Question

A particle of mass 'm' having charge '(-q)' moves in a circular orbit of radius 'r' around a fixed charge '(+Q)'. The relation between time period \(T\) and radius of the orbit 'r' is:

• A. \(r^2 = \frac{Qq}{4\pi \epsilon_0 m T^3}\)
• B. \(r^3 = \frac{Qq}{16\pi^3 \epsilon_0 m T^2}\)
• C. \(r^3 = \frac{Qq}{16\pi^3 \epsilon_0 m T}\)
• D. \(r = \frac{Qq}{16\pi^2 m \epsilon_0 T^2}\)

Hint:

In circular motion involving electrostatic forces, balance the centripetal force with the electrostatic force to find relationships between physical quantities.

Answer 1

The Correct Option is B

1. Understand the Concept

A charged particle (-q) is moving in a circular orbit around a fixed charge (+Q). The electrostatic force between the charges provides the necessary centripetal force for the circular motion.

2. Electrostatic Force

The electrostatic force (\( F_e \)) between the charges is given by Coulomb's law:

\[ F_e = \frac{kQq}{r^2} \]

where:

• \( k = \frac{1}{4\pi\epsilon_0} \) is Coulomb's constant,
• \( Q \) and \( q \) are the magnitudes of the charges,
• \( r \) is the radius of the orbit.

3. Centripetal Force

The centripetal force (\( F_c \)) required for the circular motion is:

\[ F_c = \frac{mv^2}{r} \]

where:

• \( m \) is the mass of the particle,
• \( v \) is the velocity of the particle.

4. Equate the Forces

Since the electrostatic force provides the centripetal force:

\[ F_e = F_c \] \[ \frac{kQq}{r^2} = \frac{mv^2}{r} \] 5. Relate Velocity to Time Period

The velocity (\( v \)) of the particle is related to the time period (\( T \)) and the radius (\( r \)) by:

\[ v = \frac{2\pi r}{T} \] 6. Substitute Velocity into the Equation

Substitute the expression for \( v \) into the equation from step 4:

\[ \frac{kQq}{r^2} = \frac{m\left( \frac{2\pi r}{T} \right)^2}{r} \]

Simplify:

\[ \frac{kQq}{r^2} = \frac{4\pi^2mr}{T^2} \] 7. Solve for the Relation between T and r

Substitute \( k = \frac{1}{4\pi\epsilon_0} \):

\[ \frac{Qq}{4\pi\epsilon_0 r^2} = \frac{4\pi^2mr}{T^2} \]

Rearrange to solve for \( r \):

\[ QqT^2 = 16\pi^3\epsilon_0mr^3 \] \[ r^3 = \frac{QqT^2}{16\pi^3\epsilon_0m} \] \[ r^3 = \frac{Qq}{16\pi^3\epsilon_0mT^2} \] Therefore, the relation between the time period \( T \) and the radius of the orbit \( r \) is: \[ r^3 = \frac{Qq}{16\pi^3\epsilon_0mT^2} \] The correct answer is (2) \( r^3 = \frac{Qq}{16\pi^3\epsilon_0mT^2} \).

Answer 2

To solve the problem, we need to determine the relationship between the time period (T) and the radius (r) of the orbit of a particle with mass \( m \) and charge \( -q \), moving in a circular orbit around a fixed charge \( +Q \).

1. Understanding the Motion:
This situation describes the motion of a charged particle around a central charge, which is a classical example of the Bohr model of the atom. The force acting on the particle is the electrostatic force, given by Coulomb's law:

\[ F = \frac{1}{4\pi \epsilon_0} \frac{|Qq|}{r^2} \] This electrostatic force provides the centripetal force required to keep the particle in a circular orbit. The centripetal force is given by: \[ F = \frac{mv^2}{r} \] where \( v \) is the speed of the particle. Equating the two forces gives: \[ \frac{1}{4\pi \epsilon_0} \frac{|Qq|}{r^2} = \frac{mv^2}{r} \]

2. Relating the Time Period and Radius:
The speed \( v \) of the particle can be related to the time period \( T \) by the relation: \[ v = \frac{2\pi r}{T} \] Substituting \( v \) into the force equation, we can derive the relationship between the time period \( T \) and the radius \( r \) of the orbit. After simplifying the expression, we obtain: \[ r^3 = \frac{Qq}{16\pi^2 \epsilon_0 m} T^2 \]

3. Conclusion:
The correct relationship between the time period and the radius is:

Final Answer:
The correct option is (B) \( r^3 = \frac{Qq}{16\pi^2 \epsilon_0 m} T^2 \).