq_1q_2}\).

Step 4: Take the Root
\(T=\left[\dfrac{4\pi^{2}mr^{3}}{kq_1q_2}\right]^{1/2}\).

Final Answer:
The orbital period is the square root of 4 pi squared m r cubed over k q1 q2. \[ \boxed{\text{(A) }\left[\dfrac{4\pi^{2}mr^{3}}{kq_{1}q_{2}}\right]^{\frac{1}{2}}} \]