Question

Consider a point charge +Q of mass m suspended by a massless, inextensible string of length l in free space (permittivity \( \epsilon_0 \)) as shown in the figure. It is placed at a height \( d \) (\( d > l \)) over an infinitely large, grounded conducting plane. The gravitational potential energy is assumed to be zero at the position of the conducting plane and is positive above the plane. 

• A. \( \frac{p_\theta^2}{2ml^2} - \frac{Q^2}{16\pi \epsilon_0(d - l \cos \theta)} - mg(d - l \cos \theta) \)
• B. \( \frac{p_\theta^2}{2ml^2} - \frac{Q^2}{8\pi \epsilon_0(d - l \cos \theta)} + mg(d - l \cos \theta) \)
• C. \( \frac{p_\theta^2}{2ml^2} - \frac{Q^2}{8\pi \epsilon_0(d - l \cos \theta)} - mg(d - l \cos \theta) \)
• D. \( \frac{p_\theta^2}{2ml^2} - \frac{Q^2}{16\pi \epsilon_0(d - l \cos \theta)} + mg(d - l \cos \theta) \)

Hint:

The Hamiltonian for systems involving electrostatic and gravitational forces combines both kinetic and potential energy terms. For this setup, ensure that you correctly account for the electrostatic potential due to the interaction with the conducting plane.

Answer 1

The Correct Option is D

In this problem, we need to write the Hamiltonian for a system consisting of a charged mass \( m \) suspended by an inextensible string, with a conducting plane below it. The problem involves both electrostatic and gravitational forces. Step 1: General Setup of the System.
- The position of the charge \( +Q \) is described by the angular position \( \theta \) with respect to the vertical. - The canonical momentum \( p_\theta \) is associated with this angular position, representing the rotational motion of the charge. - The kinetic energy of the system is purely due to the rotation of the charge around the pivot point: \[ T = \frac{p_\theta^2}{2ml^2}, \] where \( l \) is the length of the string. Step 2: Electrostatic Potential Energy.
The electrostatic potential energy involves the interaction between the point charge \( +Q \) and its image charge (due to the grounded conducting plane). The distance between the charge and its image is \( d - l \cos \theta \). The potential energy \( U_{\text{electrostatic}} \) is given by the formula for point charges: \[ U_{\text{electrostatic}} = -\frac{Q^2}{16\pi \epsilon_0(d - l \cos \theta)}. \] Step 3: Gravitational Potential Energy.
The gravitational potential energy \( U_{\text{grav}} \) is given by the formula for a point mass in a gravitational field: \[ U_{\text{grav}} = -mg(d - l \cos \theta), \] where \( g \) is the acceleration due to gravity, and \( d - l \cos \theta \) is the height of the charge above the conducting plane. Step 4: Constructing the Hamiltonian.
The Hamiltonian \( H \) is the sum of the kinetic energy and potential energies: \[ H = T + U_{\text{electrostatic}} + U_{\text{grav}}. \] Substituting the expressions for \( T \), \( U_{\text{electrostatic}} \), and \( U_{\text{grav}} \), we get: \[ H = \frac{p_\theta^2}{2ml^2} - \frac{Q^2}{16\pi \epsilon_0(d - l \cos \theta)} + mg(d - l \cos \theta). \] Thus, the correct Hamiltonian is: \[ H = \frac{p_\theta^2}{2ml^2} - \frac{Q^2}{16\pi \epsilon_0(d - l \cos \theta)} + mg(d - l \cos \theta), \] which matches Option (D). Final Answer: (D)