Question

A particle of charge \(-q\) and mass \(m\) moves in a circle of radius \(r\) around an infinitely long line charge of linear density \(+\lambda\). Then the time period will be given as:

• A. \( T = 2\pi r \sqrt{\frac{m}{2kq}} \)
• B. \( T^2 = \frac{4\pi m r^3}{2kq} \)
• C. \( T = \frac{1}{2\pi r} \sqrt{\frac{m}{2kq}} \)
• D. \( T = \frac{2kq}{m} \)

Answer 1

The Correct Option is A

The electric field \( E \) due to an infinitely long line charge with linear charge density \( +\lambda \) at a distance \( r \) from the line charge is given by:

\[ E = \frac{\lambda}{2 \pi \epsilon_0 r}, \]

where \( \epsilon_0 \) is the permittivity of free space.

Force on the Charged Particle: The force \( F \) acting on the particle due to the electric field is:

\[ F = -qE = -q \left( \frac{\lambda}{2 \pi \epsilon_0 r} \right). \]

Since the particle moves in a circular path, this force provides the centripetal force necessary for circular motion:

\[ F = \frac{mv^2}{r}. \]

Equating the Forces: Setting the electric force equal to the centripetal force:

\[ -q \left( \frac{\lambda}{2 \pi \epsilon_0 r} \right) = \frac{mv^2}{r}. \]

Rearranging gives:

\[ mv^2 = \frac{q \lambda}{2 \pi \epsilon_0}. \]

Finding the Time Period: The velocity \( v \) can also be expressed in terms of the radius and the time period \( T \):

\[ v = \frac{2 \pi r}{T}. \]

Substituting this expression for \( v \) into the equation:

\[ m \left( \frac{2 \pi r}{T} \right)^2 = \frac{q \lambda}{2 \pi \epsilon_0}. \]

Simplifying gives:

\[ m \times \frac{4 \pi^2 r^2}{T^2} = \frac{q \lambda}{2 \pi \epsilon_0}. \]

Rearranging for \( T^2 \):

\[ T^2 = \frac{4 \pi m r^2 \epsilon_0}{q \lambda}. \]

Final Expression: To match the answer choices, if we express \( k = \frac{1}{4 \pi \epsilon_0} \):

\[ T^2 = \frac{4 \pi m r^2}{2 k q}. \]

Thus, the time period is:

\[ T = 2 \pi r \sqrt{\frac{m}{2 k q}}. \]

Answer 2

The problem asks for the time period of a particle with charge \(-q\) and mass \(m\) moving in a circular orbit of radius \(r\) around an infinitely long line of charge with a positive linear charge density \(+\lambda\).

Concept Used:

For the particle to maintain a stable circular orbit, the electrostatic force of attraction exerted by the line charge must provide the necessary centripetal force. The solution involves the following physical principles:

1. Electric Field of an Infinite Line Charge: The magnitude of the electric field (\(E\)) at a perpendicular distance \(r\) from an infinitely long line of charge is given by: \[ E = \frac{\lambda}{2\pi\epsilon_0 r} \] Using Coulomb's constant, \(k = \frac{1}{4\pi\epsilon_0}\), we can express the electric field as: \[ E = \frac{2k\lambda}{r} \]
2. Electrostatic Force (\(F_e\)): The magnitude of the force experienced by a charge \(q\) in an electric field \(E\) is \(F_e = |q|E\).
3. Centripetal Force (\(F_c\)): For an object of mass \(m\) moving in a circle of radius \(r\) at a constant speed \(v\), the required centripetal force is: \[ F_c = \frac{mv^2}{r} \]
4. Time Period (\(T\)): The time period of one complete revolution is the circumference of the orbit divided by the orbital speed: \(T = \frac{2\pi r}{v}\).

Step-by-Step Solution:

Step 1: Determine the electrostatic force on the particle.

The electric field produced by the positively charged line at a distance \(r\) has a magnitude \(E = \frac{2k\lambda}{r}\) and is directed radially outward. The particle has a charge of \(-q\), so it experiences an attractive force (directed radially inward) towards the line charge. The magnitude of this electrostatic force is:

\[ F_e = |-q| E = q \left( \frac{2k\lambda}{r} \right) = \frac{2kq\lambda}{r} \]

Step 2: Equate the electrostatic force to the centripetal force.

This attractive electrostatic force is what keeps the particle in its circular orbit, so it must be equal to the required centripetal force.

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

Step 3: Solve for the orbital speed (\(v\)) of the particle.

We can simplify the equation from the previous step by canceling the radius \(r\) from both denominators:

\[ mv^2 = 2kq\lambda \]

Now, we solve for \(v^2\) and then take the square root to find the speed \(v\):

\[ v^2 = \frac{2kq\lambda}{m} \] \[ v = \sqrt{\frac{2kq\lambda}{m}} \]

Step 4: Calculate the time period (\(T\)) of the orbit.

The time period is given by the formula \(T = \frac{2\pi r}{v}\). We substitute the expression for \(v\) that we just derived:

\[ T = \frac{2\pi r}{\sqrt{\frac{2kq\lambda}{m}}} \]

Final Computation & Result:

To simplify the expression for the time period, we can invert the term inside the square root when moving it to the numerator:

\[ T = 2\pi r \sqrt{\frac{m}{2kq\lambda}} \]

Comparing this result with the given options, we find that it matches the second option.

The correct expression for the time period is \( T = 2\pi r \sqrt{\frac{m}{2k\lambda q}} \).