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GATE 2021 PH – Question 40

Classical Mechanics · Hamiltonian and Hamilton's equations of motion · 2 marks · Multiple choice

A point charge $+Q$ of mass $m$ is suspended by a massless inextensible string of length $l$ in free space of permittivity $\epsilon_0$. The suspension point is at height $d>l$ over an infinite grounded conducting plane. Gravitational potential energy is zero at the plane and positive above it. If $\theta$ is the angular position and $p_\theta$ its canonical momentum, choose the Hamiltonian.

Diagram for GATE 2021 PH question 40
  1. $\frac{p_\theta^2}{2ml^2}-\frac{Q^2}{16\pi\epsilon_0(d-l\cos\theta)}-mg(d-l\cos\theta)$
  2. $\frac{p_\theta^2}{2ml^2}-\frac{Q^2}{8\pi\epsilon_0(d-l\cos\theta)}+mg(d-l\cos\theta)$
  3. $\frac{p_\theta^2}{2ml^2}-\frac{Q^2}{8\pi\epsilon_0(d-l\cos\theta)}-mg(d-l\cos\theta)$
  4. $\frac{p_\theta^2}{2ml^2}-\frac{Q^2}{16\pi\epsilon_0(d-l\cos\theta)}+mg(d-l\cos\theta)$

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Correct answer: (D) $\frac{p_\theta^2}{2ml^2}-\frac{Q^2}{16\pi\epsilon_0(d-l\cos\theta)}+mg(d-l\cos\theta)$

Explanation

The charge height is $z=d-l\cos\theta$ and its kinetic energy is $ml^2\dot\theta^2/2$, giving $p_\theta=ml^2\dot\theta$. The image charge is $-Q$ a distance $2z$ away. The energy of the real charge plus induced conductor is one half the real–image pair interaction: $U_e=-Q^2/(16\pi\epsilon_0z)$. Add $U_g=mgz$ and replace the kinetic energy by $p_\theta^2/(2ml^2)$ to obtain D.

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